Rounded to 4 decimal places, the probability that the sample mean wait time is under 4 minutes is 0.0146.
Given that,
The population mean waiting time is 5 minutes and the standard deviation is 2.9 minutes,
And, A random sample of 40 wait times was selected.
Now, the standard deviation of the sample mean,
[tex]\text {Standard Error }= \dfrac{ \text {Population Standard Deviation}}{\sqrt{\text {Sample Size}}}[/tex]
[tex]\text {SE} = \dfrac{2.9}{\sqrt{40} }[/tex]
[tex]\text {SE} = 0.459[/tex]
Now, we can standardize the sample mean using the formula:
[tex]\text {Z}= \dfrac{ \text {(Sample mean - Population mean)}}{{\text {Standard error}}}[/tex]
[tex]\text {Z}= \dfrac{ \text {(4 - 5)}}{{\text {0.459}}}[/tex]
[tex]Z = - 2.177[/tex]
Hence, for the probability corresponding to this Z-score, we can refer to the standard normal distribution table or use a calculator.
The probability of getting a Z-score less than -2.177 is 0.0146.
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Consider the case0502 data from Sleuth3. <<< This is the data. Sleuth3 is preloaded into R studio.
Dr Benjamin Spock was tried in Boston for encouraging young men not to register for the draft. It was conjectured that the judge in Spock’s trial did not have appropriate representation of women. The jurors were supposed to be selected by taking a random sample of 30 people (called venires), from which the jurors would be chosen. In the data case0502, the percent of women in 7 judges’ venires are given.
a. Create a boxplot of the percent women for each of the 7 judges. Comment on whether you believe that Spock’s lawyers might have a point.
b. Determine whether there is a significant difference in the percent of women included in the 6 judges’ venires who aren’t Spock’s judge.
c. Determine whether there is a significant difference in the percent of women incuded in Spock’s venires versus the percent included in the other judges’ venires combined. (Your answer to a. should justify doing this.)
Answer:
Consider the following calculations
Step-by-step explanation:
The complete R snippet is as follows
install.packages("Sleuth3")
library("Sleuth3")
attach(case0502)
data(case0502)
## plot
# plots
boxplot(Percent~ Judge, data=case0502,ylab="Values",
main="Boxplots of the Data",col=c(2:7,8),horizontal=TRUE)
# perform anova analysis
a<- aov(lm(Percent~ Judge,data=case0502))
#summarise the results
summary(a)
### we can use the independent sample t test here
sp<-case0502[which(case0502$Judge=="Spock's"),]
nsp<-case0502[which(case0502$Judge!="Spock's"),]
## perform the test
t.test(sp$Percent,nsp$Percent)
The results are CHECK THE IMAGE ATTACHED
b)
> summary(a)
Df Sum Sq Mean Sq F value Pr(>F)
Judge 6 1927 321.2 6.718 6.1e-05 *** as the p value is less than 0.05 , hence there is a significant difference in the percent of women included in the 6 judges’ venires who aren’t Spock’s judge
Residuals 39 1864 47.8
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
c)
t.test(sp$Percent,nsp$Percent)
Welch Two Sample t-test
data: sp$Percent and nsp$Percent
t = -7.1597, df = 17.608, p-value = 1.303e-06 ## as the p value is less than 0.05 , hence we reject the null hypothesis in favor of alternate hypothesis and conclude that there is a significant difference in the percent of women incuded in Spock’s venires versus the percent included in the other judges’ venires combined
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
-19.23999 -10.49935
sample estimates:
mean of x mean of y
14.62222 29.49189
[6.18] ([1] 7.52) Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit. a) What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms? b) Find the probability that the total resistance does not exceed 5100 ohms.
Answer: a) 0.5328 b) 0.9772
Step-by-step explanation:
Given : Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms.
[tex]\mu=200[/tex] and [tex]\sigma=10[/tex]
We assume that the resistance in circuits are normally distributed.
a) Let x denotes the average resistance of the circuit.
Sample size : n= 25
Then, the probability that the average resistance for the 25 resistors is between 199 and 202 ohms :-
[tex]P(199<x<200)=P(\dfrac{199-200}{\dfrac{10}{\sqrt{25}}}<\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}<\dfrac{202-200}{\dfrac{10}{\sqrt{25}}})\\\\=P(-0.5<z<1)\ \ [\because z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}]\\\\=P(z<1)-P(z<-0.5)\\\\=P(z<1)-(1-P(z<0.5))\ \ [\because\ P(Z<-z)=1-P(Z<z)]\\\\=0.8413-(1-0.6915)\ \ [\text{By z-table}]\\\\=0.5328[/tex]
b) Total resistors = 25
Let Z be the total resistance of 25 resistors.
To find P(Z≤5100 ohms) , first we find the mean and variance for Z.
Mean= E(Y) = E(25 X)=25 E(X)=25(200)= 5000 ohm
[tex]Var(Y)=Var(25\ X)=25^2(\dfrac{\sigma^2}{n})=25^2\dfrac{(10)^2}{25}=2500[/tex]
The probability that the total resistance does not exceed 5100 ohms will be :
[tex]P(Y\leq5000)=P(\dfrac{Y-\mu}{\sqrt{Var(Y)}}<\dfrac{5100-5000}{\sqrt{2500}})\\\\=P(z\leq2)=0.9772\ \ [\text{By z-table}][/tex]
Hence, the probability that the total resistance does not exceed 5100 ohms = 0.9772
A polling company conducts an annual poll of adults about political opinions. The survey asked a random sample of 419 adults whether they think things in the country are going in the right direction or in the wrong direction, 54% said that things were going in the wrong direction. How many people would need to ve surveyed for a 90% confidence interval to ensure the margin or error would be less than 3%?
Answer: 747
Step-by-step explanation:
When prior estimate of population proportion (p) is given , then the formula to find the sample size is given by :-
[tex]n=p(1-p)(\dfrac{z^*}{E})^2[/tex]
, where z* = Critical value and E = Margin of error.
As per given , we have
p= 0.54
E= 0.03
Critical value for 90% confidence : z* = 1.645
Then, the required sample size is given by :-
[tex]n=0.54(1-0.54)(\dfrac{1.645}{0.03})^2[/tex]
[tex]n=0.2484(54.8333333333)^2[/tex]
[tex]n=746.862899999\approx747[/tex]
Hence, the number of people would be needed = 747
The mean annual tuition and fees for a sample of 12 private colleges was 36,800 with a standard deviation of 5,000 . A dotplot shows that it is reasonable to assume that the population is approximately normal. You wish to test whether the mean tuition and fees for private colleges is different from . Compute the value of the test statistic and state the number of degrees of freedom.
Answer:
The value of the test statistic and degrees of freedom is 2.148 and 11 respectively.
Step-by-step explanation:
Consider the provided information.
The mean annual tuition and fees for a sample of 12 private colleges was 36,800 with a standard deviation of 5,000 .
Thus, n = 12, [tex]\bar x=36800[/tex] σ = 5000
degrees of freedom = n-1 = 12-1 = 11
[tex]H_0: \mu = 33700\ and\ H_a: \mu \neq 33700[/tex]
Formula to find the value of z is: [tex]z=\frac{\bar x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
Where [tex]\bar x[/tex] is mean of sample, μ is the mean of population, σ is the standard deviation of population and n is number of observations.
[tex]z=\frac{36800-33700}{\frac{5000}{\sqrt{12}}}[/tex]
[tex]z=2.148[/tex]
Hence, the value of the test statistic and degrees of freedom is 2.148 and 11 respectively.
Suppose that your company has just developed a new screening test for a disease and you are in charge of testing its validity and feasibility.You decide to evaluate the test on 1000 individuals and compare the results of the new test to the gold standard.You know the prevalence of disease in your population is 30%.The screening test gave a positive result for 292 individuals.285 of these individuals actually had the disease on the basis of the gold standard determination.
Calculate the sensitivity of the new screening test.
95.0% 97.6% 99.0% 96.9%
Answer:
The sensitivity of the new screening test is 97.6%
Step-by-step explanation:
The sensitivy of a test, or true positive rate, is defined as the proportion of positive results that are correctly identified. It is complementary to the proportion of "false positives".
In this case the test gave 292 positive results. Of this 292 tests, 285 of these individuals actually had the disease.
So the sensititivity is equal to the ratio of the true positives (285) and the total positives (292)
[tex]Sensitivity=\frac{TP}{P}=\frac{285}{292}=0.976=97.6\%[/tex]
As part of a biological research project, researchers need to quantify the density of a certain type of malignant cell in blood. In order to assure the accuracy of measurement, two experienced researchers each make a sequence of separate counts of the number of such cells in the same blood sample. The 7 counts of the first researcher have a mean of 140.2 and a standard deviation of 17, while the 13 counts of the second researcher have a mean of 134.2 and a standard deviation of 15.1.
(a) Use a level 0.99 pooled variance confidence interval to compare the mean counts of the two researchers:
?≤μ1−μ2≤ ?
(b) Does the interval suggest that there is a difference in the mean counts of the two researchers?
Answer:
a) The 99% confidence interval would be given by [tex]-15.277 \leq \mu_1 -\mu_2 \leq 27.277[/tex]
b) No, since the confidence interval contains the 0 we don't have enough evidence to conclude that we have significant differences between the two means.
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
[tex]\bar X_1 =140.2[/tex] represent the sample mean 1
[tex]\bar X_2 =134.2[/tex] represent the sample mean 2
n1=7 represent the sample 1 size
n2=13 represent the sample 2 size
[tex]s_1 =17[/tex] sample standard deviation for sample 1
[tex]s_2 =15.1[/tex] sample standard deviation for sample 2
[tex]\mu_1 -\mu_2[/tex] parameter of interest.
Confidence interval
The confidence interval for the difference of means is given by the following formula:
[tex](\bar X_1 -\bar X_2) \pm t_{\alpha/2} S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}[/tex] (1)
And the pooled variance can be founded with the following formula:
[tex]s^2_p=\frac{(n_x -1)s_x^2 +(n_y-1)s_y^2}{n_x +n_y -2}[/tex]
[tex]s^2_p=\frac{(7 -1)17^2 +(13-1)15.1^2}{7 +13 -2}=248.34[/tex]
[tex]S_p =15.759[/tex] the pooled deviation
The point of estimate for [tex]\mu_1 -\mu_2[/tex] is just given by:
[tex]\bar X_1 -\bar X_2 =140.2-134.2=6[/tex]
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n_1 +n_2 -1=7+13-2=18[/tex]
Since the Confidence is 0.99 or 99%, the value of [tex]\alpha=0.01[/tex] and [tex]\alpha/2 =0.005[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,18)".And we see that [tex]t_{\alpha/2}=2.88[/tex]
The standard error is given by the following formula:
[tex]SE=S_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2}}[/tex]
And replacing we have:
[tex]SE=15.759\sqrt{\frac{1}{7}+\frac{1}{13}}=7.388[/tex]
Part a Confidence interval
Now we have everything in order to replace into formula (1):
[tex]6-2.88(15.759)\sqrt{\frac{1}{7}+\frac{1}{13}}=-15.277[/tex]
[tex]6+2.88(15.759)\sqrt{\frac{1}{7}+\frac{1}{13}}=27.277[/tex]
So on this case the 99% confidence interval would be given by [tex]-15.277 \leq \mu_1 -\mu_2 \leq 27.277[/tex]
Part b Does the interval suggest that there is a difference in the mean counts of the two researchers?
No, since the confidence interval contains the 0 we don't have enough evidence to conclude that we have significant differences between the two means.
The life span of a species of fruit fly have a bell-shaped distribution, with a mean of 33 days and a standard deviation of 4 days.
What percentage corresponds to the life span of fruit flies that are between 29 days and 37 days?
Answer:
68.26% corresponds to the life span of fruit flies that are between 29 days and 37 days
Step-by-step explanation:
Given that the life span of a species of fruit fly have a bell-shaped distribution, with a mean of 33 days and a standard deviation of 4 days.
Let X be the life span
X is N(33,4)
we can convert X into Z standard normal variate by
[tex]z=\frac{x-33}{4}[/tex]
To find the percentage corresponds to the life span of fruit flies that are between 29 days and 37 days
For this let us find probability for x lying between 29 and 37 days
[tex]29\leq x\leq 37\\=-1\leq z\leq 1[/tex]
Probability = P(|z|<1) = 0.6826
Convert to percent
68.26% corresponds to the life span of fruit flies that are between 29 days and 37 days
To find the percentage that corresponds to the life span of fruit flies between 29 days and 37 days, we need to calculate the z-scores for both values and then use the standard normal distribution table to find the area under the curve between those z-scores. The percentage is approximately 68%.
Explanation:To find the percentage that corresponds to the life span of fruit flies between 29 days and 37 days, we need to calculate the z-scores for both values and then use the standard normal distribution table to find the area under the curve between those z-scores.
To calculate the z-scores, we use the formula: z = (x - μ) / σ, where x is the value, μ is the mean, and σ is the standard deviation.
For 29 days: z = (29 - 33) / 4 = -1
For 37 days: z = (37 - 33) / 4 = 1
Looking at the standard normal distribution table or using a calculator, we find that the area between -1 and 1 is approximately 68%. Therefore, the percentage that corresponds to the life span of fruit flies between 29 days and 37 days is 68%.
We wish to obtain a 90% confidence interval for the standard deviation of a normally distributed random variable. To accomplish this we obtain a simple random sample of 16 elements from the population on which the random variable is defined. We obtain a sample mean value of 20 with a sample standard deviation of 12. Give the 90% confidence interval (to the nearest integer) for the standard deviation of the random variable. a) 83 to 307 b) 9 to 18 c) 91 to 270 d) 15 to 25 e) 20 to 34
Answer: d) 15 to 25
Step-by-step explanation:
Given : Sample size : n= 16
Degree of freedom = df =n-1 = 15
Sample mean : [tex]\overline{x}=20[/tex]
sample standard deviation : [tex]s= 12[/tex]
Significance level : [tex]\alpha= 1-0.90=0.10[/tex]
Since population standard deviation is unavailable , so the confidence interval for the population mean is given by:-
[tex]\overline{x}\pm t_{\alpha/2, df}\dfrac{s}{\sqrt{n}}[/tex]
Using t-distribution table , we have
Critical value = [tex]t_{\alpha/2, df}=t_{0.05 , 15}=1.7530[/tex]
90% confidence interval for the mean value will be :
[tex]20\pm (1.7530)\dfrac{12}{\sqrt{16}}[/tex]
[tex]20\pm (1.7530)\dfrac{12}{4}[/tex]
[tex]20\pm (1.7530)(3)[/tex]
[tex]20\pm (5.259)[/tex]
[tex](20-5.259,\ 20+5.259)[/tex]
[tex](14.741,\ 25.259)\approx(15,\ 25 )[/tex][Round to the nearest integer]
Hence, the 90% confidence interval (to the nearest integer) for the standard deviation of the random variable.= 15 to 25.
Final answer:
To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable with a sample size of 16, sample mean of 20, and sample standard deviation of 12, use the chi-square distribution to calculate the lower and upper bounds. The 90% confidence interval is approximately 86 to 283.
Explanation:
To obtain a 90% confidence interval for the standard deviation of a normally distributed random variable, we can use the chi-square distribution. Given a simple random sample of 16 elements with a sample mean of 20 and a sample standard deviation of 12, we can calculate the lower and upper bounds of the confidence interval.
Step 1: Calculate the chi-square values for the lower and upper bounds using the following formulas:
Lower bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 90% confidence level with (n-1) degrees of freedom.
Upper bound: (n-1)s² / X², where n is the sample size, s is the sample standard deviation, and X² is the chi-square value for a 10% significance level with (n-1) degrees of freedom.
Substituting the values into the formulas, we get:
Lower bound: (15)(144) / 24.996 = 86.437
Upper bound: (15)(144) / 7.633 = 283.368
Rounding to the nearest integer, the 90% confidence interval for the standard deviation of the random variable is approximately 86 to 283.
A least squares regression line was found. Using technology, it was determined that the total sum of squares (SST) was 46.8 and the sum of squares of regression (SSR) was 14.55. Use these values to calculate the percent of the variability in y that can be explained by variability in the regression model. Round your answer to the nearest integer.
Answer: 31%
Step-by-step explanation:
Formula : Percent of the variability = [tex]R^2\times100=\dfrac{SSR}{SST}\times100[/tex]
, where [tex]R^2[/tex] = Coefficient of Determination.
SSR = sum of squares of regression
SST = total sum of squares
[tex]R^2[/tex] is the proportion of the variation of Y that can be attributed to the variation of x.
As per given , we have
SSR = 14.55
SST= 46.8
Then, the percent of the variability in y that can be explained by variability in the regression model =[tex]\dfrac{14.55}{46.8}\times100=31.0897435897\%\approx31\%[/tex]
Hence, the percent of the variability in y that can be explained by variability in the regression model = 31%
Answer: 14.55/46.8= .3109
.3109x100=31.09
Step-by-step explanation:
As part of a research project on student debt at TWU, a researcher interviewed a sample of 35 students that were chosen at random concerning their monthly credit card balance. On average, these students had a balance of $2,573. The range of the data ran from a high of $22,315 to a low of $0. The median (Md) was $2,455 and the variance was $4,252. If a student selected at random had a credit card balance of $1,700; then he would have a Z Score of___________.
a. -13.4.
b. -9.73
c. 0.4
d. 9.73
Answer:
Option a is right
Step-by-step explanation:
Given that as part of a research project on student debt at TWU, a researcher interviewed a sample of 35 students that were chosen at random concerning their monthly credit card balance.
Sample average = 2573
Variance = 4252
Sample size = 35
STd deviation of X = [tex]\sqrt{4252} \\=65.21[/tex]
Score of student selected at random X=1700
Corresponding Z score = [tex]\frac{1700-2573}{65.201} \\=-13.38[/tex]
Rounding of we get Z score = -13.4
option a is right
Which of the points does NOT satisfy the inequality shaded in the diagram?
A) (4, 0)
B) (0, 0)
C) (-3, -2)
D) (-10, -1)
The point that does not satisfy the inequality is
A) (4, 0)
How to get the points that satisfy inequalityTo find points satisfying an inequality in a two-dimensional space:
Graph the Inequality: Plot the boundary line as if it were an equation (solid or dashed based on inclusion/exclusion).
Shade the Region: Determine the side of the line representing the solution set by testing a point; shade the satisfying side.
Identify Points: All points within the shaded region, including the boundary if included, satisfy the inequality.
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We know that narrower confidence intervals give us a more precise estimate of the true population proportion. Which of the following could we do to produce higher precision in our estimates of the population proportion?
A. We can select a lower confidence level and increase the sample size.
B. We can select a higher confidence level and decrease the sample size.
C. We can select a higher confidence level and increase the sample size.
D. We can select a lower confidence level and decrease the sample size.
Answer:
A. We can select a lower confidence level and increase the sample size.
Step-by-step explanation:
The length of a confidence interval is:
Direct proportional to the confidence interval. This means that the higher the confidence level, the higher the length of the interval is.
Inverse proportional to the size of the sample.This means that the higher the size of the sample, the lower, or narrower, the length of the interval is.
Which of the following could we do to produce higher precision in our estimates of the population proportion?
We want a narrower interval. So the correct answer is:
A. We can select a lower confidence level and increase the sample size.
The price to earnings ratio (P/E) is an important tool in financial work. A random sample of 14 large U.S. banks (J. P. Morgan, Bank of America, and others) gave the following P/E ratios†.24 16 22 14 12 13 17 22 15 19 23 13 11 18
The sample mean is x ≈ 17.1. Generally speaking, a low P/E ratio indicates a "value" or bargain stock.
Suppose a recent copy of a magazine indicated that the P/E ratio of a certain stock index is μ = 18.
Let x be a random variable representing the P/E ratio of all large U.S. bank stocks.
We assume that x has a normal distribution and σ = 5.1.
Do these data indicate that the P/E ratio of all U.S. bank stocks is less than 18? Use α = 0.01.(a) What is the level of significance?(b) What is the value of the sample test statistic? (Round your answer to two decimal places.)(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
Answer:
a) [tex]\alpha=0.01[/tex] is the significance level given
b) [tex]z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603[/tex]
c) Since is a one side left tailed test the p value would be:
[tex]p_v =P(Z<-0.6603)=0.2545[/tex]
Step-by-step explanation:
Data given and notation
[tex]\bar X=17.1[/tex] represent the mean P/E ratio for the sample
[tex]\sigma=5.1[/tex] represent the sample standard deviation for the population
[tex]n=14[/tex] sample size
[tex]\mu_o =18[/tex] represent the value that we want to test
[tex]\alpha=0.01[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean for the P/E ratio is less than 18, the system of hypothesis would be:
Null hypothesis:[tex]\mu \geq 18[/tex]
Alternative hypothesis:[tex]\mu < 18[/tex]
If we analyze the size for the sample is < 30 but we know the population deviation so is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:
[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)
z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
(a) What is the level of significance?
[tex]\alpha=0.01[/tex] is the significance level given
(b) What is the value of the sample test statistic?
We can replace in formula (1) the info given like this:
[tex]z=\frac{17.1-18}{\frac{5.1}{\sqrt{14}}}=-0.6603[/tex]
(c) Find (or estimate) the P-value. (Round your answer to four decimal places.)
Since is a one side left tailed test the p value would be:
[tex]p_v =P(Z<-0.6603)=0.2545[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.01[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL reject the null hypothesis, so we can conclude that the true mean for the P/E ratio is not significantly less than 18.
A lab technician is tested for her consistency by making multiple measurements of the cholesterol level in one blood sample. The target precision is a standard deviation of 1.2 mg/dL or less. If 12 measurements are taken and the standard deviation is 1.8 mg/dL, is there enough evidence to support the claim that her standard deviation is greater than the target, at = .01? (Show the answers to all 5 steps of the hypothesis test.)
Step-by-step explanation:
Given precision is a standard deviation of s=1.8, n=12, target precision is a standard deviation of σ=1.2
The test hypothesis is
H_o:σ <=1.2
Ha:σ > 1.2
The test statistic is
chi square = [tex]\frac{(n-1)s^2}{\sigma^2}[/tex]
=[tex]\frac{(12-1)1.8^2}{1.2^2}[/tex]
=24.75
Given a=0.01, the critical value is chi square(with a=0.01, d_f=n-1=11)= 3.05 (check chi square table)
Since 24.75 > 3.05, we reject H_o.
So, we can conclude that her standard deviation is greater than the target.
A computer can be classified as either cutting-edge or ancientancient. Suppose that 91% of computers are classified as ancient.
a) Two computers are chosen at random. What is the probability that both computers are ancient?
b) Eight computers are chosen at random. What is the probability that all eight computers are ancient?
c) What is the probability that at least one of eight randomly selected computers is cutting-edge? Would it be unusual that at least one of eighteight randomly selected computers is cutting-edge?
Answer:
a)P(X=2) = (2C2)(0.91)^2 (1-0.91)^{2-2}=0.8281[/tex]
b) P(X=8)=(8C8)(0.91)^8 (1-0.91)^{8-8}=0.4703[/tex]
c) [tex]P(X \geq 1)=1-P(X<1)=1-P(X=0)=1-0.4703=0.5297[/tex]
Step-by-step explanation:
A Bernoulli trial is "a random experiment with exactly two possible outcomes, "success" and "failure", in which the probability of success is the same every time the experiment is conducted". And this experiment is a particular case of the binomial experiment.
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
The complement rule is a theorem that provides a connection between the probability of an event and the probability of the complement of the event. Lat A the event of interest and A' the complement. The rule is defined by: [tex]P(A)+P(A') =1[/tex]
a. Two computers are chosen at random. What is the probability that both computers are ancient?
[tex]P(X=2)=(2C2)(0.91)^2 (1-0.91)^{2-2}=0.8281[/tex]
b. Eight computers are chosen at random. What is the probability that all eight computers are ancient?
On this case we are looking for this probability:
[tex]P(X=8)=(8C8)(0.91)^8 (1-0.91)^{8-8}=0.4703[/tex]
c. What is the probability that at least one of eight randomly selected computers is cutting-edge? Would it be unusual that at least one of eighteight randomly selected computers is cutting-edge?
Since we are interested on the cutting edge class the new probability of success would be p=1-0.91=0.09. And we want to find this probability:
[tex]P(X \geq 1)=1-P(X<1)=1-[P(X=0)][/tex]
And we can find the indiviudal probabilitiy like this:
[tex]P(X=0)=(8C0)(0.09)^0 (1-0.09)^{8-0}=0.4703[/tex]
And if we replace we got:
[tex]P(X \geq 1)=1-P(X<1)=1-P(X=0)=1-0.4703=0.5297[/tex]
The probability of both computers being ancient is about 82.81%, all eight being ancient is about 43.05%, and at least one of eight being cutting-edge is roughly 56.95%, which is not unusual.
Explanation:Probability of Selecting Ancient Computers
When dealing with probabilities of independent events, such as selecting computers at random, we can calculate the probability of multiple events occurring in sequence by multiplying the probabilities of each individual event.
The probability that both computers are ancient when two are chosen is 0.91 × 0.91 or about 0.8281 (82.81%).The probability that all eight computers are ancient when eight are chosen is 0.918 or about 0.43046721 (43.05%).To find the probability that at least one of eight computers is cutting-edge, we first calculate the probability that none are cutting-edge (all ancient) and subtract it from 1. This is 1 – 0.918 or about 0.5695 (56.95%). It would not be unusual that at least one is cutting-edge as this probability is higher than 50%.Note that these calculations assume that each computer's classification as ancient or cutting-edge is independent of the others.
Solve the following equation by taking the square root 12 - 6n2 = -420 i need help
Answer:
[tex]n=\pm 6\sqrt{2}[/tex]
Step-by-step explanation:
It may work well to divide by 6, subtract 2, and multiply by -1 before you take the square root.
[tex]12-6n^2=-420\\2-n^2=-70 \qquad\text{divide by 6}\\-n^2=-72 \qquad\text{subtract 2}\\n^2=72 \qquad\text{multiply by -1}\\\\n=\pm\sqrt{36\cdot 2} \qquad\text{take the square root}\\\\n=\pm 6\sqrt{2} \qquad\text{simplify}[/tex]
A random sample of 500 registered voters in Phoenix is asked if they favor the use of oxygenated fuels year-round to reduce air pollution. If more than 314 voters respond positively, we will conclude that at least 60% of the voters favor the use of these fuels. Round your answers to four decimal places (e.g. 98.7654).
a) Find the probability of type I error if exactly 60% of the voters favor the use of these fuelsb) What is the Type II error probability (Beta) β if 75% of the voters favor this action?
Answer:
a) 0.0853
b) 0.0000
Step-by-step explanation:
Parameters given stated that;
H₀ : p = 0.6
H₁ : p = 0.6, this explains the acceptance region as;
p° ≤ [tex]\frac{315}{500}[/tex]=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)
a).
the probability of type I error if exactly 60% is calculated as :
∝ = P (Reject H₀ | H₀ is true)
= P (p°>0.63 | p=0.6)
where p° is represented as pI in the subsequent calculated steps below
= P [tex][\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6][/tex]
= P [tex][\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ][/tex]
= P [tex][Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ][/tex]
= P [Z > 1.37]
= 1 - P [Z ≤ 1.37]
= 1 - Ф (1.37)
= 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)
≅ 0.0853
b)
The probability of Type II error β is stated as:
β = P (Accept H₀ | H₁ is true)
= P [p° ≤ 0.63 | p = 0.75]
where p° is represented as pI in the subsequent calculated steps below
= P [tex][\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75][/tex]
= P [tex][\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ][/tex]
= P[tex][Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ][/tex]
= P [Z ≤ -6.20]
= Ф (-6.20)
≅ 0.0000 (from Cumulative Standard Normal Distribution Table).
Final answer:
The probability of Type I error can be calculated using the formula P(Type I error) = P(Z > Zα), and the probability of Type II error (Beta) can be calculated using the formula Beta = P(Z < Zβ) + P(Z > Z1-β).
Explanation:
a) The probability of Type I error can be calculated using the formula:
P(Type I error) = P(Z > Zα)
where Zα is the standard score corresponding to the desired level of significance.
b) The probability of Type II error (Beta) can be calculated using the formula:
Beta = P(Z < Zβ) + P(Z > Z1-β)
where Zβ is the standard score corresponding to the desired power level and Z1-β is the standard score corresponding to the complement of the desired power level.
If the sample size is n = 75, what are the degrees of freedom for the appropriate chi-square distribution when testing for independence of two variables each with three categories? a. 4 b. 69 c. 74
Final answer:
The degrees of freedom for a chi-square test of independence with two variables each having three categories and a sample size of n = 75 is 4.
Explanation:
The question at hand involves determining the degrees of freedom (df) for a chi-square test of independence where two variables each have three categories, and the sample size is n = 75. To calculate the degrees of freedom for this scenario, you use the formula df = (r - 1)(c - 1) where r represents the number of rows (categories of one variable) and c represents the number of columns (categories of the other variable).
In this case, with both variables having three categories, we have r = 3 and c = 3, which gives us:
df = (3 - 1)(3 - 1) = (2)(2) = 4.
Therefore, the correct answer is a. 4 degrees of freedom for the chi-square distribution when testing for independence with a sample size of n = 75.
Why must integration be used to find the work required to pump water out of a tank?
A. Different volumes of water are moved different distances.
B. Water from the same horizontal planes is lifted different distances.
C. Integration is necessary because the acceleration of gravity changes at each level.
D. Integration is necessary because W = mgy.
Answer:
A. Different volumes of water are moved different distances.
Step-by-step explanation:
Integration is used to find work required to pump water out of a tank because different volumes of water are moved different distances and to sum it all we the tool required is integration. Moreover, work done is a path function or an inexact differential. It does depend upon the path followed by the process.
hence the correct answer is A.
use the intermediate value theorem to determine whether the following equation has a solution or not x^3-3x-1
Answer:
Yes, this equation has a solution. According to Intermediate Value Theorem at least one solution for [0,2]
Step-by-step explanation:
Hi there!
1) Remember a definition.
Intermediate Value Theorem:
If [tex]f[/tex] is continuous on a given closed interval [a,b], and f(a)≠f(b) and f(a)<k<f(b) then there has to be at least one number 'c' between 'a' and 'b', such that f(c)=k
----
(Check the first graph as an example)
2) The Intermediate Value Theorem can be applied to determine whether there is a solution on a given interval.
Let's choose the interval [tex][0,2][/tex]
[tex]f(x)=x^{3}-3x-1\\f(0)=(0)^{3}-3(0)-1\\f(0)=-1\\f(0)<0\\[/tex]
Proceed to the other point: 2
[tex]f(x)=x^{3}-3x-1\\f(2)=(2)^{3}-3(2)-1\\f(2)=1\\f(2)>0\\[/tex]
3) Check the 2nd Graph for a the Visual answer, of it. And the 3rd graph for all solutions of this equation.
A sample of 12 radon detectors of a certain type was selected, and each was exposed to 100 pCI/L of radon. The resulting readings were as follows: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4 Does this data suggest that the population mean reading under these conditions differs from 100? Set up an appropriate hypothesis test to answer this question.
Answer:
Null hypothesis:[tex]\mu = 100[/tex]
Alternative hypothesis:[tex]\mu \neq 100[/tex]
[tex]t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921[/tex]
[tex]p_v =2*P(t_{11}<-0.921)=0.377[/tex]
Step-by-step explanation:
1) Data given and notation
Data: 105.6, 90.9, 91.2, 96.9, 96.5, 91.3, 100.1, 105.0, 99.6, 107.7, 103.3, 92.4
We can calculate the sample mean and deviation for this data with the following formulas:
[tex]\bar X =\frac{\sum_{i=1}^n X_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (X_i- \bar X)^2}{n-1}}[/tex]
The results obtained are:
[tex]\bar X=98.375[/tex] represent the sample mean
[tex]s=0.6.109[/tex] represent the sample standard deviation
[tex]n=12[/tex] sample size
[tex]\mu_o =100[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
z would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
2) State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is equal to 100pCL/L, the system of hypothesis are :
Null hypothesis:[tex]\mu = 100[/tex]
Alternative hypothesis:[tex]\mu \neq 100[/tex]
Since we don't know the population deviation, is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
3) Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{98.375-100}{\frac{6.109}{\sqrt{12}}}=-0.921[/tex]
4) P-value
First we need to find the degrees of freedom for the statistic given by:
[tex]df=n-1=12-1=11[/tex]
Since is a two sided test the p value would given by:
[tex]p_v =2*P(t_{11}<-0.921)=0.377[/tex]
5) Conclusion
If we compare the p value and the significance level assumed [tex]\alpha=0.05[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean is not significant different from 100 at 5% of significance.
The question asked is about setting a hypothesis test to see if the mean reading from radon detectors differs from 100 pCI/L. The null and alternative hypotheses should be formed, calculations should be performed and the p-value compared with the significance level to determine if the null hypothesis should be retained or rejected.
Explanation:The subject of this question is setting up a hypothesis test for determining if the mean reading from the radon detectors differs from 100 pCI/L. The first step is to set up the null and alternate hypothesis. The null hypothesis (H0) would be that the population mean reading is 100 pCI/L (µ = 100), whereas the alternative hypothesis (H1) would propose that the mean reading differs from 100 pCI/L (µ ≠ 100).
Next, we would calculate the sample mean and sample standard deviation. Using these calculations, you can perform a t-test to compare the sample mean to the proposed population mean of 100 pCI/L. The decision of rejecting or not rejecting the null hypothesis relies on the comparison of the p-value obtained from the test statistic with the significance level.
Without the actual calculations, it is not possible to conclude whether the data suggests that the population mean reading under these conditions differs from 100 pCI/L or not. However, if the p-value is less than the significance level (commonly 0.05), we would reject the null hypothesis and conclude that the data provides enough evidence to suggest that the population mean differs from 100.
Learn more about Hypothesis Test here:https://brainly.com/question/34171008
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Suppose a realtor wants to determine the current percentage of customers who have a family of five or more. How many customers should the realtor survey in order to be 98% confident that the estimated (sample) proportion is within 2 percentage points of the true population proportion of customers who have a family of five or more?
Answer:
n=3394
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
The margin of error is the range of values below and above the sample statistic in a confidence interval.
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The population proportion have the following distribution
[tex]p \sim N(p,\sqrt{\frac{p(1-p)}{n}})[/tex]
Solution to the problem
The confidence interval for the mean is given by the following formula:
[tex]\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex]
In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2 =0.01[/tex]. And the critical value would be given by:
[tex]z_{\alpha/2}=-2.33, z_{1-\alpha/2}=2.33[/tex]
The margin of error for the proportion interval is given by this formula:
[tex] ME=z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}[/tex] (a)
And on this case we have that [tex]ME =\pm 0.02[/tex] and we are interested in order to find the value of n, if we solve n from equation (a) we got:
[tex]n=\frac{\hat p (1-\hat p)}{(\frac{ME}{z})^2}[/tex] (b)
We can use as estimation of [tex]\hat p=0.5[/tex] since we don't have any other info provided. And replacing into equation (b) the values from part a we got:
[tex]n=\frac{0.5(1-0.5)}{(\frac{0.02}{2.33})^2}=3393.06[/tex]
And rounded up we have that n=3394
Twenty years ago, entering male high school students of Central High could do an average of 24 pushups in 60 seconds. To see whether this remains true today, a random sample of 36 freshmen was chosen. Suppose their average was 22.5 with a sample standard deviation of 3.1,
(a) Test, using the p-value approach, whether the mean is still equal to 24 at the 5 percent level of significance.
(b) Calculate the power of the test if the true mean is 23.
Answer:
a) [tex]t=\frac{22.5-24}{\frac{3.1}{\sqrt{36}}}=-2.903[/tex]
[tex]p_v =2*P(t_{(35)}<-2.903)=0.0064[/tex]
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is still equal to 24 at 5% of significance.
b) Power =0.4626+0.000172=0.463
See explanation below.
Step-by-step explanation:
Part a
Data given and notation
[tex]\bar X=22.5[/tex] represent the sample mean
[tex]s=3.1[/tex] represent the sample standard deviation
[tex]n=36[/tex] sample size
[tex]\mu_o =24[/tex] represent the value that we want to test
[tex]\alpha=0.05[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to conduct a hypothesis in order to check if the mean is still equal to 24, the system of hypothesis would be:
Null hypothesis:[tex]\mu = 24[/tex]
Alternative hypothesis:[tex]\mu \neq 24[/tex]
If we analyze the size for the sample is > 30 but we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{22.5-24}{\frac{3.1}{\sqrt{36}}}=-2.903[/tex]
P-value
The first step is calculate the degrees of freedom, on this case:
[tex]df=n-1=36-1=35[/tex]
Since is a bilateral test the p value would be:
[tex]p_v =2*P(t_{(35)}<-2.903)=0.0064[/tex]
Conclusion
If we compare the p value and the significance level given [tex]\alpha=0.05[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, so we can't conclude that the true mean is still equal to 24 at 5% of significance.
Part b
The Power of a test is the probability of rejecting the null hypothesis when, in the reality, it is false.
For this case the power of the test would be:
P(reject null hypothesis| [tex]\mu=23[/tex])
If we see the null hypothesis we reject it when we have this:
The critical values from the t distribution with 35 degrees of freedom and at 5% of significance are -2.03 and 2.03. From the z score formula:
[tex]t=\frac{\bar x-\mu}{\frac{s}{\sqrt{n}}}[/tex]
If we solve for [tex]\bar x[/tex] we got:
[tex]\bar X= \mu \pm t \frac{s}{\sqrt{n}}[/tex]
Using the two critical values we have the critical values four our sampling distribution under the null hypothesis
[tex]\bar X= 24 -2.03 \frac{3.1}{\sqrt{36}}=22.951[/tex]
[tex]\bar X= 24 +2.03 \frac{3.1}{\sqrt{36}}=25.049[/tex]
So we reject the null hypothesis if [tex]\bar x<22.951[/tex] or [tex]\bar X >25.049[/tex]
So for our case:
P(reject null hypothesis| [tex]\mu=23[/tex]) can be founded like this:
[tex]P(\bar X <22.951|\mu=23)=P(t<\frac{22.951-23}{\frac{3.1}{\sqrt{36}}})=P(t_{35}<-0.0948)=0.4626[/tex]
[tex]P(\bar X >25.012|\mu=23)=P(t<\frac{25.049-23}{\frac{3.1}{\sqrt{36}}})=P(t_{35}>3.966)=0.000172[/tex]
And the power on this case would be the sum of the two last probabilities:
Power =0.4626+0.000172=0.463
Final answer:
To test whether the mean length of time students spend doing homework each week has increased, we can conduct a hypothesis test using the null hypothesis that the mean time is still 2.5 hours and the alternative hypothesis that the mean time has increased.
Explanation:
To test whether the mean length of time students spend doing homework each week has increased, we can conduct a hypothesis test. The null hypothesis, denoted as H0, would be that the mean time is still 2.5 hours. The alternative hypothesis, denoted as Ha, would be that the mean time has increased. In this case, the alternative hypothesis would be Ha: µ > 2.5, where µ represents the population mean.
To conduct the hypothesis test, we can use a t-distribution because the population standard deviation is not known. We can calculate the test statistic by using the formula: t = (x - µ) / (s/√n), where x is the sample mean, µ is the hypothesized mean, s is the sample standard deviation, and n is the sample size. Once we calculate the test statistic, we can compare it to the critical value from the t-distribution table or calculate the p-value to determine the level of significance.
find y from the picture plz
It's a six sided polygon. For any polygon the external angles add to 360 degrees. The internal angles shown are the supplements of the external angles. We have
(180 - θ₁) + (180 - θ₂) + ... + (180 - θ₆) = 360
6(180) - 360 = θ₁ + θ₂ + θ₃ + θ₄ + θ₅ + θ₆
720 = θ₁ + θ₂ + θ₃ + θ₄ + θ₅ + θ₆
The six angles add up to 720 degrees, and five of them add to
126+101+135+147+96=605
So y = 720 - 605 = 115
The degree sign is external to y so not part of the answer:
Answer: 115
Answer:y = 115 degrees
Step-by-step explanation:
The given polygon has 6 sides. It is a hexagon. The sum of the interior angles of a polygon is
180(n - 2)
Where
n is the number of sides that the polygon has. This means that n = 6
Therefore, the sum of the interior angles would be
180(6 - 2) = 720 degrees
Therefore,
126 + 101 + 135 + 147 + 96 + y = 720
605 + y = 720
Subtracting 605 from both sides of the equation, it becomes
y = 720 - 605 = 115 degrees
An urn contains 17 red marbles and 18 blue marbles. 16 marbles are chosen. In how many ways can 6 red marbles be chosen?
Answer:
Total number of ways 6 red marbles can be chosen=541549008
Step-by-step explanation:
16 marbles are chosen in which 6 are red marbles and remaining marbles ,which are 10, are blue marbles.
In order to find in how many ways 6 red marbles can be chosen we will proceed as:
Out of 17 red marbles 6 are chosen and out of 18 blue marbles 10 are chosen.
Total number of ways 6 red marbles can be chosen= [tex]17_{C_6} * 18_{C_1_0}[/tex]
Total number of ways 6 red marbles can be chosen=[tex]\frac{17!}{6!*(17-6)!} * \frac{18!}{10!*(18-10)!}[/tex]
Total number of ways 6 red marbles can be chosen= 12376*43758
Total number of ways 6 red marbles can be chosen=541549008
Answer: N = 541,549,008
Therefore the number of ways to select 6 red marbles is 541,549,008
Step-by-step explanation:
Given;
Number of red marbles total = 17
Number of blue marbles total = 18
Number of red marbles to be selected = 6
Number of blue marbles to be selected = 16 - 6 = 10
To determine the number of ways 6 red marbles can be selected N.
N = number of ways 6 red marbles can be selected from 17 red marbles × number of ways 10 blue marbles can be selected from 18 blue marbles
N = 17C6 × 18C10
N = 17!/(6! × (17-6)!) × 18!/(10! × (18-10)!)
N = 17!/(6! × 11!) × 18!/(10! × 8!)
N = 541,549,008
Therefore the number of ways to select 6 red marbles is 541,549,008
Find the percent of the data that can be explained by the regression line and regression equation given that the correlation coefficient = -.72 (Give your answer as a percent rounded to the hundredth decimal place. Include the % sign)
Answer:
51.84%
Step-by-step explanation:
The percentage of data explained by regression line is assessed using R-square. Here, in the given scenario correlation coefficient r is given. We simply take square of correlation coefficient to get r-square. r-square=(-0.72)^2=0.5184=51.84%
Find the seventh term of an increasing geometric progression if the first term is equal to 9−4sqrt5 and each term (starting with the second) is equal to the difference of the term following it and the term preceding it.
Answer:
7th term = 1.
Step-by-step explanation:
Given that, first term of increasing geometric progression is 9-4√5.
each term (starting with the second) is equal to the difference of the term following it and the term preceding it.
let first term of geometric progression be a and the increasing ratio be r.
⇒ The geometric progression is a , ar , ar² , ar³, ....... so on.
Given, each term (starting with the second) is equal to the difference of the term following it and the term preceding it.
⇒ second term = (third term - first term)
⇒ ar = (ar² - a)
⇒ r = r² - 1
⇒ r² - r -1 =0
⇒ roots of this equation is r = [tex]\frac{1+\sqrt{5} }{2}[/tex] , [tex]\frac{1-\sqrt{5} }{2}[/tex]
(roots of ax²+bx+c are [tex]\frac{-b+\sqrt{b^{2} -4ac} }{2a}[/tex] and [tex]\frac{-b-\sqrt{b^{2} -4ac} }{2a}[/tex])
and it is given, increasing geometric progression
⇒ r > 0.
⇒ r = [tex]\frac{1+\sqrt{5} }{2}[/tex].
Now, nth term in geometric progression is arⁿ⁻¹.
⇒ 7th term = ar⁷⁻¹ = ar⁶.
= (9-4√5)([tex]\frac{1+\sqrt{5} }{2}[/tex])⁶
= (0.05572809)(17.94427191) = 1
⇒ 7th term = 1.
A researcher matched 30 participants on intelligence (hence 15 pairs of participants), and then compared differences in emotional responsiveness to two experimental stimuli between each pair. For this test, what are the critical values, assuming a two-tailed test at a 0.05 level of significance?
(A) ±2.042
(B) ±2.045
(C) ±2.131
(D) ±2.145
How many degrees of freedom does the chi-square test statistic for a goodness of fit have when there are 10 categories?
a. 9
b. 7
c. 62
d. 74
Answer:
[tex] df = n-1=10-1=9[/tex]
a. 9
Step-by-step explanation:
Previous concepts
A chi-square goodness of fit test "determines if a sample data matches a population".
A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".
We assume that we have the following system of hypothesis:
H0: The data follows the distribution proposed
H1: The data not follows the distribution proposed
The statistic to check the hypothesis is given by:
[tex]\chi^2 =\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
This statistic have a Chi Square distribution distribution with k-1 degrees of freedom, where n represent the number of categories on this case k=10. And if we find the degrees of freedom we got:
[tex] df = k-1=10-1=9[/tex]
a. 9
Final answer:
The correct answer is a. 9.
The degrees of freedom for a chi-square goodness-of-fit test with 10 categories is 9, which is calculated by subtracting one from the number of categories.
Explanation:
The degrees of freedom for a chi-square goodness-of-fit test are calculated as the number of categories minus one. In the case of having 10 categories, the degrees of freedom would be 10 - 1 = 9.
Therefore, the correct answer is a. 9.
Remember, the chi-square test statistic helps us determine how well an observed distribution fits an expected distribution, and degrees of freedom are essential for determining the critical values of the test from the chi-square distribution.
For a chi-square distribution, as the degrees of freedom increase, the curve becomes more symmetrical.
Minor surgery on horses under field conditions requires a reliable short-term anesthetic producing good muscle relaxation, minimal cardiovascular and respiratory changes, and a quick, smooth recovery with minimal aftereffects so that horses can be left unattended. An article reports that for a sample of n = 75 horses to which ketamine was administered under certain conditions, the sample average lateral recumbency (lying-down) time was 18.81 min and the standard deviation was 8.4 min.
Does this data suggest that true average lateral recumbency time under these conditions is less than 20 min? Test the appropriate hypotheses at level of significance 0.10. State the appropriate null and alternative hypotheses.
Answer:
[tex]p_v =P(t_{74}<-1.227)=0.112[/tex]
If we compare the p value and a significance level for example [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.
Step-by-step explanation:
Data given and notation
[tex]\bar X=18.81[/tex] represent the average lateral recumbency for the sample
[tex]s=8.4[/tex] represent the sample standard deviation
[tex]n=75[/tex] sample size
[tex]\mu_o =20[/tex] represent the value that we want to test
[tex]\alpha[/tex] represent the significance level for the hypothesis test.
t would represent the statistic (variable of interest)
[tex]p_v[/tex] represent the p value for the test (variable of interest)
State the null and alternative hypotheses.
We need to apply a left tailed test.
What are H0 and Ha for this study?
Null hypothesis: [tex]\mu \geq 20[/tex]
Alternative hypothesis :[tex]\mu < 20[/tex]
Compute the test statistic
The statistic for this case is given by:
[tex]t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}[/tex] (1)
t-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".
Calculate the statistic
We can replace in formula (1) the info given like this:
[tex]t=\frac{18.81-20}{\frac{8.4}{\sqrt{75}}}=-1.227[/tex]
The degrees of freedom are given by:
[tex]df=n-1=75-1=74[/tex]
Give the appropriate conclusion for the test
Since is a one side left tailed test the p value would be:
[tex]p_v =P(t_{74}<-1.227)=0.112[/tex]
Conclusion
If we compare the p value and a significance level for example [tex]\alpha=0.1[/tex] we see that [tex]p_v>\alpha[/tex] so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, so we can conclude that the true mean it's not significantly less than 20 min.