A doctor wants to estimate the mean HDL cholesterol of all​ 20- to​ 29-year-old females. The number of subjects needed to estimate the mean HDL cholesterol within 3 points with 99% confidence is 160 subjects. Suppose the doctor decides that he could be content with only 90% confidence. Assuming s=14.7 based on earlier​ studies, how would this decrease in confidence affect the sample size required?

Answer:

a) Sample mean: 48.71

Sample median: 48.3

Sample variance: 5.41

Sample standard deviation: 2.37

Step-by-step explanation:

a) Sample mean:

[tex]\bar{X}=\frac{1}{N} \sum X_i=\frac{1}{80}*3896.9=48.71[/tex]

Sample median: M=48.3

Note: I order the data increasingly and take the value N / 2 = 40. In this way there are 39 values above and 39 values below the median.

Sample variance:

[tex]s^2=\frac{1}{N-1}\sum (X_i-\bar X)^2 =(\frac{1}{80-1})*427.74=5.41[/tex]

Sample standard deviation

[tex]s=\sqrt{s^2}=\sqrt{5.41}=2.37[/tex]

b)

Answer:

a) [tex]t_{crit}=1.34[/tex]

b) [tex]t_{crit}=-1.33[/tex]

c) [tex]t_{crit}=\pm 2.11[/tex]

Step-by-step explanation:

Part a

[tex]\alpha=0.1[/tex] represent the significance level

df =15

Since is a right tailed test the critical value is given by:

[tex]t_{crit}=1.34[/tex]

And we can use the following excel code to find it: "=T.INV(0.9,15)"

Part b

[tex]\alpha=0.1[/tex] represent the significance level

n=20 represent the sample size

First we need to find the degrees of freedom given by:

[tex]df=n-1=20-1=19[/tex]

Since is a left tailed test the critical value is given by:

[tex]t_{crit}=-1.33[/tex]

And we can use the following excel code to find it: "=T.INV(0.1,19)"

Part c

[tex]\alpha=0.05[/tex] represent the significance level

n=18 represent the sample size

First we need to find the degrees of freedom given by:

[tex]df=n-1=18-1=17[/tex]

The value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]

Since is a two tailed tailed we have two critical values is given by:

[tex]t_{crit}=\pm 2.11[/tex]

And we can use the following excel code to find it: "=T.INV(0.025,17)"

Answer:

[tex]p_v =P(t_{17}<-1.728)=0.051[/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 FAIL to reject the null hypothesis.

The best option for this case would be:

We would switch to α = .01 for a more powerful test.

Step-by-step explanation:

Previous concepts and data given  

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=37.8[/tex] represent the sample mean  

[tex]s=5.4[/tex] represent the sample standard deviation  

n=18 represent the sample selected  

[tex]\alpha=0.05[/tex] significance level  

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if the mean is less than 40, the system of hypothesis would be:    

Null hypothesis:[tex]\mu \geq 40[/tex]    

Alternative hypothesis:[tex]\mu < 40[/tex]    

If we analyze the size for the sample is < 30 and 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{37.8-40}{\frac{5.4}{\sqrt{18}}}=-1.728[/tex]      

P-value

First we need to calculate the degrees of freedom given by:

[tex]df=n-1=18-1=17[/tex]

Since is a left tailed test the p value would be:    

[tex]p_v =P(t_{17}<-1.728)=0.051[/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 FAIL to reject the null hypothesis.

The best option for this case would be:

We would switch to α = .01 for a more powerful test.

Since the p values is just a little higher than th significance level 0.051>0.05  but the values are two close. If we change the value of the significance by 0.01, we have that 0.051>0.01 and it's a more powerful test.

Answer:

f(x) = 3[tex](\frac{1}{2})^{x}[/tex]

Step-by-step explanation:

hope it helps!

The function that represents a vertical stretch of an exponential function is f(x) = (3)^x.

The function that represents a vertical stretch of an exponential function is f(x) = (3)x.

In this function, the base of the exponential term is 3, and the exponent, x, determines the position on the graph. When the value of x increases, the function values also increase at an exponential rate.

For example, when x = 1, f(1) = (3)1 = 3. When x = 2, f(2) = (3)2 = 9. The function values double with each increase of x.

Answer:

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/tex]  

[tex]p_v =P(z>3.785)=7.68x10^{-5}[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.  

Step-by-step explanation:

1) Data given and notation

n=51 represent the random sample taken

X=39 represent the number of boys

[tex]\hat p=\frac{39}{51}=0.765[/tex] estimated proportion of boys

[tex]p_o=0.5[/tex] is the value that we want to test

[tex]\alpha=0.01[/tex] represent the significance level

Confidence=99% or 0.99

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value (variable of interest)  

2) Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that with this method, the probability of a baby being a boy is greater than 0.5.:  

Null hypothesis:[tex]p\leq 0.5[/tex]  

Alternative hypothesis:[tex]p > 0.5[/tex]  

When we conduct a proportion test we need to use the z statisitc, and the is given by:  

[tex]z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}}[/tex] (1)  

The One-Sample Proportion Test is used to assess whether a population proportion [tex]\hat p[/tex] is significantly different from a hypothesized value [tex]p_o[/tex].

3) Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

[tex]z=\frac{0.765 -0.5}{\sqrt{\frac{0.5(1-0.5)}{51}}}=3.785[/tex]  

4) Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided [tex]\alpha=0.01[/tex]. The next step would be calculate the p value for this test.  

Since is a right tailed test the p value would be:  

[tex]p_v =P(z>3.785)=7.68x10^{-5}[/tex]  

So the p value obtained was a very low value and using the significance level given [tex]\alpha=0.01[/tex] we have [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and we can said that at 1% of significance the proportion of boys is significantly higher than 0.5.  

Answer:

[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]  

And rounded up we have that n=1291

Step-by-step explanation:

1) 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]

2) Solution to the problem

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 95% of confidence, our significance level would be given by [tex]\alpha=1-0.95=0.05[/tex] and [tex]\alpha/2 =0.025[/tex]. And the critical value would be given by:

[tex]z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96[/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 0.025[/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)  

And replacing into equation (b) the values from part a we got:

[tex]n=\frac{0.3(1-0.3)}{(\frac{0.025}{1.96})^2}=1290.78[/tex]  

And rounded up we have that n=1291

Answer: 2172

Step-by-step explanation:

Formula to find the sample size n , if the prior estimate of the population proportion(p) is known:

[tex]n= p(1-p)(\dfrac{z^*}{E})^2[/tex] , where E=  margin of error and z* = Critical z-value.

Let p be the population proportion of adults have consulted fortune tellers.

As per given , we have

p= 0.20

E= 0.02

From z-table , the z-value corresponding to 98% confidence interval = z*=2.33

Then, the required sample size will be :

[tex]n= 0.20(1-0.20)(\dfrac{2.33}{0.02})^2[/tex]

[tex]n= 0.20(0.80)(116.5)^2[/tex]

[tex]n= 2171.56\approx2172[/tex]

Hence, the required sample size = 2172

Answer:

First question:

Top 6%: 4.87 ounces

Bottom 6%: 4.75 ounces

Second question:

Top 7%: Score of 649.4.

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

For the first problem, we have that:

[tex]\mu = 4.81, \sigma = 0.04[/tex]

Top 6%

The value of X when Z has a pvalue of 0.94. This is [tex]Z = 1.555[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.555 = \frac{X - 4.81}{0.04}[/tex]

[tex]X - 4.81 = 1.555*0.04[/tex]

[tex]X = 4.8722[/tex]

Bottom 6%

The value of X when Z has a pvalue of 0.06. This is [tex]Z = 1.555[/tex]

For the second problem, we have that:

[tex]\mu = 496, \sigma = 109[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.555 = \frac{X - 4.81}{0.04}[/tex]

[tex]X - 4.81 = -1.555*0.04[/tex]

[tex]X = 4.7477[/tex]

Top 7%

The value of X when Z has a pvalue of 0.93. This is [tex]Z = 1.475[/tex]

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]1.475 = \frac{X - 496}{104}[/tex]

[tex]X - 496 = 104*1.475[/tex]

[tex]X = 649.4[/tex]

Answer:

a) P(24) = 1025.

b) P(24) = 191.

Step-by-step explanation:

This population can be modeled by the following exponential model.

[tex]P(t) = P_{0}e^{rt}[/tex]

In which P(t) is the population after t hours, [tex]P_{0}[/tex] is the initial population and r is the decimal growth rate.

The initial number of bacteria in the culture is 28. This means that [tex]P_{0} = 28[/tex].

Population after 24 hours.

(a) No antibiotic is present, so the relative growth rate is 15%.

So r = 0.15.

[tex]P(t) = P_{0}e^{rt}[/tex]

[tex]P(24) = 28e^{0.15*24} = 1024.75[/tex]

(b) An antibiotic is present in the culture, so the relative growth rate is reduced to 8%.

So r = 0.08.

[tex]P(t) = P_{0}e^{rt}[/tex]

[tex]P(24) = 28e^{0.08*24} = 190.99[/tex]

Answer:Jarvis's class average is 75

Step-by-step explanation:

The total possible average score for the math course is 100

a) If the teacher rates homework at 10%, it means that the total possible score for homework

is 10/100 × 100 = 10

If his homework average is 93, then his score would be

(93×10)/100 = 9.3

b) If the teacher rates quizzes at 30%, it means that the total possible score for quizzes

is 30/100 × 100 = 30

If his quiz average is 82, then his score would be

(82×30)/100 = 24.6

c) If the teacher rates test at 40%, it means that the total possible score for quizzes

is 40/100 × 100 = 40

If his quiz average is 72, then his score would be

(72×40)/100 = 28.8

d) If the teacher rates final exam at 20%, it means that the total possible score for quizzes

is 20/100 × 100 = 20

If his final exam is 60, then his score would be

(60×20)/100 = 12

Jarvis's class average would be

9.3 + 24.6 + 28.8 + 12 = 74.7

Approximately 75

Answer:

First

Step-by-step explanation:

Since, the probability that the first person pick the paper marked "you lose" is 1/200, which is smaller than the probability of who draws later.

Taking your turn to draw last is the most strategic choice to minimize the risk of being the unfortunate person to draw the "you lose" paper.

The optimal strategy in this scenario is to draw last.

Here's the reason:

When drawing first, you have a 1/200 chance of drawing the "you lose" paper initially.

However, if you draw last, you get to observe the outcomes of all the previous draws.

If none of the previous guests has drawn the "you lose" paper, then the odds of it being in the hat when it's your turn are 1/200.

In contrast, if someone before you draws the "you lose" paper, you won't have to draw at all since the game ends. This means that by drawing last, you have the best chance of avoiding the "you lose" paper if others before you didn't draw it.

Thus, taking your turn to draw last is the most strategic choice to minimize the risk of being the unfortunate person to draw the "you lose" paper.

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Answer:

An independent samples t-test

Step-by-step explanation:

We have two different groups and we want to test if the scores are equal or not , so the best appropiate mthod would be an independent t test. Here we show the steps to conduct this test.

Data given and notation  

[tex]\bar X_{1}[/tex] represent the mean for group men  

[tex]\bar X_{2}[/tex] represent the mean for group women  

Assuming these values for the remaining data:

[tex]s_{1}[/tex] represent the sample standard deviation for men

[tex]s_{2}[/tex] represent the sample standard deviation for women

[tex]n_{1}[/tex] sample size for the group men  

[tex]n_{2}[/tex] sample size for the group women  

t would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value  

Concepts and formulas to use  

Suppose that we need to conduct a hypothesis in order to check if the mean are equal or not, the system of hypothesis would be:  

H0:[tex]\mu_{1} = \mu_{2}[/tex]  

H1:[tex]\mu_{1} \neq \mu_{2}[/tex]  

For this case is better apply a t test to compare means since we don't know the population deviations, and the statistic is given by:  

[tex]z=\frac{\bar X_{1}-\bar X_{2}}{\sqrt{\frac{s^2_{1}}{n_{1}}+\frac{s^2_{2}}{n_{2}}}}[/tex] (1)  

t-test: Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other.  

Calculate the statistic  

We just need to replace in formula (1) and find the calculated value.    

Find the critical value

In order to find the critical value we need to take in count that we are conducting a two tailed test, and we need a significance level provided in order to find the critical region

Statistical decision

If our calculates value [tex]t_{calculated}>t_{critical}[/tex] or [tex]t_{calculated}<t_{critical}[/tex] we reject the null hypothesis. In other case we FAIL to reject the null hypothesis.

Final answer:

To find the variance of a binomial distribution with n = 800 and p = 0.87, use the formula Variance = n × p × (1-p). The variance in this case is approximately 90.64.

Explanation:

To find the variance of a binomial distribution, we use the formula:

Variance = n × p × (1-p)

Where n is the number of trials and p is the probability of success.

So, in this case, with n = 800 and p = 0.87, we can calculate the variance as:

Variance = 800 × 0.87 ×  (1-0.87)

Variance = 800 × 0.87 × 0.13

Variance = 90.64

Therefore, the variance of the binomial distribution is approximately 90.64.

Answer:

The 99% confidence interval would be given (0.004;0.436).

Answer:

[tex]P(600<\bar X<700)=0.89347[/tex]

Step-by-step explanation:

Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the total amounts spent by students vacationing for a week in Florida of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(650,120)[/tex]  

Where [tex]\mu=650[/tex] and [tex]\sigma=120[/tex]

And let [tex]\bar X[/tex] represent the sample mean, the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu,\frac{\sigma}{\sqrt{n}})[/tex]

On this case  [tex]\bar X \sim N(650,\frac{120}{\sqrt{15}})[/tex]

We are interested on this probability

[tex]P(600<\bar X<700)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

If we apply this formula to our probability we got this:

[tex]P(600<\bar X<700)=P(\frac{600-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{700-\mu}{\frac{\sigma}{\sqrt{n}}})[/tex]

[tex]=P(\frac{600-650}{\frac{120}{\sqrt{15}}}<Z<\frac{700-650}{\frac{120}{\sqrt{15}}})=P(-1.614<Z<1.614)[/tex]

And we can find this probability on this way:

[tex]P(-1.614<Z<1.614)=P(Z<1.614)-P(Z<-1.614)[/tex]

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.  

[tex]P(-1.614<Z<1.614)=P(Z<-1.614)-P(Z<-1.614)=0.94674-0.05326=0.89347[/tex]

To find the probability, calculate the z-scores for $600 and $700, and find the area under the normal curve between these two z-scores.

To find the probability that a simple random sample (SRS) of 15 students will spend an average of between $600 and $700, we need to calculate the z-scores for these two values.

First, we find the z-score for $600:

z = (600 - 650) / (120 / √15) = -1.5496

Next, we find the z-score for $700:

z = (700 - 650) / (120 / √15) = 1.5496

We then use a standard normal table or calculator to find the area under the normal curve between these two z-scores. The probability is the difference between these two areas.

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Answer: value of the speed, llv(t)ll = 4.0

Step-by-step explanation:

Given;

time interval t1 =5, t2= 7

Length of path ( distance) L = 8

Speed = distance travelled/ time taken

Speed = dL/dt

Speed = 8/(t2-t1) = 8/(7-5)

Speed = 8/2

Speed = 4.0

Since it's moving at constant speed the speed = 4.0

Answer:

(B) The citizens of the city, because they lose help they could have used to buy a home.

Step-by-step explanation:

Nul and alternative hypotheses are:

Type II error occurs when one fails to reject null hypothesis when the null hypothesis was wrong.

In this case Type II error happens when the conclusion is the rate of home ownership is not increasing after tax cut, where actually it is.

With this conclusion city council does not continue tax cut, and citizens of the city is harmed because they lose help they could have used to buy a home.

Final answer:

The citizens of the city would be harmed by a Type II error as they would miss out on the help to buy homes.

Explanation:

Who would be harmed by a Type II error?

(B) The citizens of the city, because they lose help they could have used to buy a home.

A Type II error in this context would harm the citizens of the city as they would miss out on the intended help in buying homes due to the failure to detect an increase in the rate of home ownership.

Answer:

The boys need to complete the run in 385.9 seconds or less in order to earn a certificate of recognition from the fitness association.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 450

Standard Deviation, σ = 50

We are given that the distribution of time for fitness test is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

We have to find the value of x such that the probability is 0.10

P(X<x) = 0.10

[tex]P( X < x) = P( z < \displaystyle\frac{x - 450}{50})= 0.10[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(z\leq -1.282) = 0.10[/tex]

[tex]\displaystyle\frac{x - 450}{50} = -1.282\\\\x = 385.9[/tex]

Hence, boys need to complete the run in 385.9 seconds or less in order to earn a certificate of recognition from the fitness association.

The boys need to beat a time of 514 seconds, which corresponds to the 90th percentile of the normal distribution with a mean of 450 seconds and a standard deviation of 50 seconds, to be in the fastest 10% and earn a certificate of recognition.

To determine the time the boys need to beat to be in the fastest 10% for the mile run, we need to find the corresponding z-score for the 90th percentile (since 100% - 10% = 90%) in a standard normal distribution. The z-score that correlates with the 90th percentile is approximately 1.28. We can then use the z-score formula:

Z = (X - μ) / σ

Where X is the time to beat, μ is the mean, and σ is the standard deviation. Substituting the values we know (μ = 450 seconds, σ = 50 seconds, and Z = 1.28):

1.28 = (X - 450) / 50

Multiplying both sides by 50 gives us:

64 = X - 450

Adding 450 to both sides gives:

X = 514 seconds

Therefore, boys need to beat a time of 514 seconds (or 8 minutes and 34 seconds) to earn a certificate of recognition.

Answer:

Option b) significantly different from 24

Step-by-step explanation:

We are given the following in the question:

Population mean, μ = 24

Sample mean, [tex]\bar{x}[/tex] = 25

Sample size, n = 25

Alpha, α = 0.05

Population standard deviation, σ = 2

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 24\text{ years}\\H_A: \mu \neq 24\text{ years}[/tex]

We use Two-tailed z test to perform this hypothesis.

Formula:

[tex]z_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]z_{stat} = \displaystyle\frac{25 - 24}{\frac{2}{\sqrt{16}} } = 2[/tex]

Now, we calculate the p-value from the standard normal table.

P-value = 0.0455

Since the p-value is less than the significance level, we fail to accept the null hypothesis and accept the alternate hypothesis.

Thus, we conclude that mean age is significantly different from 24.

At 5% level of significance, it can be concluded that the mean age is B. Significantly different from 24.

From the information given, the university has had an average age of 25 years. The critical value at 5% level with the degree of freedom is 1.753 while the test statistic will be:

= (25 - 24)/2 × ✓26

= 1/2 × 4 = 2

Since the test statistic is greater than 1.753, one should reject the null hypothesis. Therefore, it shows that that the mean age is significantly different from 24.

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Answer:

Confidence interval for the intercept:

[tex]37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889[/tex]

Confidence interval for the slope:

[tex]33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517[/tex]

Step-by-step explanation:

We start defining our equation's terms, starting from the linear regression model [tex]\hat{y} =\hat{\beta}_{0} + \hat{\beta}_{1}x[/tex]

In this model [tex]{\beta}_{0} [/tex] is the intercept estimator and [tex]{\beta}_{1} [/tex] is the slope estimator.

in the equation y = 37.67 + 33.18x]

[tex]{\beta}_{0} = 37.67 [/tex] and [tex]{\beta}_{1} = 33.81[/tex]

Then we have the standard errors (se) for each estimator:

[tex]se(\hat{\beta_{0}}}) =  5.69[/tex] and [tex]se(\hat{\beta_{1}}}) =  7.94[/tex]

The sample number is 49 species (here we assume that all the individuals of a the same species are summarised with a central tendency measure, i.e. mean, median or mode, if each species contained more than one individual).

The formula for the 96% confidence interval for the intercept [tex]{\beta}_{0} [/tex] we have:

[tex]\hat{\beta_0} - t_{\alpha/2,n-2} se(\beta_0) \leq  \beta _0 \leq \hat{\beta_0} + t_{\alpha/2,n-2}[/tex]

Where  [tex]t_{\alpha/2,n-2}[/tex] represents the p-value in a t distribution when α=0.04 (so that we have a confidence interval of 96%, or 0.96), two-tailed, and n-2 degrees of freedom. In this example, n-2 = 47, and the t-value (47 degrees of freedom, 0.04 significance level, two tails) is ± 2.1123.

We input these values into our formula:

[tex]37.67-2.1123(5.69) \leq \beta_0 \leq 37.67+2.1123(5.69)\\25.651 \leq \beta_0 \leq 49.6889[/tex]

Similarly, the 96% confidence interval for the slope  [tex]{\beta}_{1} [/tex] is:

[tex]\hat{\beta_1} - t_{\alpha/2,n-2} se(\beta_1) \leq  \beta _1 \leq \hat{\beta_1} + t_{\alpha/2,n-2}[/tex]

Where [tex]t_{\alpha/2,n-2} =± 2.1123[/tex]

And into the formula:

[tex]33.18-2.1123(7.94) \leq \beta_1 \leq 33.18+2.1123(7.94)\\16.4083 \leq \beta_1 \leq 49.9517[/tex]

The confidence interval does not include 0, so there is enough evidence saying that there is enough correlation between the metatarsal-to-femur ratio and maximal sprint speed in kilometers per hour. This study shows that measuring the lengths of metatarsal 3 and femur in mammals is a reliable predictor of maximal speed for cursorial mammals.

The 96% confidence interval for the slope of the population regression line is calculated using the formula b ± t(α/2, df) * SE(b). Plugging in the given values (slope= 33.18, standard error= 7.94) along with t(0.02, 47) which is approximately ± 2.01, we get the interval (16.98, 49.38).

In statistics, the confidence interval for the slope of the population regression line can be estimated by using the following formula: b ± t(α/2, df) * SE(b), where b is the slope, t is the t-score, df represents the degrees of freedom, α is the significance level, and SE(b) is the standard error of the slope.

In this case, we are asked to construct a 96% confidence interval for the slope, so α = 1 - 0.96 = 0.04. Since we are dealing with a two-tailed test, we will distribute α across the two tails, giving us α/2 = 0.02 for each half. To calculate our t-score, we will use the t distribution table and look for the value that corresponds with df = n - 2 (since we have two parameters, the intercept and slope) and α/2 = 0.02. Given we have 49 species, the df = 49 - 2 = 47, the t-score is approximately ± 2.01.

With this information, we can substitute the values into the formula, where b = 33.18 and SE(b) = 7.94, and compute the confidence interval for the slope, giving us 33.18 ± 2.01 * 7.94 = (16.98, 49.38). Thus, we estimate with 96% confidence that the true population slope lies between 16.98 and 49.38.

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Answer:

6 mg of the metal needs to be added.

Step-by-step explanation:

Let the amount (in mg) of metal that needs to be added by y.

Therefore, the amount of silver in the above metal is 0.04y.

Prior to mixing, 11 mg of a metal contained 38% of silver (Given).

Therefore, the amount of silver before= [tex]\frac{38}{100}*11[/tex]= 4.18 mg

The total amount of silver after mixing, 4.18 + 0.04y mg

The total amount of metal after mixing, 11 + y mg

New percentage of silver = 26% .

Thus, [tex]\frac{4.18+0.04y}{11+y}*100= 26[/tex]

[tex](4.18 +0.04 y) * 100 = 26 *(11 +y)[/tex]

[tex]418+4y=286+26y[/tex]

[tex]132=22y[/tex]

y=6 mg

Therefore, the amount of metal that needs to be added is 6 mg.

Answer:

Step-by-step explanation:

Suppose we a point [tex]P(x,y,z)[/tex] such that its distance from either the point [tex]A(3,4,-5)[/tex] or [tex]B(-2,1,4)[/tex] is the same.

Using this information we can formula:

distance AP = distance BP

first, let's find the distance from AP, using the distance formula.

[tex]r = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2 + (z_1 - z_2)^2}[/tex]

[tex]AP = \sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2}[/tex]

similarly, we can find the distance BP

[tex]BP = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}[/tex]

since both distances are exactly the same we can equate them

[tex]AP = BP[/tex]

[tex]\sqrt{(3 - x_2)^2 + (4 - y_2)^2 + (-5 - z_2)^2} = \sqrt{(-2 - x_2)^2 + (1 - y_2)^2 + (4 - z_2)^2}[/tex]

we can simplify it a bit squaring both sides, and getting rid of the subscripts.

[tex](3 - x)^2 + (4 - y)^2 + (-5 - z)^2 = (-2 - x)^2 + (1 - y)^2 + (4 - z)^2[/tex]

what we have done here is formulated an equation which consists of any point P that will have the same distance from (3,4,-5) and (-2,1,4).

To put it more concretely,

This is the equation of the the plane from that consists of all points (P) from which the distance from both (3,4,-5) and (-2,1,4) are equal.

Answer:

[tex] \frac{1}{X} \sim log N(-\mu , \sigma^2)[/tex]

Step-by-step explanation:

For this case we know that the distribution for the random variable is given by:

[tex]X \sim logN(\mu ,\sigma^2)[/tex]

The density function for the log normal random variable is given by:

[tex] f)x,\sigma) = \frac{1}{x \sigma \sqrt{2\pi}}e^{- \frac{ln x^2}{2\sigma^2}}[/tex]

And we want to find the distribution for the random variable [tex]\frac{1}{X}[/tex]

In order to find this distribution we can use the cumulative distribution function like this:

Let [tex] Y= \frac{1}{X}[/tex], if we solve for X from this transformation we got:

[tex] X= \frac{1}{Y}[/tex]

And then we have this:

[tex] F_Y (y) = P(Y \leq y) = P(X \leq \frac{1}{y}) = F_X (\frac{1}{y})[/tex]

And we can find the density function as the derivate of the distribution function like this:

[tex] f_Y (y) = F'_Y (y) = -\frac{1}{y^2} f_Y(\frac{1}{y})[/tex]

[tex] f_Y (y)= -\frac{1}{y^2} \frac{1}{\frac{1}{y} \sigma \sqrt{2\pi}} e^{- \frac{ln(\frac{1}{y})^2}{2\sigma^2}}[/tex]

But we see that we don't have an specified form for the distribution

If we assume that X follows a normal distribution [tex] X\sim N (\mu_z,\sigma^2_z)[/tex] and we use the transformation [tex]X=e^Y[/tex] we see that X follows a log normal distribution. And we see that:

[tex]\frac{1}{X}= \frac{1}{e^Y}=e^{-Y}[/tex]

And if we find the distribution of [tex]e^{-y}[/tex] we got this:

[tex] f_Y (y) = F'_Y (y) = -e^{-y} f_Y(e^{-y})[/tex]

[tex] f_Y (y)= -e^{-y} \frac{1}{e^{-y} \sigma \sqrt{2\pi}} e^{- \frac{ln(e^{-y})^2}{2\sigma^2}}[/tex]

[tex] f_Y (y) = -\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{y^2}{2\sigma^2}}[/tex]

And then we see that [tex]Y= \frac{1}{X} \sim log N(-\mu , \sigma^2)[/tex]

Answer:

integers are closed under division

Step-by-step explanation:

Integers are not closed under division.Let us consider two integers i and j.

Then [tex]\[\frac{i}{j}\][/tex] need not necessarily be an integer. For example, 2 and 3 are integers but [tex]\[\frac{2}{3}\][/tex] is not an integer. So the first option , namely, integers are closed under division is not true.

On the other hand, the remaining three options given are correct.

Answer:

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Step-by-step explanation:

Values (x) ∑(Xi-X)^2

----------------------------------

29.1                    0.1702

28.5                  1.0252

28.8                  0.5077

29.4                   0.0127

29.8                  0.0827

29.8                  0.0827

30.1                   0.3452

30.6                   1.1827

----------------------------------------

236.1                 3.4088

Mean = 236.1 / 8 = 29.51

[tex]S_{x}=\sqrt{3.4088/(8-1)}=0.6978[/tex]

Statement of the null hypothesis:

H0: u ≥ 30 the mean wedding cost is not less than $30,000

H1: u < 30 the mean wedding cost is less than $30,000

Test Statistic:

[tex]t=\frac{X-u}{S/\sqrt{n}}=\frac{29.51-30}{0.6978/\sqrt{8}}= \frac{-0.49}{0.2467}=-1.9861[/tex]

Test criteria:

SIgnificance level = 0.05

Degrees of freedom = df = n - 1 = 8 - 1 = 7

Reject null hypothesis (H0) if

[tex]t<-t_{0.05,n-1}\\ t<-t_{0.05,8-1}\\ t<-t_{0.05,7}[/tex]

Finding in the t distribution table α=0.05 with df=7, we have

[tex]t_{0.05,7}=2.365[/tex]

[tex]t>-t_{0.05,7}[/tex] = -1.9861 > -2.365

Result: Fail to reject null hypothesis

Conclusion: Do no reject the null hypothesis

u ≥ 30 the mean wedding cost is not less than $30,000

There is no sufficient evidence to support the claim that wedding cost is less than $30000.

Hope this helps!

The function f(x) = x // 2 (integer division) is an example of a function from N to N that is surjective because every natural number is covered, but not injective because different numbers can result in the same output.

To provide an example of a function f from the natural numbers N to N that is surjective but not injective, consider the function f(x) = x // 2, where '//' denotes integer division. For the function to be surjective, each element y in N must have at least one x such that f(x) = y. This is indeed the case here since for any y > 0, we can choose x = 2y or x = 2y + 1, and f(x) will equal y. To show that it is not injective, we can find two different numbers, x1 and x2, such that f(x1) = f(x2). For instance, if x1 = 4 and x2 = 5, both f(x1) and f(x2) equal 2, thus violating the definition of injectivity. Hence, f(x) = x // 2 is surjective because every y in N is an image of some x, but it is not injective because at least two different values in the domain map to the same value in the codomain.

Answer:

[tex]P(\bar X < 5000)=P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]

Step-by-step explanation:

1) Previous concepts

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 Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Let X the random variable that represent the mass of minerals of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(\mu,\sigma)[/tex]  

Where [tex]\mu=?[/tex] and [tex]\sigma=?[/tex]

From the central limit theorem we know that the distribution for the sample mean [tex]\bar X[/tex] is given by:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]

2) Solution to the problem

For this case we know this condition given :

[tex]P(X<5000)=0.6[/tex]

We can use the Z score given by this formula:

[tex]Z=\frac{X-\mu}{\sigma}[/tex]

And using this formula we got:

[tex]P(Z<\frac{5000-\mu}{\sigma})=0.6[/tex]

And we can find a value on the normal standard distribution that accumulates 0.6 of the are aon the left and 0.4 of the area on the right, on this case the value is Z=0.2533. And we can use the following excel code to find it :"=NORM.INV(0.6,0,1)"

So then we can do this:

[tex]0.2533=\frac{5000-\mu}{\sigma}[/tex]  (1)

By the other hand when we find the z score for the sample mean we have this:

[tex]Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And we want to find this probability:

[tex]P(\bar X < 5000)[/tex]

And if we use the z score formula we got:

[tex]P(Z< \frac{5000 -\mu}{\frac{\sigma}{\sqrt{16}}})=P(Z<\sqrt{16} \frac{5000-\mu}{\sigma})[/tex]  (2)

And replacing condition (1) into equation (2) we got:

[tex]P(Z<4 (0.2533))=P(Z<1.0132)=0.845[/tex]

And we can use the following excel code to find it: "=NORM.DIST(1.0132,0,1,TRUE)"

Answer:

Step-by-step explanation:

The test statistic, t, in this hypothesis testing scenario, is calculated by using the formula t = ([tex]\overline{X}[/tex] - μ₀) / (σ / √n). Substituting given values into the formula, the answer is b) -4.076.

In this hypothetical testing question, the test statistic, t, is calculated by dividing the difference between the sample mean and the null hypothesis mean by the standard error.

By dividing the standard deviation by the square root of the sample size, the standard error is determined. Here, the sample mean is 68.5, the null hypothesis mean is 74, the standard deviation is 6.4, and the sample size is 20.

The formula to calculate t is:

t = ([tex]\overline{X}[/tex] - μ₀) / (σ / √n)

Applying the given figures to the formula, the calculation is:

t = (68.5 - 74) / (6.4 / √20) = -4.076

So, the correct answer is b) -4.076.

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Answer:

The 90% confidence interval would be given by (57.006;62.994)  

Step-by-step explanation:

Previous concepts

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".  

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".

[tex]\bar X=60[/tex] represent the sample mean  

[tex]\mu[/tex] population mean (variable of interest)  

[tex]\sigma=10[/tex] represent the population standard deviation  

n=30 represent the sample size  

Assuming the X follows a normal distribution  

[tex]X \sim N(\mu, \sigma=10)[/tex]

The sample mean [tex]\bar X[/tex] is distributed on this way:

[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})[/tex]  

The confidence interval on this case is given by:

[tex]\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}[/tex] (1)

The next step would be find the value of [tex]\z_{\alpha/2}[/tex], [tex]\alpha=1-0.90=0.1[/tex],[tex]\alpha/2 =0.05[/tex] and [tex]z_\alpha/2=1.64[/tex]  

Using the normal standard table, excel or a calculator we see that:  

[tex]z_{\alpha/2}=1.64[/tex]

Since we have all the values we can replace:

[tex]60 - 1.64\frac{10}{\sqrt{30}}=57.006[/tex]  

[tex]60 + 1.64\frac{10}{\sqrt{30}}=62.994[/tex]  

So on this case the 90% confidence interval would be given by (57.006;62.994)  

Final answer:

To construct the 90% confidence interval for the average number of days it takes to find a job using the website's service, the formula for a confidence interval is applied using the given sample mean of 60 days, population standard deviation of 10 days, and a sample size of 30 customers. The calculated interval is between 57 and 63 days.

Explanation:

To construct a 90% confidence interval for the population mean number of days it takes to find a job using the website's service, we need to use the formula:

CI = \(\bar{x} \pm z*\frac{\sigma}{\sqrt{n}}\)

Where:

\(\bar{x}\) is the sample mean (60 days)

\(z\) is the z-score that corresponds to the desired confidence level (For 90%, \(z=1.645\) from the z-table)

\(\sigma\) is the population standard deviation (10 days)

\(n\) is the sample size (30 customers)

Plugging the values into the formula gives:

CI = 60 \pm 1.645 * (\frac{10}{\sqrt{30}})

Calculating the margin of error:

ME = 1.645 * (\frac{10}{\sqrt{30}}) \approx 3.00

Now compute the confidence interval:

CI = [60 - 3.00, 60 + 3.00]

CI = [57.00, 63.00]

So, we are 90% confident that the population mean number of days it takes to find a job using the website's service is between 57 and 63 days.

Answer:

APR = 2.41% x 12 = 28.92%

Step-by-step explanation:

Her APR is 28.92%.

Sheila's APR is calculated by multiplying the monthly periodic rate of 2.41% by 12, yielding an APR of 28.92%.

The question refers to the process of calculating an Annual Percentage Rate (APR) from a given monthly periodic rate. Sheila's monthly periodic rate is 2.41%. The APR can be calculated by multiplying this monthly rate by the number of months in a year, which is 12.

To find Sheila's APR, we perform the following calculation:

APR = Monthly Periodic Rate × Number of Periods in a Year = 2.41% × 12 = 28.92%.

Hence, Sheila's APR is 28.92%.

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