A manufacturer of chocolate candies uses machines to package candies as they move along a filling line. Although the packages are labeled as 8 ounces, the company wants the packages to contain a mean of 8.17 ounces so that virtually none of the packages contain less than 8 ounces. A sample of 50 packages is selected perodically, and the packaging process is stopped if there is evidence that the mean amount packaged is different from 8.17 ounces. Suppose that in a particular sample of 50 packages the mean amount dispensed is 8.159 ounces, with a sample standard deviation of 0.051 ounce.

a. Is there evidence that the population mean amount is different from *.17 ounces? (Use a 0.05 level of significance.)
b. Determine the p-value and interpret its meaning.

Answer:

The 80% confidence interval is given by (13.622;16.378)  

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".  

Assuming the X the random variable that represent the number of feet it takes the brakes to stop a car going 50 miles per hour follows a normal distribution  

[tex]X \sim N(\mu, \sigma)[/tex]  

[tex]\bar X =15[/tex] represent the sample mean

[tex]s=2.28[/tex] represent the sample deviation

n=6 sample size selected

Confidence =0.8 or 80%

2) Confidence interval

In order to find the critical value is important to mention that we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 80% of confidence, our significance level would be given by [tex]\alpha=1-0.80=0.2[/tex] and [tex]\alpha/2 =0.1[/tex]. The degrees of freedom are given by:  

[tex]df=n-1=6-1=5[/tex]  

We can find the critical values in excel using the following formulas:  

"=T.INV(0.1,5)" for [tex]t_{\alpha/2}=-1.48[/tex]  

"=T.INV(1-0.1,5)" for [tex]t_{1-\alpha/2}=1.48[/tex]  

The critical value [tex]tc=\pm 1.48[/tex]  

Now we can calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

[tex]m=t_c \frac{s}{\sqrt{n}}[/tex]  

[tex]m=1.48 \frac{2.28}{\sqrt{6}}=1.378[/tex]  

The interval for the mean is given by this formula:  

[tex]\bar X \pm t_{c} \frac{s}{\sqrt{n}}[/tex]  

And calculating the limits we got:  

[tex]15 -1.48 \frac{2.28}{\sqrt{6}}=13.622[/tex]  

[tex]15 +1.48 \frac{2.28}{\sqrt{6}}=16.378[/tex]  

The 80% confidence interval is given by (13.622;16.378)  

Answer:

c.  There is no difference in the length between male and female babies

Step-by-step explanation:

When a confidence interval contains a zero, then there is no statistically significant different between the two group means because this means that there is a zero in the difference between the groups and a zero in the difference means there is no difference.

Answer:

Step-by-step explanation:

Given that a university found that 20% of its students withdraw without completing the introductory statistics course.

Each student is independent of the other and there are only two outcomes

X no of students in the registered 20 is binomial with p = 0.2

a)the probability that two or fewer will withdraw.

=[tex]P(X\leq 2)\\=0.2061[/tex]

b. Compute the probability that exactly four will withdraw.

=[tex]P(X=4) = 0.2182[/tex]

c. Compute the probability that more than three will withdraw.

[tex]=P(X>3)\\\\=1-F(2)\\= 1-0.4115\\=0.5885[/tex]

d. Compute the expected number of withdrawals.

E(x) = np = 4

Answer:

Type I: 1.9%, Type II: 1.6%

Step-by-step explanation:

given null hypothesis

H0=the individual has not taken steroids.

type 1 error-falsely rejecting the null hypothesis

⇒   actually the null hypothesis is true⇒the individual has not taken steroids.

 but we rejected it ⇒our prediction is the individual has taken steroids.

typr II error- not rejecting null hypothesis when it has to be rejected

⇒actually null hypothesis is false ⇒the individual has taken steroids.

but we didnt reject⇒the individual has not taken steroids.

let us denote

the individual has taken steroids by 1

the individual has not  taken steroids.by 0

                            predicted

                              1       0

   actual          1    98.4%  1.6%

                        0   1.9%   98.1%

so for type 1 error  

 actual-0

predicted-1

therefore from above table we can see that probability of Type I error is 1.9%=0.019

so for type II error

   actual-1

predicted-0

therefore from above table we can see that probability of Type I error is 1.6%=0.016

Final answer:

The probability of a Type I error in the described scenario is 1.9%, and the probability of a Type II error is 1.6%.

Explanation:

The question asks about the probabilities of Type I and Type II errors in the context of a screening process for detecting a rare disease. A Type I error occurs when the test incorrectly indicates that the disease is present when it actually is not. The probability given for a Type I error is 1.9%, since this is the rate at which the test wrongly gives a positive reaction in individuals without the disease. A Type II error occurs when the test fails to indicate that the disease is present when it actually is. The probability given for a Type II error is the complement of the test's sensitivity, which means 100% - 98.4%, equaling 1.6%.

Therefore:

Probability of a Type I error: 0.019 (or 1.9%)

Probability of a Type II error: 0.016 (or 1.6%)

Answer:

[tex]t=\frac{75.5-69}{\frac{7.007}{\sqrt{6}}}=2.272[/tex]  

[tex]p_v =2*P(t_{(5)}>2.272)=0.072[/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.  

We can say that at 5% of significance the true mean for the Co concentracion it's not significant different from 69.

Step-by-step explanation:

Data given and notation  

Data: 77, 82, 72, 68, 69, 85

The mean and sample deviation can be calculated from 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)}{n-1}}[/tex]

[tex]\bar X=75.5[/tex] represent the sample mean  

[tex]s=7.007[/tex] represent the sample standard deviation  

[tex]n=6[/tex] sample size  

[tex]\mu_o =69[/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 population mean is different from 69, the system of hypothesis are :  

Null hypothesis:[tex]\mu = 69[/tex]  

Alternative hypothesis:[tex]\mu \neq 69[/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".  

Calculate the statistic  

We can replace in formula (1) the info given like this:  

[tex]t=\frac{75.5-69}{\frac{7.007}{\sqrt{6}}}=2.272[/tex]  

P-value  

We need to calculate the degrees of freedom first given by:  

[tex]df=n-1=6-1=5[/tex]  

Since is a two tailed test the p value would given by:  

[tex]p_v =2*P(t_{(5)}>2.272)=0.072[/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.  

We can say that at 5% of significance the true mean for the Co concentracion it's not significant different from 69.

Answer: fail to reject the seven-minute average waiting time claim.

Step-by-step explanation:

As per given ,

Objective for test : the average waiting time is seven minutes or more.

Then ,

[tex]H_0: \mu=7\\\\H_a: \mu>7[/tex]

Since alternative hypothesis is right-tailed thus the test is an right-tailed test.

In a right tailed test , the rejection area lies on the right side of the critical value.

It means that if the observed z-value is greater than the critical value then it will fall into the rejection region other wise not.

i.e. If the value of your test statistic is less than the critical value, the correct decision is we fail to reject null hypothesis.

i.e. fail to reject the seven-minute average waiting time claim.

The complete statement would become:

If the value of your test statistic is less than the critical value, the correct decision is to fail to reject the seven-minute average waiting time claim.

The correct decision is to fail to reject the seven-minute average waiting time claim.

The correct decision is to fail to reject the seven-minute average waiting time claim.

To determine whether the average waiting time exceeds seven minutes, we conduct a hypothesis test using the sample mean and standard deviation. We can use a t-test to compare the sample mean of 8.15 minutes to the claimed average of seven minutes.

If the value of the test statistic is less than the critical value for the chosen level of significance (such as 0.05), we fail to reject the null hypothesis and conclude that there is not enough evidence to support that the average waiting time exceeds seven minutes.

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Answer:

There is a probability of 90.49% of needing more than 80 rolls to reach 300.

Step-by-step explanation:

The Central Limit Theorem tells us that the sampling distribution, as the size of the sample gets larger, approaches the normal distribution. This is independent of the population distribution of the random variable.

Then we can use the normal distribution parameters to model this problem.

Let Y be the sum of 80 dices.

[tex]Y=\sum_{i=1}^{80}X_i\\\\E(Y)=80*E(X_i)=80*[(1/6)*(1+2+3+4+5+6)]=80*3.5=280\\\\V(Y)=80V(X_i)\\\\V(Y)=80*(1/6)[(1-3.5)^2+(2-3.5)^2+(3-3.5)^2+(4-3.5)^2+(5-3.5)^2+(6-3.5)^2]\\\\V(Y)=80*(1/6)*17.5=233.33[/tex]

[tex]\sigma=\sqrt{V(Y)}=\sqrt{233.33}=15.28[/tex]

We can use the z-value to calculate the probability of getting 300 or more in 80 rolls.

[tex]z=(X-\mu)/\sigma=(300-280)/15.28=1.31\\\\\\P(X>300)=P(z>1.31)=0.0951\\\\P(X<300)=1-0.0951=0.9049[/tex]

There is a probabilitity P=0.9049 that the 80 rolls will not reach a 300 score. Thus we can conclude that there is a probability of 90.49% of needing more than 80 rolls to reach 300.

Answer:

0.9430

Step-by-step explanation:

The other answer is almost right but answers the wrong question. The question asks for the probability that at least 80 rolls are necessary not more than 80. So we can use the same process but use where we are using the sum of 79 die. The other answer also misses continuity correction. We must use continuity correction since we're approximating a discrete variable with a continuous one through CLT so P(X <= 300) -> P(X < 300.5). With only these changes, we get 0.943.

Answer:

Step-by-step explanation:

Given that 34​% of U.S. adults say they are more likely to make purchases during a sales tax holiday.

You randomly select 10 adults.

Let X be the no of adults in the selection of 10 who say they are more likely to make purchases during a sales tax holiday.

Each person is independent of the other and also there are two outcomes

Hence X is binomial with n =10 and p = 0.34

q = 0.66

P(X=r) [tex]=10Cr (0.34)^r(0.68)^{10-r}[/tex]

The probability that the number of adults who say they are more likely to make purchases during a sales tax holiday is​

(a) exactly​ two,

=[tex]P(X=2)\\= 0.1873[/tex]

(b) more than​ two,

=[tex]P(X>2)\\= 1-F(2)\\=1-0.2838\\= 0.7162[/tex]

(c) between two and​ five, inclusive.

=[tex]P(2\leq x\leq 5)\\= F(5)-F(1)\\=0.9164-0.0965\\=0.8199[/tex]

The question is about binomial probability involving a scenario of U.S adults making a purchase during a tax sale. Depending on the number of adults to be considered (2, more than 2, between 2 & 5), the probabilities are calculated using the formula for binomial probability taking into account the probability of single trial success and the number of trials.

This question for the subject of Mathematics involves the concept of Binomial Probability

. To solve this question, we use the formula for binomial probability which is: P(k;n,p) = C(n, k) * (p^k) * ((1-p)^(n-k)) where C(n,k) refers to the combination of n things taken k at a time, n is the number of trials, k is the number of successes, and p is the probability of success on a single trial.

Given in the question, p = 0.34 (probability of an adult making a purchase during a tax sale), n = 10 (number of adults selected).

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Answer:

D) the decrease in the prey population due to interactions with the predator

Step-by-step explanation:

We have that:

x is the predator population

y is the prey population.

The coefficient c appears in the equation of dy/dt. So this coefficient is related to the population of the prey. It appears with a minus sign, this means that the prey population decreases. It also multiplies x, so it means that it is related to the predator population.

The correct answer is:

D) the decrease in the prey population due to interactions with the predator

Answer:

84.13%

Step-by-step explanation:

Population mean (μ) = 500 hours

Standard deviation (σ) = 50 hours

Assuming a normal distribution, for any given number of hours 'X', the z-score is determined by:

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

For X=550

[tex]z=\frac{550-500}{50}\\z=1[/tex]

For a z-score of 1, 'X' corresponds to the 84.13-th percentile of a normal distribution.

Therefore, the probability of that a randomly chosen light bulb lasts less than 550 hours, P(X<550), is 84.13%.

So lets write down what we have a(t) = 66 and v(0) = 0 and s(0) = 0

a) Determine the position function.

To do this we have to integrate our acceleration function twice, or we have to integrate the acceleration function to get our velocity function and then integrate that to get our position function.

So:

[tex]\int\limits {a(t)} \, dt = \int\limits {66} \, dt[/tex]

= 66t + c

This means that v(t) = 66t + c

We know that v(0) = 0, so:

v(0) = 66(0) + c

c = 0

So v(t) = 66t (This will be helpful for us later)

Now we have to integrate again.

[tex]\int\limits {66t} \, dt[/tex]

[tex]= 33t^2 + c[/tex]

*note that both of these integrands are done with the reverse power rule for integration*

So we can say that [tex]s(t) = 33t^2 + c[/tex]

But we know that s(0) = 0

So

s(0) = 33(0)^2 + c

c = 0

So... s(t) = 33t^2

b) Now we can answer this question using our position function!

All we have to do is plug in t=5 for s(t)

So...

s(5) = 33(5)^2

s(5) = 825 ft

c) So this is essentially the same problem as b except now we are solving for the time instead of the distance.

We know that in 1 mile there are 5280 ft

So in 1/3 a mile there are 5280(1/3) ft or 1760 ft

Now we can set 1760 equal to s(t) and solve for t

1760 = 33t^2

t^2 = 53.33

[tex]t = \sqrt{53.33}[/tex]

t = 7.30s (we only consider positive answers for time)

d) This is the same question as c just a different distance so:

Setting 300 for s(t)

300 = 33t^2

t^2 = 9.091

[tex]t = \sqrt{9.091}[/tex]

t = 3.015s

e) So for this question we have to approach it in terms of the velocity function not the position function. Then we will solve for the time it took to travel with that velocity and then plug that time value into the position function.

So:

v(t) = 66t

We know that in this case v(t) = 178 ft/s

So: 178 = 66t

t = 2.6969 // t = 2.7s

Now we can use this time in our position function to solve for the distance traveled.

s(t) = 33t^2

s(2.7) = 33(2.7)^2

s(2.7) - 240.57 ft

Hope this helped!

To solve the problem we must know about the concept of Acceleration.

Acceleration is defined as the rate of change of velocity of an object with respect to time.

[tex]a = \dfrac{dv}{dt}[/tex]

Velocity is defined as the rate of change of position of an object with respect to time.

[tex]v=\dfrac{dx}{dt}[/tex]

As we know that acceleration is written as,

[tex]a = \dfrac{dv}{dt}\\\\\int dv =\int a\ dt[/tex]

substitute the value a(t) = 66,

[tex]\int dv =\int (66)\ dt[/tex]

[tex]v = 66t + c[/tex]

As the condition given v(0) = 0,

[tex]v = 66t + c\\\\0 = 66(0)+c\\\\c = 0[/tex]

Therefore, the velocity of the racer can be written as v=66t.

The velocity is written as,

[tex]v=\dfrac{dx}{dt}\\\\\int dx= \int v\ dt \\\\\int dx= \int (66t)\ dt\\\\x = 33t^2+c[/tex]

Substitute the given value x(0) = 0

[tex]0 = 33(0)^2+c\\\\c=0[/tex]

Thud, the position of the racer can be given by the function s = 33t².

A.) The position function t is greater than or equal to 0.

The position function t greater than or equal to 0 can be given by the function s = 33t².

B.) Distance traveled by the racer in 5 seconds.

To find the distance traveled by the racer in 5 seconds substitute the value of t in the position function,

s = 33(5)²

s = 825 m

C.) Time is taken by the racer to travel 1/3 mi.

We already have the position function,

s = 1/3 mi = 536.448 m

s = 33t²

536.448 = 33t²

t = 4.0319 second

D.) Time is taken by the racer to travel 300 ft

We already have the position function,

s = 300 ft = 91.44 m

s = 33t²

91.44 = 33t²

t = 1.6646 second

E.) Distance traveled by the race when it reaches a speed of 178 ft/s.

178 ft/s = 54.2544 m/s

v = 66t

54.2544 = 66t

t = 0.822 sec

Distance traveled in t = 0.8220 sec,

s = 33t²

s = 33(0.822)²

s = 22.2975 m

Learn more about Acceleration:

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Answer:

Part 1

a)[tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{100}}=0.0956[/tex]  

b) [tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{363}}=0.0502[/tex]

c) [tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{1551}}=0.0243[/tex]

Part 2

We can see that if we increase the sample size the margin of error decrease and that makes sense since n is on the denominator in the formula for the margin of error and if we increase the denominator the result needs to decrease.

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]

Part 1

For this case in order to find the margin of error we need to assume a confidence level fixed, let's assume 95% for example. The Margin of error is given by this formula:

[tex]ME= z_{\alpha/2}\sqrt{\frac{p(1-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 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]\hat p = 0.61[/tex] and we are interested in order to find the value of ME.

So we can replace for each case and see what we got:

a)[tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{100}}=0.0956[/tex]  

b) [tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{363}}=0.0502[/tex]

c) [tex] ME=1.96 \sqrt{\frac{0.61 (1-0.61)}{1551}}=0.0243[/tex]

Part 2

We can see that if we increase the sample size the margin of error decrease and that makes sense since n is on the denominator in the formula for the margin of error and if we increase the denominator the result needs to decrease.

Final answer:

The margin of error decreases as the sample size increases, demonstrating the importance of a larger sample size for more accurate polling results.

Explanation:

A 2002 poll reported that 61% of people worried that they would be exposed to SARS. To find the approximate margin of error for different sample sizes (n), we use the formula for the margin of error at the 95% confidence level, which conveniently simplifies to roughly 1 over the square root of the sample size (n) for large sample sizes. However, a more precise calculation involves the formula ME = Z * √(p(1-p)/n), where Z is the Z-score corresponding to the desired confidence level, p is the proportion, and n is the sample size.

As n increases, the margin of error decreases, illustrating the inverse relationship between sample size and the margin of error. This is because with larger sample sizes, we get a more accurate representation of the population, thus increasing the reliability of our poll's results.

The final answer is approximately X≈43.76.

To find the 15th percentile of a normal distribution, you can use the Z-score formula and then use the standard normal distribution table (Z-table) or a calculator to find the corresponding Z-score.

The formula to convert a value from a normal distribution to a standard normal distribution (Z-score) is:

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

Where:

X is the value in the original distribution.

μ is the mean of the original distribution.

σ is the standard deviation of the original distribution.

Z is the Z-score.

Given:

μ=50

σ=6

We want to find the 15th percentile, which corresponds to the value of

X for which 15% of the data falls below it.

First, we find the Z-score corresponding to the 15th percentile using the standard normal distribution table:

[tex]$Z_{15}=$ Z-score for 15 th percentile[/tex]

Then, we rearrange the formula to solve for X:

[tex]$X=\mu+Z_{15} \times \sigma$[/tex]

Let's calculate:

Find the Z-score for the 15th percentile using the Z-table. The closest value is approximately -1.04.

Substitute the values into the formula to find X:

[tex]$\begin{aligned} & X=50+(-1.04) \times 6 \\ & X=50-6.24 \\ & X \approx 43.76\end{aligned}$[/tex]

So, the 15th percentile of the normal distribution with mean μ=50 and standard deviation σ=6 is approximately 43.76.

Answer:

Option A) reject null hypothesis if z is greater than 1.645

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 250

p = 30% = 0.3

Alpha, α = 0.05

Number of women belonging to union , x = 75

First, we design the null and the alternate hypothesis  

[tex]H_{0}: p = 0.3\\H_A: p > 0.3[/tex]

This is a one-tailed(right) test.

Rejection Region:

[tex]z_{critical} \text{ at 0.05 level of significance } = 1.645[/tex]

So, the rejection region will be

[tex]z > 1.64[/tex]

That is we will reject the null hypothesis if the calculated z-statistic is greater than 1.645

Option A) reject null hypothesis if z is greater than 1.645

Final answer:

The rejection region for the given hypothesis test with a significance level of 0.05 is when the z-score is greater than 1.645.

Explanation:

To find the rejection region for this hypothesis test, we need to use the given significance level (alpha, a) of 0.05 to determine the critical z-value. In a one-tailed test, because we are looking for the proportion that exceeds 30%, we focus on the right tail of the normal distribution. Referencing the normal distribution table, a z-value with 0.05 to its right is approximately 1.645. Hence, we reject the null hypothesis if our test statistic z is greater than 1.645.

Utilizing the sample data where 75 out of 250 business students have PCs at home, we would calculate the test statistic and compare it to the critical value. If our calculated z-score exceeds 1.645, then we would reject the null hypothesis and conclude that more than 30% of business students have PCs at home, leading the lab to reconsider its proposed expansion.

The research firm should use the following procedure to analyze the data for the mean difference in salary within married couples:

a) One-sample t procedure, matched pair

It is because, The "matched pair" aspect indicates that each husband's salary is paired with his respective wife's salary. This pairing is essential because the focus is on comparing the salaries within each couple.

The one-sample t procedure is appropriate in this scenario because it compares the mean salary difference within each couple to determine if there is a significant difference between what husbands and wives earn.

This procedure is suitable when the same sample is measured twice (in this case, the salaries of husbands and wives in each couple) and the goal is to compare the means of the differences within the pairs.

By using the one-sample t procedure with matched pairs, the research firm can effectively analyze the data and draw conclusions regarding the mean salary difference within married couples.

The complete question: A research firm wants to determine whether there is a difference in married couples between what the husband earns and what the wife earns. The firm takesa random sample of married couples and measures the annual salary of each husband and wife. What procedure should the firm use to analyze the data for the mean difference in salary within married couples?

a)One-sample t procedure, matched pair

b)Two-sample t procedure

c)One-sample z procedure, matched pair

d)Two-sample z procedure

e)Not enough information to

Answer:

[tex]\frac{37}{389}[/tex]

Step-by-step explanation:

Given that a  medical study investigated the effect of calcium and vitamin supplements on the risk of older Americans for broken bones. A total of 389 older Americans who lived at home and were in good health were studied over a three-year period. While all of the 389 people took in at least 700 milligrams of calcium and 200 units of vitamin D through their normal diet, 187 of them were given additional supplements containing 500 milligrams of calcium citrate and 70 units of vitamin D daily. Of the 187 who took additional supplements, 11 of them suffered broken bones over the three-year period. Of the 202 older Americans who did not take the additional supplement, 26 of them suffered broken bones over the study period.

                     Group I           Group II           Total

n                      187                  202                389

favour x             11                    26                  37

The fraction of older Americans who were included in the study suffered broken bones during the three-year period

=Total x/total n

= [tex]\frac{37}{389}[/tex]

The range within which a student's typing speed will be in two months with a 95.45% probability is 49.66 to 65.34 words per minute.

To find the range within which a student's typing speed will be in two months with a 95.45% probability, we need to calculate the prediction interval. The regression model equation is given as Y' = 7x3 + 5x2 + 3x + 11, where x3 represents hours of instruction per week, x2 represents hours of practice per week, and x represents hours of typing per week necessary for school or work.

Since the student is taking 2 hrs of typing instruction per week (x3 = 2), practicing 5 hrs per week (x2 = 5), and typing 2.5 hours per week for work (x = 2.5), we can substitute these values into the regression equation to find the predicted typing speed (Y').

Using the given equation and substituting the values, we get:

Y' = 7(2) + 5(5) + 3(2.5) + 11

Y' = 14 + 25 + 7.5 + 11 = 57.5 words per minute

Since the standard error of the estimate is 4, the prediction interval can be calculated by adding or subtracting 1.96 times the standard error from the predicted value. Therefore, the range for a 95.45% probability is:

57.5 - (1.96 x 4) to 57.5 + (1.96 x 4) = 57.5 - 7.84 to 57.5 + 7.84 = 49.66 to 65.34 words per minute.

Answer:

1.8 cycles

Step-by-step explanation:

The average number of clock cycles per instruction is given by the sum of the product of each possible number of cycles by its likelihood.

1 cycle: 50%

2 cycles : 25%

3 cycles : 20%

4 cycles : 5%

[tex]Avg = 0.5*1+0.25*2+0.20*3+0.05*4\\Avg= 1.8\ cycles[/tex]

The average number of clock cycles per instruction is 1.8.

In a program with varying instruction types requiring different clock cycles to execute, the average number of clock cycles per instruction is calculated as 1.8. The calculation includes 25% of instructions requiring 2 cycles, 20% requiring 3 cycles, 5% requiring 4 cycles, with the rest needing only 1 cycle. For a Pentium chip which executes 100 million instructions per second with each instruction requiring one clock cycle, the actual execution rate for the average instruction mix would be over 55 million instructions per second.

To calculate the average number of clock cycles per instruction, we first multiply each type of instruction by the number of clock cycles they require, and then sum these up. The calculation would go as follows:

We then sum up these results: 0.5 + 0.6 + 0.2 + 0.5 = 1.8 clock cycles per instruction on average.

In relation to the Pentium chip information, this would mean that a single Pentium chip could execute an average instruction mix at a rate of over 55 million instructions per second, given that it can execute 100 million instructions per second if each instruction only required one clock cycle.

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Answer:

[tex]z=\frac{13-20}{4}=-1.75[/tex]

Assuming:

H0: [tex]\mu \geq 20[/tex]

H1: [tex]\mu <20[/tex]

[tex]p_v = P(Z<-1.75) = 0.0401[/tex]

Step-by-step explanation:

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".

Let X the random variable of interest (number of correct answers in the test), on this case we now that:

[tex]X \sim Binom(n=100, p=0.2)[/tex]

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]

We need to check the conditions in order to use the normal approximation.

[tex]np=100*0.2=20 \geq 10[/tex]

[tex]n(1-p)=20*(1-0.2)=16 \geq 10[/tex]

So we see that we satisfy the conditions and then we can apply the approximation.

If we appply the approximation the new mean and standard deviation are:

[tex]E(X)=np=100*0.2=20[/tex]

[tex]\sigma=\sqrt{np(1-p)}=\sqrt{100*0.2(1-0.2)}=4[/tex]

So we can approximate the random variable X like this:

[tex]X\sim N(\mu =20, \sigma=4)[/tex]

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 letter [tex]\phi(b)[/tex] is used to denote the cumulative area for a b quantile on the normal standard distribution, or in other words: [tex]\phi(b)=P(z<b)[/tex]

The z score is given by this formula:

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

If we replace we got:

[tex]z=\frac{13-20}{4}=-1.75[/tex]

Let's assume that we conduct the following test:

H0: [tex]\mu \geq 20[/tex]

H1: [tex]\mu <20[/tex]

We want to check is the score for the student is significantly less than the expected value using random guessing.

So on this case since we have the statistic we can calculate the p value on this way:

[tex]p_v = P(Z<-1.75) = 0.0401[/tex]

The difference of proportions of high-school and college students reading newspapers regularly can be assessed using a 95% confidence interval, and the significance of this difference can be verified using a z-test considering appropriately formulated null and alternative hypotheses. The questions involve application of fundamental concepts of statistics and hypothesis testing.

Given the data, the proportions of high-school students (p₁) and college students (p₂) who read newspapers regularly are 287/500 (0.574) and 252/420 (0.6) respectively.

A: The 95% confidence interval for the difference between the proportions is calculated by [p₁-p₂ ± Z*√((p₁*(1-p₁)/n₁) + (p₂*(1-p₂)/n₂))], where Z is the z-value (1.96 for 95% confidence), n₁ and n₂ are the sample sizes. Plug in the given numbers to get the interval.

B: Based on the 95% confidence interval, we can judge whether 0 is in this interval. If so, we can't conclude that there's a significant difference between the two proportions. If not, there is a significant difference.

C: The null hypothesis (H₀) is p₁ - p₂ = 0, indicating no difference. And the alternative hypothesis (Hₐ) is p₁ - p₂ < 0, suggesting there's a significant difference and p₁ is less than p₂.

D: To find p ˆ, the pooled estimate is (x₁+x₂) / (n₁+n₂), where x₁ and x₂ are the counts of successes (those who read newspapers) in each group. Plug in the numbers to do the math.

E: Preconditions for a z-procedure: 1. The samples are random. 2. Both sample sizes are sufficiently large (n>30) to apply the Central Limit Theorem. 3. The events are independent.

F: SEp ˆ, the pooled estimate of the standard errors, is √(p ˆ*(1-p ˆ)*(1/n₁+1/n₂))

G: The test statistic is (p₁-p₂) / SEp ˆ, and this z-value can be used to find the P-value in a standard normal distribution table. If the P-value is small, we reject the null hypothesis in favor of the alternative.

H: If the P-value<0.05, then the conclusion would be that the proportion of high school students who read the newspaper regularly is significantly less than the proportion of college students.

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The confidence interval and hypothesis testing don't provide enough evidence to suggest a significant difference in the proportions of high-school and college students who read newspapers regularly. The requirements for using a z-interval and z-procedure were satisfied, suggesting that the statistical methods used were appropriate. The test statistic and P-value confirm that the null hypothesis cannot be rejected.

This question pertains to comparing proportions using hypothesis testing and confidence interval estimation. Before we perform these statistical analyses, we need to ensure that the conditions for each are satisfied.

A. To construct a 95% confidence interval, and to confirm that the conditions for a z-interval are satisfied, we consider the following:

Calculations work out to a confidence interval of ±3.83%. Therefore, according to our analysis, we are 95% confident that the true difference in proportions of high-school students and college students who read newspapers regularly lies in this range.

B. Given our confidence interval, we don't have enough evidence to suggest a significant difference between the two proportions of high-school and college students who read newspapers regularly.

C. The null hypothesis (H0) would be: p1=p2, indicating no difference between the proportions. The alternative hypothesis (Ha), would be: p1

D. To calculate pooled estimate of the population proportions (denoted as p ˆ), the formulas and calculations result in p ˆ = 0.574.

E. Sample size for both samples are larger than 30, implying that we can safely use a z-procedure for the hypothesis test. Also, the samples are independent and randomly selected, satisfying the necessary conditions.

F. The pooled estimate of the standard errors (SEp ˆ), works out to 0.028.

G. The calculated test statistic (z) is 0.83 and the P-value is 0.20.

H. Given a significance level, α = .05, the P-value > α. Therefore, we fail to reject the null hypothesis. This means we do not have enough evidence to suggest that the proportion of high-school students who read newspapers regularly is less than the proportion of college students who do.

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A circle centered at [tex]O(a,b)[/tex] with radius [tex]R[/tex] (the length of [tex]OP[/tex]) has equation

[tex](x-a)^2+(y-b)^2=R^2[/tex]

which can be parameterized by

[tex]\vec c(t)=\langle x(t),y(t)\rangle=\langle a+R\cos t,b+R\sin t\rangle[/tex]

with [tex]0\le t\le2\pi[/tex].

The tangent line to [tex]\vec c(t)[/tex] at a point [tex]P(x_0,y_0)[/tex] is [tex]\frac{\mathrm dy}{\mathrm dx}[/tex] with [tex]x=x_0[/tex] and [tex]y=y_0[/tex]. By the chain rule (and this is where we use implicit differentiation),

[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\frac{\mathrm dy}{\mathrm dt}}{\frac{\mathrm dx}{\mathrm dt}}=\dfrac{R\cos t}{-R\sin t}=-\dfrac{\cos t}{\sin t}[/tex]

At the point [tex]P[/tex], we have

[tex]x_0=a+R\cos t\implies\cos t=\dfrac{x_0-a}R[/tex]

[tex]y_0=b+R\sin t\implies\sin t=\dfrac{y_0-b}R[/tex]

so that the slope of the line tangent to the circle at [tex]P[/tex] is

[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{\frac{x_0-a}R}{\frac{y_0-b}R}=-\dfrac{x_0-a}{y_0-b}[/tex]

Meanwhile, the slope of the line through the center [tex]O(a,b)[/tex] and the point [tex]P(x_0,y_0)[/tex] is

[tex]\dfrac{b-y_0}{a-x_0}[/tex]

Recall that perpendicular lines have slopes that are negative reciprocals of one another; taking the negative reciprocal of this slope gives

[tex]-\dfrac1{\frac{b-y_0}{a-x_0}}=-\dfrac{a-x_0}{b-y_0}=-\dfrac{x_0-a}{y_0-b}[/tex]

which is exactly the slope of the tangent line.

Answer:

0.2638,0.5902,0.1250,0.5902

Step-by-step explanation:

Given that in a certain ​ state, 32.2​% of all community college students belong to ethnic minorities.

i.e. probability for any random college student to belong to ethnic minorities

=0.322

This is constant as each student is independent of the other

X no of college students  belong to ethnic minorities in the sample of 10 is

Bin (10, 0.322)

a) Exactly 3 belong to an ethnic minority.

[tex]=P(X=3)\\=0.2638[/tex]

b. Three or fewer belong to an ethnic minority.

=[tex]P(X\leq 3)\\=0.5902[/tex]

c. Exactly 5 do not belong to an ethnic minority.

[tex]P(X'=5)=P(X=5)\=0.1250[/tex]

d. Six or more do not belong to an ethnic minority.

[tex]P(X'\geq 6)\\=P(X<4)\\=0.5902[/tex]

Answer:

a) p=0.39, where p the parameter of interest represent the true proportion of adults that would erase all their personal information online if they could

b) Null hypothesis:[tex]p = 0.39[/tex]

Alternative hypothesis:[tex]p \neq 0.39[/tex]

Step-by-step explanation:

A hypothesis is defined as "a speculation or theory based on insufficient evidence that lends itself to further testing and experimentation. With further testing, a hypothesis can usually be proven true or false".  

The null hypothesis is defined as "a hypothesis that says there is no statistical significance between the two variables in the hypothesis. It is the hypothesis that the researcher is trying to disprove".

The alternative hypothesis is "just the inverse, or opposite, of the null hypothesis. It is the hypothesis that researcher is trying to prove".

On this case the claim that they want to test is: "The true proportion of adults that would erase all their personal information online if they could is 0.39 or 39%". So we want to check if the population proportion is different from 0.39 or 0.39%, so this needs to be on the alternative hypothesis and on the null hypothesis we need to have the complement of the alternative hypothesis.

Part a. Express the original claim in symbolic form. Let the parameter represent the adults that would erase their personal information.

p=0.39, where p the parameter of interest represent the true proportion of adults that would erase all their personal information online if they could

Part b. Identify the null and alternative hypotheses.

Null hypothesis:[tex]p = 0.39[/tex]

And for the alternative hypothesis we have    

Alternative hypothesis:[tex]p \neq 0.39[/tex]  

If all the students at the university spent 2.2 hours per day studying, with a standard deviation of 2 hours, and we find the probability of observing a sample mean of 1.8 hours studying has an extremely low probability, we say that the observed time is statistically significant.

Statistical significance does not imply practical or substantive significance but indicates strong evidence against the null hypothesis.

The number of students randomly sampled at the studied university = 200

The average time spent by the sampled students per day on social media = 2.3 hours

The average time spent by the sampled students per day studying = 1.8 hours

Population mean on studying = 2.2 hours

Assumed level of statistical significance = 5%

Thus, if the probability of observing a sample mean of 1.8 hours of studying is extremely low given the population mean of 2.2 hours, we would say that the observed average study time of 1.8 hours is statistically significantly different from the population mean. This suggests that the difference is unlikely to have occurred by chance alone.

The correct answer is:

b. statistically significant.

A sample mean of 1.8 hours studying is significantly different from the population mean of 2.2 hours, indicating an important deviation.

Certainly! In statistical terms, "statistically significant" means that an observed result is unlikely to have occurred by chance alone.

In this scenario:

- The population mean (average time spent on studying) is 2.2 hours per day.

- The sample mean (average time spent on studying) is 1.8 hours per day.

- The standard deviation of the population is 2 hours.

To determine whether the observed sample mean of 1.8 hours studying is statistically significant, we can use hypothesis testing or calculate the z-score and find the corresponding p-value.

A low probability associated with the observed sample mean suggests that it's unlikely to occur under the assumption that the population mean is 2.2 hours. This indicates that the difference between the sample mean and the population mean is not likely due to random chance, but rather reflects a true difference in the population.

Therefore, we conclude that the observed time spent on studying (1.8 hours) is statistically significant, as it deviates significantly from the expected population mean of 2.2 hours.

The complete question is here:

Researchers doing a study comparing time spent on social media and time spent on studying randomly sampled 200 students at a major university. They found that students in the sample spent an average of 2.3 hours per day on social media and an average of 1.8 hours per day on studying. If all the students at the university in fact spent 2.2 hours per day on studying, with a standard deviation of 2 hours, and we find the probability of observing a sample mean of 1.8 hours studying has an extremely low probability, we say that observed time is:

a. statistically unlikely.

b. statistically significant.

c. statistically wrong.

d. statistically rare.

Answer:

The solutions are linearly independent because the Wronskian is not equal to 0 for all x.

The value of the Wronskian is [tex]\bold{W=-3e^{7x}}[/tex]

Step-by-step explanation:

We can calculate the Wronskian using the fundamental solutions that we are provided and their corresponding the derivatives, since the Wroskian is defined as the following determinant.

[tex]W = \left|\begin{array}{cc}y&z\\y'&z'\end{array}\right|[/tex]

Thus replacing the functions of the exercise we get:

[tex]W = \left|\begin{array}{cc}e^{5x}&e^{2x}\\5e^{5x}&2e^{2x}\end{array}\right|[/tex]

Working with the determinant we get

[tex]W = 2e^{7x}-5e^{7x}\\W=-3e^{7x}[/tex]

Thus we have found that the Wronskian is not 0, so the solutions are linearly independent.

Final answer:

The Wronskian of [tex]y = e^{5x[/tex] and [tex]z = e^{2x[/tex] is calculated as [tex]-3e^{7x[/tex], which is nonzero for all x, thereby showing that these solutions are linearly independent. This property is vital in forming a solution space for differential equations.

Explanation:

Suppose you solved a second-order equation by rewriting it as a system and found two scalar solutions:  [tex]y = e^{5x[/tex] and  [tex]z = e^{2x[/tex]. To demonstrate that these solutions are linearly independent, we consider the corresponding vector solutions y1 and y2, and use the Wronskian for this purpose.

Calculating the Wronskian

The Wronskian of two functions f and g is defined as:

[tex]W(f,g) = det [[ f, g ],[ f', g' ]][/tex]

For our functions  [tex]y = e^{5x[/tex] and  [tex]z = e^{2x[/tex], the derivatives are [tex]5e^{5x} and 2e^{2x[/tex], respectively. Plugging these into the Wronskian formula, we get:

[tex]W(y,z) = det [[ e^{5x}, e^{2x} ],[ 5e^{5x}, 2e^{2x} ]][/tex]=[tex](e^{5x})(2e^{2x}) - (5e^{5x})(e^{2x}) = -3e^{7x[/tex]

This result is nonzero for all values of x, indicating that the solutions are linearly independent. The concept of linear independence is crucial in the study of differential equations, as it ensures that the solutions can form a basis for the solution space of the differential equation.

Answer: 12

Step-by-step explanation:

Given : Students at a liberal arts college study for an average of 10 hours per week with a standard deviation of 2 hours per week.

[tex]\mu=10\text{ hours}[/tex] and [tex]\sigma=2\text{ hours}[/tex]

The distribution of their study time happens to be uni-modal, symmetric and bell shaped i.e. Normally distributed.

According to the Empirical rule , about 68% of the population lies within one standard deviation from mean .

i.e. Approximately 68% of students study between [tex]\mu-\sigma[/tex] and  [tex]\mu+\sigma[/tex]  hours a week.

i.e. Approximately 68% of students study between [tex]10-2[/tex] and  [tex]10+2[/tex]  hours a week.

i.e. Approximately 68% of students study between 8 and 12 hours a week.

Hence, the value of B = 12.

The question is about a normal distribution in statistics where 68% of data falls within one standard deviation from the mean. Given the mean study, hours is 10 per week and the standard deviation is 2 hours, the value of B representing the upper limit of the 68% range is 12 hours per week.

The subject matter of this problem is based on the principles of statistics, particularly the concept of a normal distribution which is characterized by being uni-modal, symmetric, and bell-shaped. In a normal distribution, approximately 68% of data falls within one standard deviation from the mean.

In this question, we are given that the average study hours are 10 per week (the mean), and the standard deviation is 2 hours. A study time of 8 hours a week represents one standard deviation below the mean. Therefore, one standard deviation above the mean will represent 12 hours a week (mean + standard deviation: i.e. 10 + 2).

Thus, in terms of the normal distribution of study hours, the value of B is 12 hours. That means that approximately 68% of students at this particular liberal arts college study between 8 and 12 hours a week.

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Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

The independent variable is annual tuition. The average salary for a graduate with an annual tuition of $30,000 is $138.7 thousand. The slope of the regression model is -0.91.

The independent variable of a regression model is the variable that is being manipulated or controlled by the researcher. In this case, the independent variable is the annual tuition of the universities and colleges.

According to the given linear regression model, when the annual tuition is $30,000, the average mid-career salary for a graduate is calculated as follows: y = -0.91(30) + 161 = $138.7 thousand.

The slope of the regression model is the coefficient of the independent variable, which is -0.91.

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The independent variable is the annual tuition, and the dependent variable is the average mid-career salary of graduates. The average salary for a graduate with an annual tuition of $30,000 is estimated to be $137.9 thousand. The slope of the regression model is -0.91, indicating a decrease in salary for every increase in tuition.

The independent variable in this linear regression model is annual tuition, denoted by x. The average mid-career salary of graduates, denoted by y, is the dependent variable, meaning it depends on the value of the independent variable.

To find the average salary for a graduate of a college or university with an annual tuition of $30,000, we substitute x = 30 into the regression model. Plugging in the value, we get: ˆy = -0.91(30) + 161. Solving this equation, we find ˆy = $137.9 thousand.

The slope of the regression model is the coefficient of the independent variable, which is -0.91. This means that for every increase of $1,000 in annual tuition, the average mid-career salary of graduates is estimated to decrease by $910.

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The fireworks display is 7621.17 feet  north of the person at point A

As we know 1 mile = 5280 feet

So, 6 miles = 6 × 5280

= 31,680 feet.

Calculate the time it takes for sound to travel from the fireworks display to each person.

Person at point A hears the burst after 4 seconds.

Person at point B hears the burst after 4 + 4 = 8 seconds.

Now, let's find the distance each person is from the fireworks display.

Distance = Speed × Time

For person A:

Distance of A = Speed of sound × Time_A

Distance of A = 1100 feet/second × 4 seconds

Distance of A = 4400 feet

For person B:

Distance of B = Speed of sound × Time_B

Distance of B = 1100 feet/second × 8 seconds

Distance of B = 8800 feet

Use the Pythagorean theorem to find the distance north of person A where the fireworks display occurred.

Let the distance north be x feet.

According to the problem, we have a right-angled triangle formed by points A, B, and the fireworks display location.

The distance between A and B (6 miles or 31,680 feet) is the hypotenuse of the triangle.

Using the Pythagorean theorem:

Distance of A² + x² = Distance of B²

(4400)² + x² = (8800)²

19360000 + x² = 77440000

x² = 77440000 - 19360000

x² = 58080000

x = √58080000

x = 7621.17 feet

So, the fireworks display occurred approximately 7621.17 feet north of the person at point A.

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The fireworks display took place approximately 30,762 feet north of the person at point A. This was calculated by using the information about the extra distance the sound traveled to reach Person B and applying the Pythagorean theorem.

It seems like this question is about the speed of sound and distances of objects based on perception. We are given that the sound from the fireworks travels at 1100 feet per second and the time difference in hearing it between two people is four seconds. Therefore, the sound traveled an extra 4400 feet (4 seconds * 1100 feet/second) to reach the second person.

Let's consider the triangle formed by Person A, Person B, and the fireworks. Since Person A and Person B are 6 miles apart (which is 31680 feet), and we know the extra distance the sound traveled to reach Person B, we can calculate how far the fireworks occurred north of Person A (using Pythagorean theorem). Thus, the distance of the fireworks display north of person A is: sqrt((31680 feet)^2 - (4400 feet)^2), which equals approximately 30,762 feet.

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Answer:

Step-by-step explanation:

Given that the housing market has recovered slowly from the economic crisis of 2008.​ Recently, in one large​ community, realtors randomly sampled 38 bids from potential buyers to estimate the average loss in home value.

s = sample std deviation = 3000

Sample mean = 9379

Sample size n = 38

df = 37

Std error of sample mean = [tex]\frac{s}{\sqrt{n} } \\=486.66[/tex]

confidence interval 95% = Mean ± t critical * std error

=Mean ±1.687*486.66 = Mean ±821.003

=(8557.997, 10200.003)

a) If std deviation changes to 9000 instead of 3000, margin of error becomes 3 times

Hence 2463.008

b) The more the std deviation the more the width of confidence interval.