Q: A national retail company has historically observed that 14% of its employees participate in an optional professional development program. If a random sample of 400 employees is selected, what is the correct standard deviation of the sample proportion of participants?

$$\sqrt{\frac{0.14 \times 0.86}{400}}$$

Q: A national online retailer wants to estimate the proportion of satisfied customers with 90% confidence and a margin of error of 5%. If the estimated proportion is 0.75, which formula should be used to determine the minimum sample size required?

$$n = \left( \frac{z_{0.05} \sqrt{0.75 \times 0.25}}{0.05} \$$right)^2$$

Q: For the scenario where 14% of employees participate in a program, and a sample of 400 is taken, which normal approximation would you use to find the probability that more than 40 employees participate?

$$P\left( \hat{p} \u003e \frac{40}{400} \right)$$ where $$\hat{p} \sim N(0.14, \sqrt{\frac{0.14 \times 0.86}{400}})$$

Q: A 95% confidence interval for a proportion is constructed as $$\hat{p} \pm z_{0.025} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. $$For the cybersecurity training data ($$\hat{p} = 0.80$$, $$n = 200$$), what is the margin of error?

$$1.96 \times \sqrt{\frac{0.8 \times 0.2}{200}}$$

Q: In the financial advisor scenario, what is the standard error of the sample proportion for a sample of 250 clients, assuming $$p = 0.20$$?

$$\sqrt{\frac{0.20 \times 0.80}{250}}$$

Q: A company wants to be 90% confident that its estimate of customer satisfaction is within 5% of the true proportion. If $$z_{0.05} = 1.645$$ and $$p = 0.75$$, what is the minimum sample size required?

$$n = \left( \frac{1.645 \times \sqrt{0.75 \times 0.25}}{0.05} \$$right)^2$$ \approx 169$$

Q: Suppose in the financial advisor scenario, you want to find the probability that between 18% and 25% of the sample experience positive returns. Which of the following expressions correctly represents this probability using the normal approximation?

$$P\left( 0.18 \u003c \hat{p} \u003c 0.25 \right)$$ where $$\hat{p} \sim N(0.20, \sqrt{\frac{0.20 \times 0.80}{250}})$$

Question 1

A national retail company has historically observed that 14% of its employees participate in an optional professional development program. If a random sample of 400 employees is selected, what is the correct standard deviation of the sample proportion of participants?

Accepted Answer

$$\sqrt{\frac{0.14 	imes 0.86}{400}}$$

Question 2

A financial advisor claims that 20% of clients who follow their investment strategy experience positive returns over a year. In a random sample of 250 clients, what is the expected number of clients with positive returns?

Accepted Answer

$$250 	imes 0.20 = 50$$

Question 3

In a sample of 200 employees who completed a cybersecurity training, 160 avoided clicking on simulated phishing links. What is the point estimate for the proportion of all trained employees who would avoid clicking phishing links?

Accepted Answer

$$\frac{160}{200} = 0.80$$

Question 4

A national online retailer wants to estimate the proportion of satisfied customers with 90% confidence and a margin of error of 5%. If the estimated proportion is 0.75, which formula should be used to determine the minimum sample size required?

Accepted Answer

$$n = \left( \frac{z_{0.05} \sqrt{0.75 	imes 0.25}}{0.05} \$$right)^2$$

Question 5

A retail chain wants to compare return rates for Nike and Adidas shoes. Out of 200 Nike customers, 150 returned their shoes; out of 180 Adidas customers, 120 returned theirs. What is the point estimate for the difference in return rates (Nike minus Adidas)?

Accepted Answer

$$\frac{150}{200} - \frac{120}{180} = 0.75 - 0.6667 = 0.0833$$

Question 6

For the scenario where 14% of employees participate in a program, and a sample of 400 is taken, which normal approximation would you use to find the probability that more than 40 employees participate?

Accepted Answer

$$P\left( \hat{p} \u003e \frac{40}{400} ight)$$ where $$\hat{p} \sim N(0.14, \sqrt{\frac{0.14 	imes 0.86}{400}})$$

Question 7

A 95% confidence interval for a proportion is constructed as $$\hat{p} \pm z_{0.025} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}. $$For the cybersecurity training data ($$\hat{p} = 0.80$$, $$n = 200$$), what is the margin of error?

Accepted Answer

$$1.96 	imes \sqrt{\frac{0.8 	imes 0.2}{200}}$$

Question 8

In the financial advisor scenario, what is the standard error of the sample proportion for a sample of 250 clients, assuming $$p = 0.20$$?

Accepted Answer

$$\sqrt{\frac{0.20 	imes 0.80}{250}}$$

Question 9

A company wants to be 90% confident that its estimate of customer satisfaction is within 5% of the true proportion. If $$z_{0.05} = 1.645$$ and $$p = 0.75$$, what is the minimum sample size required?

Accepted Answer

$$n = \left( \frac{1.645 	imes \sqrt{0.75 	imes 0.25}}{0.05} \$$right)^2$$ \approx 169$$

Question 10

For the difference in return rates between Nike and Adidas shoes, which formula gives the standard error for the difference in sample proportions?

Accepted Answer

$$\sqrt{\frac{p_1$(1-p_1$)}{n_1$} + \frac{p_2$(1-p_2$)}{n_2$}}$$

Question 11

A 99% confidence interval for the difference in return rates between Nike and Adidas shoes is to be constructed. Which $$z-$$value should be used?

Accepted Answer

$$z_{0.005} = 2.576$$

Question 12

Suppose in the financial advisor scenario, you want to find the probability that between 18% and 25% of the sample experience positive returns. Which of the following expressions correctly represents this probability using the normal approximation?

Accepted Answer

$$P\left( 0.18 \u003c \hat{p} \u003c 0.25 ight)$$ where $$\hat{p} \sim N(0.20, \sqrt{\frac{0.20 	imes 0.80}{250}})$$

Question 13

In the context of constructing a confidence interval for a single proportion, which of the following is NOT an assumption required for the normal approximation to be valid?

Accepted Answer

The sample is drawn from a normally distributed population.

Question 14

If a 95% confidence interval for the proportion of employees who avoid phishing links is $$(0.74, 0.86)$$, which of the following is the best interpretation?

Accepted Answer

We are 95% confident that the true proportion of all trained employees who would avoid clicking phishing links is between 0.74 and 0.86.

Question 15

A company wants to compare the return rates of two brands using a 99% confidence interval. If the interval does not contain zero, what can be concluded?

Accepted Answer

There is significant evidence at the 1% level that the return rates differ between the two brands.

Question 16

In the scenario where 160 out of 200 employees avoided phishing links, which of the following is the correct formula for the 95% confidence interval for the true proportion?

Accepted Answer

$$0.80 \pm 1.96 	imes \sqrt{\frac{0.80 	imes 0.20}{200}}$$

Question 17

If a company wants to reduce the margin of error in its confidence interval for customer satisfaction from 5% to 2.5%, what should happen to the sample size (all else equal)?

Accepted Answer

The sample size should be quadrupled.

Step-by-Step Guidance for STAT 101 Assignment #6 (Probability & Confidence Intervals)

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