Q: A national retail company historically observes that 14% of its employees participate in an optional professional development program. If a random sample of 400 employees is selected, what is the probability that more than 40 employees participate in the program? (Use the normal approximation to the binomial distribution.)

First, calculate the mean $$\mu = np = 400 \times 0.14 = 56. $$The standard deviation $$\sigma = \sqrt{np(1-p)} = \sqrt{400 \times 0.14 \times 0.86} \approx 6.92. $$The probability that more than 40 employees participate is $$P(X \u003e 40) = P\left(Z \u003e \frac{40.5 - 56}{6.92}\right) = P(Z \u003e -2.24) \approx 0.9875.$$

Question 1

A national retail company historically observes that 14% of its employees participate in an optional professional development program. If a random sample of 400 employees is selected, what is the probability that more than 40 employees participate in the program? (Use the normal approximation to the binomial distribution.)

Accepted Answer

First, calculate the mean $$\mu = np = 400 	imes 0.14 = 56. $$The standard deviation $$\sigma = \sqrt{np(1-p)} = \sqrt{400 	imes 0.14 	imes 0.86} \approx 6.92. $$The probability that more than 40 employees participate is $$P(X \u003e 40) = P\left(Z \u003e \frac{40.5 - 56}{6.92}ight) = P(Z \u003e -2.24) \approx 0.9875.$$

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 adults, what is the probability that between 18% and 25% of the sample experience positive returns? (Use the normal approximation to the binomial.)

Accepted Answer

Let $$p = 0.20$$, $$n = 250. $$Mean $$\mu = np = 50$$, standard deviation $$\sigma = \sqrt{np(1-p)} = \sqrt{250 	imes 0.2 	imes 0.8} \approx 6.32. $$For 18%: $$X_1 = 0.18$$ 	imes 250 = 45$$, for 25%: X_2$ = 0.25 	imes 250 = 62.5$$. $$P(45 \u003c X \u003c 62.5) = P\left(\frac{45.5 - 50}{6.32} \u003c Z \u003c \frac{62.5 - 50}{6.32}ight) = P(-0.71 \u003c Z \u003c 1.98) \approx 0.836$$

Question 3

A company rolled out a new cybersecurity training module. In a random sample of 200 employees who completed the training, 160 reported not clicking any simulated phishing links. Construct a 95% confidence interval for the true proportion of all trained employees who would avoid clicking phishing links.

Accepted Answer

Sample proportion $$\hat{p} = \frac{160}{200} = 0.8. $$Standard error $$SE = \sqrt{\frac{0.8 	imes 0.2}{200}} \approx 0.0283. $$For 95% confidence, $$z^* = 1.96. CI$$: $$0.8 \pm 1.96 	imes 0.0283 = (0.745, 0.855).$$

Question 4

A national online retailer wants to estimate the proportion of customers satisfied with their shopping experience. Management wants to be 90% confident that the estimate is within a margin of error of 5%. If approximately 75% of customers are satisfied, what is the minimum sample size required?

Accepted Answer

Margin of error $$E = 0.05$$, $$z^* = 1.645$$ for 90% confidence, $$p = 0.75$$. $$n = \frac{p(1-p)(z^*)^2$}{E^2$} = \frac{0.75 	imes 0.25 	imes (1.645)^2$}{0.05^2$} \approx 203$$

Question 5

A retail chain wants to determine if the return rates for Nike and Adidas shoes differ significantly. Out of 200 Nike customers, 150 returned their shoes. Out of 180 Adidas customers, 120 returned their shoes. Construct a 99% confidence interval for the difference in return rates between Nike and Adidas shoes.

Accepted Answer

Nike: $$\hat{p}_1 = \frac{150}{200} = 0.75$$, Adidas: $$\hat{p}_2 = \frac{120}{180} = 0.667. $$Standard error $$SE = \sqrt{\frac{0.75 	imes 0.25}{200} + \frac{0.667 	imes 0.333}{180}} \approx 0.048$$. $$z^* = 2.576$$ for 99% CI. CI: $$0.75 - 0.667 \pm 2.576 	imes 0.048 = (0.083 - 0.124, 0.083 + 0.124) = (-0.041, 0.207).$$

Question 6

Which of the following best describes the purpose of constructing a confidence interval for a population proportion?

Accepted Answer

To estimate the range of values within which the true population proportion is likely to fall, with a specified level of confidence.

Question 7

A company wants to reduce successful phishing incidents. After training, 160 out of 200 employees avoided clicking simulated phishing links. Which assumption is necessary for the normal approximation to be valid when constructing a confidence interval for the proportion?

Accepted Answer

Both $$np \geq 10$$ and $$n(1-p) \geq 10$$ must be satisfied.

Question 8

In the context of estimating a proportion, what effect does increasing the sample size have on the margin of error, assuming all other factors remain constant?

Accepted Answer

The margin of error decreases as sample size increases.

Question 9

A financial advisor wants to estimate the proportion of clients experiencing positive returns. If the advisor wants a smaller margin of error, which of the following should they do?

Accepted Answer

Increase the sample size.

Question 10

A retail chain finds that 75% of Nike customers and 66.7% of Adidas customers return their shoes. Which formula is used to calculate the standard error for the difference in proportions?

Accepted Answer

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

Question 11

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

Accepted Answer

$$n = \frac{p(1-p)(z^*)^2$}{E^2$}$$

Question 12

A company wants to construct a 95% confidence interval for the proportion of employees who avoid phishing links. Which value should be used for $$z^*$$?

Accepted Answer

$$z^* = 1.96$$

Question 13

A financial advisor samples 250 clients and finds 20% experience positive returns. What is the standard error for the sample proportion?

Accepted Answer

$$SE = \sqrt{\frac{0.2 	imes 0.8}{250}} \approx 0.0253$$

Question 14

A company wants to estimate the proportion of employees participating in a program. If the sample proportion is 0.14 and the sample size is 400, what is the 95% confidence interval for the proportion?

Accepted Answer

Standard error $$SE = \sqrt{\frac{0.14 	imes 0.86}{400}} \approx 0.0174$$. $$z^* = 1.96. CI$$: $$0.14 \pm 1.96 	imes 0.0174 = (0.106, 0.174).$$

Question 15

A retail chain wants to compare return rates for Nike and Adidas shoes. Which statistical test is most appropriate to determine if the difference in return rates is significant?

Accepted Answer

A two-proportion z-test.

Question 16

A company samples 200 employees and finds 80% avoided phishing links. What is the margin of error for a 95% confidence interval?

Accepted Answer

Margin of error $$E = z^* 	imes SE = 1.96 	imes \sqrt{\frac{0.8 	imes 0.2}{200}} \approx 0.0555$$

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