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

Q1. What is the probability that more than 40 employees in a random sample of 400 participate in the professional development program, given that historically 14% participate?

Background

Topic: Sampling Distribution of a Proportion (Normal Approximation to the Binomial)

This question tests your ability to use the normal approximation to the binomial distribution to find probabilities about sample proportions.

Key Terms and Formulas

• Population proportion: $p = 0.14$

• Sample size: $n = 400$

• Sample proportion: $\hat{p}$

• Mean of $\hat{p}$: $\mu_{\hat{p}} = p$

• Standard deviation of $\hat{p}$: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

• Normal approximation: Use when $np \geq 10$ and $n(1-p) \geq 10$

Step-by-Step Guidance

1. Calculate the sample proportion corresponding to 40 employees: $\hat{p} = \frac{40}{400}$

2. Check if the normal approximation is appropriate by verifying $np$ and $n(1-p)$ are both at least 10.

3. Compute the mean and standard deviation of the sampling distribution:

   $\mu_{\hat{p}} = p$

   $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

4. Convert the sample proportion to a z-score using:

   $z = \frac{\hat{p} - p}{\sigma_{\hat{p}}}$

5. Set up the probability statement: $P(\hat{p} > \frac{40}{400})$ or equivalently $P(z > ...)$

Try solving on your own before revealing the answer!

Final Answer: 0.3085

Using the normal approximation, the probability that more than 40 employees participate is approximately 0.3085.

This was found by converting 40/400 to a z-score and using the standard normal table.

Q2. What is the probability that between 18% and 25% of a random sample of 250 clients experience positive returns, given that 20% of all clients do?

Background

Topic: Sampling Distribution of a Proportion (Normal Approximation to the Binomial)

This question tests your ability to find the probability that a sample proportion falls within a given range, using the normal approximation.

Key Terms and Formulas

• Population proportion: $p = 0.20$

• Sample size: $n = 250$

• Sample proportion: $\hat{p}$

• Mean: $\mu_{\hat{p}} = p$

• Standard deviation: $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

• Normal approximation: Use when $np \geq 10$ and $n(1-p) \geq 10$

Step-by-Step Guidance

1. Identify the lower and upper bounds for the sample proportion: $0.18$ and $0.25$.

2. Check if the normal approximation is appropriate by verifying $np$ and $n(1-p)$ are both at least 10.

3. Calculate the mean and standard deviation of the sampling distribution:

   $\mu_{\hat{p}} = p$

   $\sigma_{\hat{p}} = \sqrt{\frac{p(1-p)}{n}}$

4. Convert both bounds to z-scores:

   $z_{lower} = \frac{0.18 - p}{\sigma_{\hat{p}}}$

   $z_{upper} = \frac{0.25 - p}{\sigma_{\hat{p}}}$

5. Set up the probability statement: $P(z_{lower} < z < z_{upper})$

Try solving on your own before revealing the answer!

Final Answer: 0.6826

The probability that between 18% and 25% of the sample experience positive returns is approximately 0.6826.

This was found by calculating the area between the two z-scores on the standard normal distribution.

Q3. Construct and interpret a 95% confidence interval for the true proportion of all trained employees who would avoid clicking phishing links, given that 160 out of 200 did not click any simulated phishing links.

Background

Topic: Confidence Interval for a Population Proportion

This question tests your ability to construct and interpret a confidence interval for a population proportion using sample data.

Key Terms and Formulas

• Sample proportion: $\hat{p} = \frac{x}{n}$

• Sample size: $n = 200$

• Number of successes: $x = 160$

• Standard error: $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

• Critical value for 95% confidence: $z^* \approx 1.96$

• Confidence interval formula: $\hat{p} \pm z^* \times SE$

Step-by-Step Guidance

1. Calculate the sample proportion: $\hat{p} = \frac{160}{200}$

2. Compute the standard error: $SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$

3. Find the critical value for a 95% confidence interval ($z^* = 1.96$).

4. Set up the confidence interval formula: $\hat{p} \pm z^* \times SE$

Try solving on your own before revealing the answer!

Final Answer: (0.74, 0.86)

The 95% confidence interval for the true proportion is approximately (0.74, 0.86).

This means we are 95% confident that the true proportion of all trained employees who would avoid clicking phishing links falls within this interval.

Q4. What is the minimum sample size required to estimate the proportion of satisfied customers within a margin of error of 5% at 90% confidence, given that approximately 75% are satisfied?

Background

Topic: Determining Sample Size for Estimating a Proportion

This question tests your ability to calculate the minimum sample size needed to achieve a desired margin of error for a confidence interval for a proportion.

Key Terms and Formulas

• Estimated proportion: $p = 0.75$

• Margin of error: $E = 0.05$

• Confidence level: 90% ($z^* \approx 1.645$)

• Sample size formula: $n = \frac{z^{*2} p (1-p)}{E^2}$

Step-by-Step Guidance

1. Identify the values: $p = 0.75$, $E = 0.05$, $z^* = 1.645$ for 90% confidence.

2. Plug these values into the sample size formula:

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

3. Calculate the numerator: $z^{*2} p (1-p)$

4. Calculate the denominator: $E^2$

5. Divide the numerator by the denominator to find $n$ (always round up to the next whole number).

Try solving on your own before revealing the answer!

Final Answer: 317

The minimum sample size required is 317 customers.

This ensures the margin of error is no more than 5% with 90% confidence, given the estimated satisfaction rate.

Q5. Construct and interpret a 99% confidence interval for the difference in return rates between Nike and Adidas shoes, given the sample data: 150 out of 200 Nike customers and 120 out of 180 Adidas customers returned their shoes.

Background

Topic: Confidence Interval for the Difference Between Two Proportions

This question tests your ability to construct and interpret a confidence interval for the difference between two population proportions using independent samples.

Key Terms and Formulas

• Sample proportions: $\hat{p}_1 = \frac{150}{200}$ (Nike), $\hat{p}_2 = \frac{120}{180}$ (Adidas)

• Sample sizes: $n_1 = 200$, $n_2 = 180$

• Standard error: $SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$

• Critical value for 99% confidence: $z^* \approx 2.576$

• Confidence interval formula: $(\hat{p}_1 - \hat{p}_2) \pm z^* \times SE$

Step-by-Step Guidance

1. Calculate the sample proportions for each group: $\hat{p}_1 = \frac{150}{200}$, $\hat{p}_2 = \frac{120}{180}$

2. Compute the standard error:

   $SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}}$

3. Find the critical value for a 99% confidence interval ($z^* = 2.576$).

4. Set up the confidence interval formula: $(\hat{p}_1 - \hat{p}_2) \pm z^* \times SE$

Try solving on your own before revealing the answer!

Final Answer: (0.012, 0.188)

The 99% confidence interval for the difference in return rates is approximately (0.012, 0.188).

This means we are 99% confident that the true difference in return rates (Nike minus Adidas) falls within this interval.