The image provides a question related to statistics, involving confidence intervals and sample size estimation. Here's how to approach parts (b), (c), and (d):

(b) Confidence Interval for Proportion

The 99% confidence interval is already provided as:

$0.1537 \leq p \leq 0.4296$

This means that with 99% confidence, the true proportion of customers who own only a cat falls within this range.

(c) Sample Size for Estimating Population Mean

The problem asks for the sample size needed to estimate the population mean amount spent with 95% confidence, a margin of error ($E$) of ±$1.50, and a standard deviation ($\sigma$) of $10.

The formula for sample size $n$ is:

$n = \left( \frac{Z \cdot \sigma}{E} \right)^2$

Where:

• $Z$: Z-score for 95% confidence (1.96)
• $\sigma$: Standard deviation ($10)
• $E$: Margin of error ($1.50)

Substitute the values: $n = \left( \frac{1.96 \cdot 10}{1.50} \right)^2$

$n = \left( \frac{19.6}{1.50} \right)^2$

$n = \left( 13.07 \right)^2 = 171$

Thus, 171 customers need to be sampled.

(d) Sample Size for Proportion with 99% Confidence

The goal is to determine the required sample size for estimating the proportion of customers who own only a cat, with a margin of error ($E$) of ±0.035 and 99% confidence.

The formula for sample size $n$ for proportions is:

$n = \frac{Z^2 \cdot \hat{p} \cdot (1 - \hat{p})}{E^2}$

Where:

• $Z$: Z-score for 99% confidence (2.576)
• $\hat{p}$: Estimated proportion of success (use midpoint of the confidence interval: $\frac{0.1537 + 0.4296}{2} = 0.29165$)
• $E$: Margin of error (0.035)

Substitute the values: $n = \frac{2.576^2 \cdot 0.29165 \cdot (1 - 0.29165)}{0.035^2}$

$n = \frac{6.635 \cdot 0.29165 \cdot 0.70835}{0.001225}$

$n = \frac{6.635 \cdot 0.20668}{0.001225} = \frac{1.3712}{0.001225} = 1119.5$

Rounding up, 1,120 customers need to be selected.

Final Answers:

• (b): $0.1537 \leq p \leq 0.4296$
• (c): 171
• (d): 1,120

Let me know if you'd like more detailed steps or clarifications!

Follow-up Questions:

1. What is the role of the Z-score in determining sample size?
2. How is the margin of error used to set the bounds for the confidence interval?
3. What would happen if the confidence level was reduced to 90%?
4. Why do we round up the sample size to the next whole number?
5. What assumptions are made about the population when using these formulas?

Tip:

Always double-check whether the confidence level and margin of error match the requirements in a problem before proceeding with calculations.