To solve this hypothesis testing problem regarding the customer satisfaction rate, let’s break it down step by step.

Problem Details:

• Claimed satisfaction rate ($p_0$): 58% or 0.58
• Sample size ($n$): 179
• Number of satisfied customers ($x$): 114
• Significance level ($\alpha$): 0.01

We are tasked to:

1. Compute the test statistic ($z$).
2. Compute the $p$-value.
3. Determine if there is sufficient evidence to reject the null hypothesis ($H_0$).

Step 1: Formulate the Hypotheses

• Null hypothesis ($H_0$): $p = 0.58$ (the satisfaction rate is 58%).
• Alternative hypothesis ($H_a$): $p \neq 0.58$ (the satisfaction rate is different).

Step 2: Compute the Test Statistic

The test statistic for a proportion is given by: $z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}$ Where:

• $\hat{p} = \frac{x}{n}$ is the sample proportion.

Calculate $\hat{p}$:

$\hat{p} = \frac{114}{179} \approx 0.63799$

Plug into the formula for $z$:

$z = \frac{0.63799 - 0.58}{\sqrt{\frac{0.58(1-0.58)}{179}}}$ $z = \frac{0.05799}{\sqrt{\frac{0.58 \cdot 0.42}{179}}}$ $z = \frac{0.05799}{\sqrt{0.0023627}} \approx \frac{0.05799}{0.04861} \approx 1.193$

So, the test statistic is approximately: $z \approx 1.19$

Step 3: Compute the $p$-value

This is a two-tailed test, so the $p$-value is: $p = 2 \cdot P(Z > |z|)$ Using a standard normal table or calculator: $P(Z > 1.19) \approx 0.1170$ $p = 2 \cdot 0.1170 = 0.234$

Step 4: Compare $p$-value with $\alpha$

• $p$-value $\approx 0.234$
• Significance level $\alpha = 0.01$

Since $p > \alpha$, we fail to reject the null hypothesis. There is not sufficient evidence to conclude that the satisfaction rate is different from 58%.

Final Answers:

1. Test statistic ($z$): 1.19
2. $p$-value: 0.234
3. Conclusion: Fail to reject $H_0$. No sufficient evidence to suggest the satisfaction rate is different from 58%.

Would you like a deeper explanation of any of these steps or calculations?
Here are 5 follow-up questions you could explore:

1. How is the critical value method related to this test?
2. What would the conclusion be if $\alpha = 0.05$ instead?
3. How would a one-tailed test differ in this situation?
4. Can this test be applied to a much smaller sample size? Why or why not?
5. How does the choice of significance level ($\alpha$) affect hypothesis testing?

Tip: Always double-check your $p$-value using statistical software or a normal distribution calculator for precise results.