From the provided image, this is a hypothesis testing problem. I will guide you step-by-step to solve it.

Problem Outline

We are testing if the mean customer satisfaction rating differs from 85 using the provided sample data.

(a) State the Null and Alternative Hypotheses:

1. Null Hypothesis (H₀): $\mu = 85$ (The mean customer satisfaction rating is 85).
2. Alternative Hypothesis (H₁): $\mu \neq 85$ (The mean customer satisfaction rating is not 85).

(b) Perform the Hypothesis Test:

1. Given Information:

   • Sample size ($n$) = 42
   • Sample mean ($\bar{x}$) = 87.1
   • Population standard deviation ($\sigma$) = 5.2
   • Significance level ($\alpha$) = 0.05 (two-tailed test)

2. Test Statistic Formula: $z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}}$ Substituting:

   • $\bar{x} = 87.1$
   • $\mu = 85$
   • $\sigma = 5.2$
   • $n = 42$

   $z = \frac{87.1 - 85}{\frac{5.2}{\sqrt{42}}}$

   First, calculate the standard error: $SE = \frac{\sigma}{\sqrt{n}} = \frac{5.2}{\sqrt{42}}$

   Next, calculate $z$.

3. Determine Critical Value(s):

   • Since this is a two-tailed test at $\alpha = 0.05$, the critical $z$-values are $-1.96$ and $1.96$.

4. Compare $z$-value to Critical Values:

   • If $z$ falls outside the range $[-1.96, 1.96]$, we reject $H_0$.

(c) Conclusion:

• Based on whether the computed $z$-value is significant, determine if there is enough evidence to reject $H_0$.

Let me calculate these values for you.import math

Given values

x_bar = 87.1 # sample mean mu = 85 # population mean under null hypothesis sigma = 5.2 # population standard deviation n = 42 # sample size

Step 1: Calculate the standard error

SE = sigma / math.sqrt(n)

Step 2: Calculate the z-score

z = (x_bar - mu) / SE

Step 3: Critical values for a two-tailed test at alpha = 0.05

critical_value = 1.96

z, SE, critical_value(2.6172222051262444, 0.8023774198028779, 1.96)(2.6172222051262444, 0.8023774198028779, 1.96)success### Results:

1. Standard Error (SE): $0.802$
2. Computed $z$-value: $2.617$
3. Critical $z$-values: $\pm 1.96$

Decision:

• The computed $z$-value ($2.617$) falls outside the range $[-1.96, 1.96]$.
• This means we reject the null hypothesis ($H_0$).

Conclusion:

At the $0.05$ level of significance, there is enough evidence to support the claim that the mean customer satisfaction rating differs from 85.

Would you like a deeper explanation or help with any specific part? Here are related questions for further learning:

1. What is the importance of the critical value in hypothesis testing?
2. How would the conclusion change if this was a one-tailed test?
3. Why is the standard deviation divided by the square root of $n$ to compute the standard error?
4. How does increasing the sample size affect the test result?
5. How does the significance level ($\alpha$) impact hypothesis testing?

Tip: Always check the direction of the test (one-tailed vs. two-tailed) when comparing the $z$-value to critical values.