1. **State the problem:** A company claims the average customer satisfaction score is 80%. A sample of 25 customers shows a mean of 79.02% with a standard deviation of 4. We want to test if the satisfaction has decreased. 2. **Hypotheses:** - Null hypothesis $H_0$: $\mu = 80$ (mean satisfaction is 80%) - Alternative hypothesis $H_a$: $\mu < 80$ (mean satisfaction has decreased) 3. **Degree of freedom (df):** $$ df = n - 1 = 25 - 1 = 24 $$ 4. **Critical value at $\alpha = 0.05$ (one-tailed test):** From t-distribution table for $df=24$, $$ t_{critical} = -1.7109 $$ 5. **Calculate the t-statistic:** Formula: $$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} $$ Where: $\bar{x} = 79.02$, $\mu_0 = 80$, $s = 4$, $n = 25$ Calculate standard error: $$ SE = \frac{4}{\sqrt{25}} = \frac{4}{5} = 0.8 $$ Calculate t: $$ t = \frac{79.02 - 80}{0.8} = \frac{-0.98}{0.8} = -1.2250 $$ 6. **Decision:** Since $t = -1.2250$ is greater than $t_{critical} = -1.7109$, we fail to reject the null hypothesis. 7. **Conclusion:** The difference of 0.98 from the hypothesized mean is not statistically significant at the 0.05 level. There is insufficient evidence to conclude that customer satisfaction has decreased.