1. **State the problem:** We want to test the marketing manager's claim that the average monthly data usage of prepaid customers is 5 GB based on a sample of 50 customers with mean 4.6 GB and standard deviation 1.2 GB at a 5% significance level. 2. **State the null and alternative hypotheses:** - Null hypothesis $H_0$: $\mu = 5$ GB (the average usage is 5 GB) - Alternative hypothesis $H_a$: $\mu \neq 5$ GB (the average usage is not 5 GB) 3. **Identify the appropriate statistical test:** Since the population standard deviation is unknown and the sample size is 50 (which is large), we use a one-sample t-test to compare the sample mean to the claimed population mean. 4. **Compute the test statistic:** The test statistic for a one-sample t-test is $$ t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}} $$ where $\bar{x} = 4.6$, $\mu_0 = 5$, $s = 1.2$, and $n = 50$. Calculate: $$ t = \frac{4.6 - 5}{1.2 / \sqrt{50}} = \frac{-0.4}{1.2 / 7.071} = \frac{-0.4}{0.1697} \approx -2.357 $$ 5. **Determine the critical value:** At a 5% significance level for a two-tailed test with $df = n-1 = 49$, the critical t-value is approximately $\pm 2.009$ (from t-distribution tables). 6. **Make a decision:** Since the calculated test statistic $-2.357$ is less than $-2.009$, it falls in the rejection region. 7. **Interpretation:** We reject the null hypothesis and conclude that there is sufficient evidence at the 5% significance level to say the average monthly data usage is different from 5 GB for TNM Malawi prepaid customers.