In an online shop, the time taken, in hours, to sell a watch is a normally distributed continuous random variable $X$. The standard deviation of $X$ is 0.68 hours and the expected value of $X$ is 1.75 hours. After an aggressive advertising campaign, the total time taken to sell 8 watches is found to be 11 hours. Test, at 5% level of significance, whether there is evidence that the mean time taken to sell a watch has decreased. State an assumption that you have used in your calculation.

1. Define the population mean and state the null and alternative hypotheses.

Let $\mu$ be the population mean time taken, in hours, to sell a watch.
$H_0: \mu = 1.75$
$H_1: \mu < 1.75$

2. State the test statistic and parameters.

Under $H_0$, the test statistic is $Z = \frac{\bar{X} - \mu}{\sigma / \sqrt{n}} \sim N(0, 1)$
Given $n = 8$, the sample mean $\bar{x} = \frac{11}{8} = 1.375$, and population standard deviation $\sigma = 0.68$.
Level of significance $\alpha = 0.05$.

3. Calculate the value of the test statistic and the corresponding $p$-value.

$z = \frac{1.375 - 1.75}{0.68/\sqrt{8}} \approx -1.5598$
Using a graphing calculator,
$p\text{-value} = P(Z \le -1.5598) \approx 0.0594$ (to 3 s.f.)

4. Compare the $p$-value to the significance level and state the conclusion in context.

Since the $p\text{-value} = 0.0594 > 0.05$, we do not reject $H_0$.
There is insufficient evidence at the 5% level of significance to conclude that the mean time taken to sell a watch has decreased after the advertising campaign.

5. State an assumption made in the calculation.

Assumption: The standard deviation of the time taken to sell a watch remains unchanged at 0.68 hours after the advertising campaign.
(Alternatively: The times taken to sell each of the 8 watches are independent of one another.)