One sample z-test is performed if the sample size is large enough (n > 30) and we want to know if the sample comes from the specific population.

For this case, we're specified that:

Population mean = $\mu$ = $150
Population standard deviation = $\sigma$ = $30.20
Sample mean = $\overline{x}$ = $160
Sample size = n = 40 > 30
Level of significance = $\alpha$ = 2.5% = 0.025
We want to determine if the average customer spends more in his store than the national average.

Forming hypotheses:

Null Hypothesis: Nullifies what we're trying to determine. Assumes that the average customer doesn't spend more in the store than the national average. Symbolically, we get: $H_0: \mu_0 \leq \mu = 150$
Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically $H_1: \mu_0 > \mu = 150$

where $\mu_0$ is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

$z = \dfrac{\overline{x} - \mu_0}{\sigma/\sqrt{n}} = \dfrac{160 - 150}{30.20/\sqrt{40}} \approx 2.094$

The test is single tailed, (right tailed).

The critical value of z at level of significance 0.025 is 1.96

Since we've got 2.904 > 1.96, so we reject the null hypothesis.

(as for right tailed test, we reject null hypothesis if the test statistic is > critical value).

Thus, we accept the alternate hypothesis that customer spends more in his store than the national average.

Learn more about one-sample z-test here:

brainly.com/question/21477856

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