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A manager of a grocery store wants to determine if consumers are spending more than the national average. The national average is $150. 00 with a standard deviation of $30. 20. The manager collects 40 random receipts and finds that the average is $160. Complete a hypothesis test with a significance level of 2. 5% to determine if the average customer spends more in his store than the national average. Which of the following is a valid conclusion for the manager based on this test? The customers spend more than the national average in his store. The manager should decrease prices in his store. The customers do not spend more than the national average in his store. The customers in his store just come from a rich neighborhood.

Sagot :

The valid conclusions for the manager based on the considered test is given by: Option

When do we perform one sample z-test?

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 = [tex]\mu[/tex] = $150
  • Population standard deviation = [tex]\sigma[/tex] = $30.20
  • Sample mean = [tex]\overline{x}[/tex] = $160
  • Sample size = n = 40 > 30
  • Level of significance = [tex]\alpha[/tex] = 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: [tex]H_0: \mu_0 \leq \mu = 150[/tex]
  • Alternate hypothesis: Assumes that customer spends more in his store than the national average. Symbolically [tex]H_1: \mu_0 > \mu = 150[/tex]

where [tex]\mu_0[/tex] is the hypothesized population mean of the money his customer spends in his store.

The z-test statistic we get is:

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

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:

https://brainly.com/question/21477856

Answer:

A

Step-by-step explanation:

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