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il63 [147K]
3 years ago
7

What’s the correct answer for this question?

Mathematics
2 answers:
Maurinko [17]3 years ago
3 0

Answer:

B

Step-by-step explanation:

I put the answer in an atachement

pentagon [3]3 years ago
3 0

Answer:

B.

Step-by-step explanation:

In the attached file

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

<h3>When do we perform one sample z-test?</h3>

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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