Question

A sporting goods store believes the average age of its customers is 36 or less. A random sample of 38 customers was​ surveyed, and the average customer age was found to be 38.9 years. Assume the standard deviation for customer age is 9.0 years. Using α=0.01​, complete parts a and b below.

a. Does the sample provide enough evidence to refute the age claim made by the sporting goods​ store?

Determine the null and alternative hypotheses.

H0​: μ ▼ less than < greater than > less than or equals ≤ greater than or equals ≥ not equals ≠ equals =

H1​: μ ▼ not equals ≠ less than or equals ≤ greater than or equals ≥ less than < greater than > equals =

The​ z-test statistic is . ​(Round to two decimal places as​ needed.)

The critical​ z-score(s) is(are) . ​(Round to two decimal places as needed. Use a comma to separate answers as​ needed.)

Because the test statistic ▼ is greater than the critical value/falls within the critical values/is less than the critical value/does not fall within the critical values, ▼ reject/do not reject the null hypothesis.

b. Determine the​ p-value for this test.

The​ p-value is . ​(Round to three decimal places as​ needed.)

Answer 1

= 36, n= 38, = 38.9, = 9, α=0.01​

Ho: 36

Ha: > 36

calculate z test statistics

z= 1.986

The z-test statistic is= 1.99

now calculate z critical value for right tailed test with α=0.01​

using normal z table we get

The critical​ z-score is = 2.33

Because the test statistic is less than the critical do not reject the null hypothesis.

b)

calculate P-Value

P-Value = 1 - P(z < 1.99)

using normal z table we get

P(z < 1.99) = 0.9767

P-Value = 1 -  0.9767

P-Value = 0.023