Question

A sporting goods store believes the average age of its customers is

35

or less. A random sample of

42

customers was​ surveyed, and the average customer age was found to be

39.2

years. Assume the standard deviation for customer age is

9.0

years. Using

alphaαequals=0.10

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.

Upper H 0H0​:

muμ

▼

less than or equals≤

greater than or equals≥

not equals≠

greater than>

equals=

less than<

nothing

Upper H 1H1​:

muμ

▼

less than<

equals=

less than or equals≤

greater than or equals≥

not equals≠

greater than>

nothingThe​ z-test statistic is

nothing.

​(Round to two decimal places as​ needed.)

The critical​ z-score(s) is(are)

nothing.

​(Round to two decimal places as needed. Use a comma to separate answers as​ needed.)

Because the test statistic

▼

does not fall within the critical values,

is greater than the critical value,

is less than the critical value,

falls within the critical values,

▼

do not reject

reject

the null hypothesis.

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

The​ p-value is

nothing.

​(Round to three de

Answer 1

Solution:-

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 35
Alternative hypothesis: u < 35

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample z-test.

Analyze sample data. Using sample data, we compute the standard error (SE), z statistic test statistic (z).

SE = s / sqrt(n)

S.E = 1.38873
z = (x - u) / SE

z = 3.02

z_Critical = - 1.28

Rejection region is z < -1.28

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Interpret results. Since the z-value (3.02) does not lies in the rejection region, hence we failed to reject the null hypothesis.

Because the test statistic is greater than the critical value, do not reject the null hypothesis.

b)

The observed sample mean produced a t statistic test statistic of -0.894.

P-value = P(z < 3.02)

Use the z-value calculator for finding p-values.

P-value = 0.999

Interpret results. Since the P-value (0.999) is greater than the significance level (0.01), we cannot reject the null hypothesis.