Homework: Homework 2 40f 10 (4 complete) Score: 0.78 of 1 pt Section 11.7 Exerci

Question

Homework: Homework 2 40f 10 (4 complete) Score: 0.78 of 1 pt Section 11.7 Exercise 17-T 25 randomly solected customers and found the ages shown (in yoars). The mean is 29.44 and the standard deviation is 10.77. The owner wants to know if the mean age of all customers is 35 years old. Use the given information to complete parts a through f a) What is the null Ho: I b) Is the alternative one or two-sided The alternative hypothesis is two-sided. c) What is the value of the test statistic? The test statistic is-258 (Round to two decimal places as needed.) d) What is the P-value of the test statistic? P-value (Round to four decimal places as needed) Enter your answer in the answer box and then click Check Answer Clear All

Explanation / Answer

Solution:-

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

Null hypothesis: = 35

Alternative hypothesis: 35

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample mean is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 2.154

DF = n - 1 = 25 - 1

D.F = 24

t = (x - ) / SE

t = - 2.58

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

Since we have a two-tailed test, the P-value is the probability that the t statistic having 24 degrees of freedom is less than - 2.58 or greater than 2.58.

Now check the t to p-value calculator for the t value less than - 2.58, P(t < - 2.58) = 0.008215

Now check the t to p-value calculator for the t value greater than 2.58, P(t > 2.58) = 0.008215

Thus, the P-value = 0.008215 + 0.008215 = 0.01643

Interpret results. Since the P-value (0.01643) is less than the significance level (0.05), we have to reject the null hypothesis.

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