Ages of cable TV shoppers. Cable TV’s Home Shopping Network (HSN) reports that the average age of its shoppers is 52 years. Suppose you want to test the null hypothesis,\({H_0}:\mu = 52\), using a sample of\(n = 50\) cable TV shoppers.

a. Find the p-value of a two-tailed test if\(\overline x = 53.3\)and\(s = 7.1\)

b. Find the p-value of an upper-tailed test if\(\overline x = 53.3\)and\(s = 7.1\)

c. Find the p-value of a two-tailed test if\(\overline x = 53.3\)and\(s = 10.4\)

d. For each of the tests, parts a–c, give a value of\(\alpha \)that will lead to a rejection of the null hypothesis.

e. If\(\overline x = 53.3\), give a value of s that will yield a two-tailed p-value of 0.01 or less.

It is given that the sample mean is 53.3 and standard deviation is 7.1. Also, the sample size is 50.

Apvaluealso known as the probability value, indicates how likely it is that your data occurred under the null hypothesis. This is accomplished by computing the probability of your test statistic, which is the number calculated by a statistical test based on your data.

a.

Here, \(\overline x = 53.3\) and \(s = 7.1\)

Also, the null and the alternative hypotheses are

\(\begin{aligned}H { _0}:\mu = 52\\H{ _a}:\mu \ne 52\end{aligned}\)

Since, test statistic is calculated using the following formula,

\(z = \frac{{\overline x - \mu }}{{\frac{s}{{\sqrt n }}}}\)

Therefore,

\(\begin{aligned}z &= \frac{{5.33 - 52}}{{\frac{{7.1}}{{\sqrt {50} }}}}\\ &= \frac{{1.3}}{{1.0041}}\\ &= 1.294\end{aligned}\)

Now, from the standard normal table, p-value for two tailed test is 0.1954.