Question

A sample of 16 items provides a sample standard deviation of 9.5. Test the following hypotheses using = .05. H0 : 2 50 Ha : 2 > 50 Calculate the value of the test statistic (to 2 decimals).

Answer 1

Answer:

We conclude that the population standard deviation is greater than 50.

Step-by-step explanation:

We are given that a sample of 16 items provides a sample standard deviation of 9.5.

Let $\sigma^{2}$ = population standard deviation

So, Null Hypothesis, $H_0$ : $\sigma^{2} \leq$ 50     {means that the population standard deviation is less than or equal to 50}

Alternate Hypothesis, $H_A$ : $\sigma^{2}$ > 50      {means that the population standard deviation is greater than 50}

The test statistics that will be used here is One-sample chi-square test statistics;

                        T.S.  =  $\frac{(n-1) \times s^{2} }{\sigma^{2} }$  ~ $\chi^{2}__n_-_1$

where, s = sample standard deviation = 9.5

           n = sample of items = 16

So, the test statistics =  $\frac{(16-1) \times 9.5^{2} }{50 }$  ~ $\chi^{2}__1_5$

                                    =  27.08

The value of chi-square test statistics is 27.08.

Now, at 5% level of significance the chi-square table gives a critical value of 25.00 at 15 degrees of freedom for the right-tailed test.

Since the value of our test statistics is more than the critical value of chi as 27.08 > 25.00, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the population standard deviation is greater than 50.