Question

The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces. We are interested in testing to determine whether the variance of the population is significantly more than 0.003.
At 95% confidence, the null hypothesis:

a. should be rejected
b. should not be rejected
c. should be revised
d. None of these alternatives is correct

Answer 1

Answer:

Option b. should not be rejected

Step-by-step explanation:

We are given that the contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounces.

We have to test whether the variance of the population is significantly more than 0.003, i.e.;

  Null Hypothesis, $H_0$ : $\sigma$ = $\sqrt{0.003}$

Alternate Hypothesis, $H_1$ : $\sigma > \sqrt{0.003}$

The test statistics used here for testing variance is;

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

where, s = sample standard deviation = 0.06

           n = sample size = 26 cans

So, Test statistics = $\frac{(26-1)0.06^{2} }{0.003 }$ ~ $\chi^{2}__2_5$

                            = 30

So, at 5% level of significance chi square table gives critical value of 37.65 at 25 degree of freedom. Since our test statistics is less than the critical so we have insufficient evidence to reject null hypothesis.

Therefore, we conclude that null hypothesis should not be rejected and variance of population is 0.003.