Answer to Question #264388 in Statistics and Probability for pk003

Question

In 64 randomly selected hours of production, the mean and the Standard Deviation of the number of acceptable pieces produced by a automatic stamping machine are

X| = 1,038

s = 146

At the .05 level of significance, does this enable us to reject the null hypothesis u = 1000 against the alternative hypothesis u > 1000

Answer 1

$H_0: \mu=1000 \\ H_1: \mu > 1000$

This corresponds to a right-tailed test.

The t-statistic is computed as follows:

$t = \frac{\bar{x} - \mu}{ s / \sqrt{n}} \\ t = \frac{1038-1000}{146 / \sqrt{64}} = 2.082$

The p-value for 63 (n-1) degrees of freedom and for right tailed test is p = 0.0207 (from t distribution table).

Consider 0.05 as the level of significance.

Since p = 0.0207 < 0.0 5 , the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean μ is greater than 1000, at the 0.05 significance level.