Question

The following data (in pounds), which were selected randomly from a normally distributed population of values, represent measurements of a machine part that is supposed to weigh, on average, 8.5 pounds.

8.7	8.6	8.3	8.2	8.5	8.6	8.4	8.3	8.4	8.7
8.9	8.5	8.2	8.5	8.5	8.3	8.4	8.9	8.5	8.7

Use these data and α = .01 to test the hypothesis that the parts average 8.5 pounds.

Answer 1

The null hypothesis for the test will be,

$H_{0}: \mu=8.5$

against the alternative hypothesis

$H_{1}: \mu \neq 8.5$

This is two tailed test.

As per given sample data,

sample mean, $\bar{x}=\sum \frac{x_{i}}{n}=\frac{x_{1}+x_{2}+...+x_{20}}{n}=\frac{8.7+8.6+...+}{20}=8.505$

sample standard deviation, $s=\sqrt{\frac{\sum \left (x_{i}-\bar{x} \right )^{2}}{n}}=0.2038$

sample size, n = 20.

The t test with (n - 1) = 19 degree of freedom will be applied here.

Therefore, the test statistic will be obtained as below:

$t = \frac{\bar{x}-\mu}{s/\sqrt{n}}=\frac{\left (8.505-8.5 \right )}{0.2038/\sqrt{20}}=\frac{0.005}{0.0456}=0.1097$

The p-value for the two tailed test can be obtained using the Excel formula TDIST(t, df, tails) as below:

p value = TDIST (0.1097, 19, 2) = 0.9138.

The level of significance is α = .01.

As can be observed, p- value for the test is not less than the level of significance, therefore, we cannot reject the null hypothesis and hence it is concluded that parts average 8.5 pounds.