Questions: What is your decision regarding the null hypothesis at 5% level of significance?

Solution Steps

The Standard Error \( SE \) is calculated using the formula: \[ SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{2.5}{5} + \frac{2.5}{5}} = 1.0 \]

The test statistic \( t \) is computed as follows: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{7.0 - 12.0}{1.0} = -5.0 \]

The degrees of freedom \( df \) are calculated using the formula: \[ df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}} = \frac{1.0}{0.125} = 8.0 \]

The p-value \( P \) is calculated as: \[ P = 2(1 - T(|t|)) = 2(1 - T(5.0)) = 0.0011 \]

Since the p-value \( 0.0011 \) is less than the significance level \( \alpha = 0.05 \), we reject the null hypothesis.

The analysis indicates that there are more defective parts produced.

Final Answer