Question

The true average diameter of ball bearings of a certain type is supposed to be \(0.5 \mathrm{in}\). What conclusion is appropriate when testing \(H_{0}: \mu=0.5\) versus \(H_{a}: \mu \neq 0.5\) in each of the following situations: a. \(n=13, t=1.6, \alpha=.05\) b. \(n=13, t=-1.6, \alpha=.05\) c. \(n=25, t=-2.6, \alpha=.01\) d. \(n=25, t=-3.6\)

Short answer

Based on the statistical analysis and hypothesis testing for each situation, we can conclude: a) We cannot reject the null hypothesis that the average diameter of ball bearing is 0.5 in. b) We again cannot reject the null hypothesis that the average diameter of the ball bearing is 0.5 in. c) We do not reject the null hypothesis that the average diameter of ball bearing is 0.5in. d) We reject the null hypothesis that the average diameter of ball bearing is 0.5in.

Step-by-step solution

Understanding Hypothesis Testing

Hypothesis testing is a statistical method that is used in making statistical decisions using experimental data. In this exercise, we are testing two hypotheses. The null hypothesis \(H_{0}: \mu = 0.5\) suggests the average diameter of ball bearing is 0.5 in. The alternative hypothesis \(H_{a}: \mu \neq 0.5\) suggests that the average diameter is not 0.5 in.

Computing Critical Value for each situation

For hypothesis testing, we first calculate a 't' score and then find a critical value associated to our significance level α which in our situation a and b is 0.05 and in c is 0.01. This critical value is then compared to our calculated 't' statistic. If absolute value of our calculated 't' statistic is greater than the critical value, we reject the null hypothesis. The degree of freedom for t-distribution is \(n-1\). For each situation, the critical values (two-tailed) for degrees of freedom 12 and 24 and significance level 0.05 and 0.01 are approximately 2.179, 2.064, and 2.797, respectively according to t-distribution table.

Comparing the test statistic with the Critical Value

Next, we compare the test statistic (t) with the critical value for each situation. a. The absolute value of the test statistic \(|1.6|\) is less than the critical value 2.179, so we cannot reject the null hypothesis. b. The absolute value of the test statistic \(|-1.6|\) is still less than the critical value 2.179, so we again cannot reject the null hypothesis. c. The absolute value of the test statistic \(|-2.6|\) is less than the critical value 2.797, so we do not reject the null hypothesis. d. There is no specific significance level mentioned in this case. However, if we assume the significance level as 0.05 (or even 0.01), the absolute value of the test statistic \(|-3.6|\) is more than the critical value of 2.064 (or 2.797), so we do reject the null hypothesis in this case.