Question

(1 point) A certain manufactured product is supposed to contain 23% potassium by weight. A sample of 10 specimens of this product had an average of 23.24 and and a standard deviation of 8 = 0.24. If the mean percentage is found to differ from 23, the manufacturing process will be recalibrated. Using a = 0.01, test an appropriate null and alternate hypotheses. (a) the degree of freedom (b) the critical t value (c) the test statistic The final conclusion is A. There is not sufficient evidence to reject the null hypothesis that u =23 B. We can reject the null hypothesis that =23

Answer 1

Given. Sample size (n)= 10 Sample mean (7) = 23.24 Sample standard deviation(s) = 0.24 Population mean(u) =23 Set up the null and the alternative hypothesis. Ho : u = 23 H 1 = 23 The significance level, a =0.01. Degrees of freedom: df=n-1 = 10-1

b) Critical value: The critical value at the given level of significance with 9 degrees of freedom is, 3.250 Decision rule: Reject H , if the t > 3.250

Calculate test statistic: Under Ho, the test statistic for one sample t test is, _23.24-23 0.24 VIO = 3.162 Conclusion: Here, the test statistic value (3.162) is less than the critical value (3.250). So, fail to reject null hypothesis at the given level of significance. There is not sufficient evidence to reject the null hypothesis that u = 23.