----------------------------------- A pitching machine is calibrated to deliver a fastball at a speed of 98 miles per hour. Every day, a technician samples the speed of twenty-five fastballs in order to determine if the machine needs adjustment. Today, the sample showed a mean speed of 99 miles per hour with a standard deviation of 1.75 miles per hour. At a 95 percent confidence level, what is the z-value in relation to the critical value? The: A) critical value exceeds the z-value by 1.3 standard deviations. B) z-value exceeds the critical value by 1.5 standard deviations. C) critical value exceeds the z-value by 0.7 standard deviations. D) z-value exceeds the critical value by 0.9 standard deviations.
D is correct. Can you explain though? Thanks.
Agree with kevin002. correct answer is D) H0: population mean equals 98 H1: population mean does not equal 98 z(critical) = 1.96 at 95% confidence level (alpha = 0.05, two-tailed test) z = (sample mean - population mean)/(s/(n^1/2)) = 99 - 98/(1.75/5) = 2.85 z - z(critical) = 2.85 - 1.96 = 0.89
Yup. So given a 95% confidence interval, and inferring that it’s a two tailed test (because the mean is either equal to 98 or not equal to 98), you know that the critical value will be +/- 1.96 (from the z-table). Note: You are not given the population std dev., so typically you would use t-distribution, but clearly none of the choice involve t-distribution. So you know you have to use the z-table. Then you calculate the test statistic which is: ((X-u) / (Sigma / (sqrt of n)) = (99-98) / (1.75/5) = 2.857 Then you just compare the test statistic (i.e. 2.857) with the critical values. 2.857-1.96 = 0.897 Hope this helps
Yes, that’s very helpful. When I practice each section (soon aftre reviewing it), it is easy to see where the problem is from. But when I combine all subject areas, I miss it. Scheweser QBank does not have very good explanations of the solutions but I do appreciate having the opportunity to review a few everyday. Might help me retain more! Thanks.