Some kind of Z test

Question

Historically, a manufacturer finds that 1% of integrated circuits of a certain type fail during burn-in testing. Burn-in testing results (pass/fail) on the latest shipment of 10,500 circuits are to be analyzed using hypothesis testing to compare the failure rate of the this shipment with the historical norm. Of these circuits, 125 fail. Is this good evidence that the failure rate differs from the historical rate? To decide, conduct an hypothesis test at the 0.01 significance level

(1)The observed value of the (unstandardized) test statistic is 

(2) To four decimal places, the p-value for the continuity-corrected large sample test is

(3) Does the test reject the null hypothesis at the 0.01 level?

Answer 1

Hello Stephaie! You are correct in stating that this is some kind of Z test! Lol. Specifically, this is a one sample z test, as this is a proportion and all proportions are normally distribution or z distributions. The information provided is p = .01, p hat = 125/10500 or .0119. To calculate the z test statistic the numerator is( p hat - p) or .0119 - .01 and the denominator is the square root of (p * 1-p)/n or (.01 * .99)/10500. Combing the numerator and denominator provides a z test statistic of 1.957...question 1.

To get the p-value I use the TI 84 Plus, but the tables provide the same information. This is a two tail test, as no direction is provided (< or >) and the word differs is provided. I found the probability of the right tail using normalcdf(1.957, 9999, 0, 1) which is .02517 for the right tail only. Since this is a two tails test, you'd have to multiply that .02517 x 2 to account for the probability of the left tail. You can do this due to symmetry of the distribution. So, your answer for the p-value for your two tail test is .0513...question 2.

The conclusion: Since the p-value (.0503) is greater than the alpha level (.01), we'd fail to reject the null. There is no convincing evidence that the failure rate differs from the historical rate of .01.....question 3. I hope this solution is descriptive enough for you to replicate in the future. Also, you can verify the calculations by using the 1PropZTest function in your calculator by going to Stats, TESTS, 5, and entering the information.

Have a good one now,

Dr. Peoples