Question

A publisher reports that 42% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually different from the reported percentage. A random sample of 120 found that 32% of the readers owned a personal computer. Find the value of the test statistic. Round your answer to two decimal places.

Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.9 parts/million (ppm). A researcher believes that the current ozone level is not at a normal level. The mean of 26 samples is 4.5ppm with a variance of 0.64. Assume the population is normally distributed. A level of significance of 0.05 will be used. Find the value of the test statistic. Round your answer to three decimal places.

A sample of 1300 computer chips revealed that 40% of the chips do not fail in the first 1000 hours of their use. The company's promotional literature states that 42% of the chips do not fail in the first 1000 hours of their use. The quality control manager wants to test the claim that the actual percentage that do not fail is different from the stated percentage. Determine the decision rule for rejecting the null hypothesis, H0, at the 0.02 level.

Reject H0 if, ___ ____ ____

Answer 1

& Given that P=0.42, n=120, P = 0.32 Test statistic (G) P-P = 0.32-0.42 0.42 (100.42) 120 o.ousi I [2.22 Given that n=26, x=4.5, S²=0.64, l=6.9 S=0.8 Test statistic = t = x-le s/n = 4.5-4.9 0.81/26 = -2.550 Decision sole: At x=0.02 lois (two failed) Z-C.V = & 2.33 so deject Ho if Z>2.33 (oo) < -2.33/ *