Question

A manager believes that the shelf life of apple juice is normally distributed. A sample of 30 containers of juice was taken and the shelf life was recorded. You are given the results below. The average shelf life in the sample was 23.07 days with a standard deviation of 4.29 days. 15 21 24 17 21 24 19 21 25 20 20 20 21 21 21 21 22 22 22 22 22 22 22 25 27 29 30 31 32 33 a. State the null and alternative hypotheses. b. Compute the test statistic for the goodness of fittest. c. At 95% confidence using the p-value approach, test the hypotheses. What do you conclude about the distribution? that telenone calls come into the switchboard at 10-minute intervals,

Answer 1

Assume μ0​ = 25

The provided sample mean is X = 23.07 and the known population standard deviation is o = 4.29, and the sample size is n = 30. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho:p> 25 Ha: u <25 This corresponds to a left-tailed test, for which a z-test for one mean, with known population standard deviation will be used. (2) Rejection Region Based on the information provided, the significance level is a = 0.05, and the critical value for a left- tailed test is zc = -1.64. The rejection region for this left-tailed test is R = {z:z< -1.64}

(3) Test Statistics The Z-statistic is computed as follows: L = X-Mo o/n 23.07 - 25 4.29/30 = -2.464 (4) Decision about the null hypothesis Since it is observed that z = -2.464 < ze = -1.64, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p= 0.0069, and since p=0.0069 <0.05, it is concluded that the null hypothesis is rejected. (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 25, at the 0.05 significance level.

The provided sample mean is X = 23.07 and the population standard deviation is o = 4.29. The size of the sample is n = 30 and the required confidence level is 95%. Based on the provided information, the critical z-value for a = 0.05 is ze = 1.96. The 95% confidence for the population mean is computed using the following expression Ñ _ 2c XO F + 2 xo 2 vn Therefore, based on the information provided, the 95 % confidence for the population mean u is CI= (2:07 – 1.96 .4.29,23.07 + 1,96%0.29) 1.96 x 4.29 1.96 x 4.29 30 30 = (23.07 – 1.535, 23.07 + 1.535) = (21.535, 24.605)