Question

Sally has now graduated and become director of quality control for a major cookie conglomerate. The company uses an average of 2000 chocolate chips per batch of cookies, with a known standard deviation of 42 chips.

a) Sally wishes to test the hypothesis that µ=2000 chips per batch against the alternative that µ is different from 2000. She conducts a random sample of 50 batches and finds a mean of 1988 chips. What test statistic will be used, and what is its value?

b) Find the P-value and interpret your answer from part (a): for what α is the mean significantly different from 2000?

c) Find a 95% confidence interval for µ and compare your results with your conclusions above.

Answer 1

The provided sample mean is X = 1988 and the known population standard deviation is o 42, and the sample size is n= 50. - (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: M 2000 Ha: u + 2000 This corresponds to a two-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 two-tailed test is ze = 1.96. The rejection region for this two-tailed test is R = {z: 121 > 1.96} (3) Test Statistics The Z-statistic is computed as follows: X - μo Z = 1988 – 2000 42/V50 -2.02 olſn (4) Decision about the null hypothesis Since it is observed that |z| = 2.02 > ze = 1.96, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p = 0.0434, and since p= 0.0434 < 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 different than 2000, at the 0.05 significance level. Confidence Interval The 95% confidence interval is 1976.358 < d < 1999.642.

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