Question

A cereal company claims that mean weight of cereal boxes is at most 16.1 ounces. Suppose that a plant manager wishes to test whether the true mean weight of cereal boxes is greater than 16.1 ounces. Suppose that for this problem the population standard deviation is 0.4 and the population distribution is normal. The manager obtain a random sample of size 25 and finds a mean of 16.3 ounces. Using p value approach test the claim of company at significance level at 1%.

Answer key: Right sided test.

Z stat = 2.5

P value (z < 2.5) = 0.0062

α= 0.01. Since α > p value. Reject null hypothesis

(Please show the work step by step with figure, decision rule and decision.)

Answer 1

Solution : In the given problem, we need test whether the population mean (1) is greater than 16-1 (ounces). For the testing procedure a kandom sample of size 25 was taken. In the Sample, the mean observed was. 16.3 counces). Also the population standard deviation was given of 014 ounces The null and alternative hypothesis are, Ho : 4 = 16.1 Vs H₂ : M7 16.1 16.3 ounces ounces So Mo = 1611 (hypothesized mean) Sample size (n) = 25 and population was normally distributed Sample mean (2 population standard deviation () = 0.4 Hence using Z test for mean testing we get, the calculated value of test statistic, Zo= 2.5 p-value = p (2720) = P(2725) = 0.0062 -(foz sight tailed test) the significance level (d) = d.op. = P(Z) Zd) ay pvalue reject Hou This can be shown in the following figures Z-distribution turve Region: acceptance 12 significance Rejection Regjon level (a) pvalue Za zo Since test statistic (z.) falls in sojection region We seject null hypothesis (Ho). Since z 0