Question

A researcher performs a hypothesis test to test the claim that for a particular manufacturer, the mean weight of cereal in its 18 ounce boxes is less than 18 ounces. Use the given computer output to draw a conclusion about the cereal box weight at a significance level of 0.05.

Test of Ho: μ = 18 vs Ha: μ < 18

n=21    Mean =17.883    StDev = 0.212    t_calc = -2.53

• A. Fail to reject the null- there is strong evidence that the mean weight is 18 ounces.

• B. Fail to reject the null- there is strong evidence that the mean weight is less than 18 ounces.

• C. Reject the null- there is strong evidence that the mean weight is 18 ounces.

• D. There is insufficient information to draw a conclusion.

• E. Reject the null- there is strong evidence that the mean weight is less than 18 ounces.

Answer 1

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (17.883 - 18)/(0.212/sqrt(21))
t = -2.53

P-value Approach
P-value = 0.0199
As P-value < 0.05, reject the null hypothesis.

E. Reject the null- there is strong evidence that the mean weight is less than 18 ounces.