Select Page

Manager claims average sales for her shop is $1800 a day during winter months. 10 winter days selected at random, and the mean of the sales is$1830. The standard deviation of the population is \$200. Can one reject the claim at a significance level of .05? Find the 95% confidence interval of the mean. Does the confidence interval interpretation agree with the hypothesis test results? Explain. Assume the variable is normally distributed.

Solution: We need to test the following null hypothesis: In order to test the null hypothesis we use a two-tailed z-test (because we know the standard deviation of the population). The z-statistics is computed as The critical values for the 0.05 significance level is . This means that we fail to reject the null hypothesis. In other words, we don’t have enough evidence to reject the claim.

The 95% confidence interval for the mean is  Since , the interpretation of confidence interval agrees with the fact that we failed to reject the null hypothesis. In other words, given the confidence interval , the value 1800 is not an “exceptionally strange value” to support the rejection of the null hypothesis.

GO TO NEXT PROBLEM >>