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.

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