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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