# statistics

posted by on .

The MacBurger restaurant chain claims that the waiting time of customers for service is
normally distributed, with a mean of 3 minutes and a standard deviation of 1 minute. The
quality-assurance department found in a sample of 50 customers at the Warren Road
MacBurger that the mean waiting time was 2.75 minutes. At the .05 significance level, can
we conclude that the mean waiting time is less than 3 minutes?

• statistics - ,

You can use a one-sample z-test on this data.

Null hypothesis:
Ho: µ = 3 -->meaning: population mean is equal to 3 minutes
Alternate hypothesis:
Ha: µ < 3 -->meaning: population mean is less than 3 minutes

Using the z-test formula to find the test statistic:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
z = (2.75 - 3)/(1/√50)

I'll let you finish the calculation.

Check a z-table for your critical or cutoff value at .05 level of significance for a one-tailed test (the alternate hypothesis is showing a specific direction; therefore, the test is one-tailed). Compare the test statistic you calculated to the critical value from the table. If the test statistic exceeds the critical value, reject the null and conclude that the mean waiting time is less than 3 minutes. If the test statistic does not exceed the critical value from the table, you cannot reject the null and conclude a difference.

I hope this will help.

• statistics - ,

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