You need to set up hypotheses, calculate the z-test statistic (since this is a z-test), then compare to the critical value from a z-table to determine whether or not to reject the null hypothesis.
Hypotheses:
Ho: µ = 22 -->this is the null hypothesis.
Ha: µ does not equal 22 -->this is the alternate or alternative hypothesis.
This would be a two-tailed or nondirectional test because the alternative hypothesis doesn't specify a specific direction.
The reason we know this is a two-tailed test is because the problem asks if there is evidence that the mean time is different, which means the results could be in either tail of the distribution.
Therefore, using a z-test formula:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
With your data:
z = (25.36 - 22)/(7.24/√50)= 0.656
Now we will need to find the critical value at 0.10 level of significance using a z-table. Since this is a two-tailed test, we split the 0.10 into 0.05 and 0.05 for both tails of the distribution curve.
z=0.398 this is the critical value
since the observed value (calculated from a formula) exceeds our critical value we have to reject the null hypothesis and accept the alternative hypothesis
hence the mean shopping time at supermarket is different from the claimed 22 minutes.
one question per post please
An article claimed that the typical supermarket trip takes a mean of 22 minutes. suppose that in an effort to test this claim, you select a sample of 50 shoppers at a local supermarket. the mean shopping time for the sample of 50 shoppers is 25.36 min. with a standard deviation of 7.24 minutes. using a 0.10 level of significance is there evidence that the mean shopping time at the local supermarket is different from the claimed value of 22 minutes?
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