An article (N. Hellmich. "'Supermarket Guru' Has a Simple Mantra,” USA Today, June 19, 2002, p. 70) claimed that the typical super market 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 minutes with a standard deviation of 7.24 minutes. Using the 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?
Z = (mean1 - mean2)/Standard Error of difference between means.
SEdiff = √(SE^2 of mean1 + SE^2 of mean2)
SE^2 = SD^2/(n-1)
Since you only have SE for one mean, use that as your best estimate.
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion = to Z score.
To analyze whether there is evidence that the mean shopping time at the local supermarket is different from the claimed value of 22 minutes, you can perform a hypothesis test using the t-test.
To begin, let's state the null and alternative hypotheses:
Null Hypothesis (H0): The mean shopping time at the local supermarket is equal to 22 minutes.
Alternative Hypothesis (H1): The mean shopping time at the local supermarket is different from 22 minutes.
Next, we can determine the test statistic to use. Since the population standard deviation is unknown, we use the t-distribution.
The test statistic for this scenario is the t-test statistic, which is calculated using the formula:
t = (sample mean - hypothesized mean) / (sample standard deviation / sqrt(sample size))
Given the sample information provided:
Sample mean (x̄) = 25.36 minutes
Hypothesized mean (μ0) = 22 minutes
Sample standard deviation (s) = 7.24 minutes
Sample size (n) = 50 shoppers
We can now calculate the t-test statistic:
t = (25.36 - 22) / (7.24 / sqrt(50))
Next, we determine the critical value or p-value associated with our chosen level of significance (α). In this case, the level of significance is 0.10.
If you choose to use critical values, you will need to find the t-critical value with (1-α/2) degrees of freedom (where α/2 is to account for a two-tailed test since the alternative hypothesis is two-sided).
Alternatively, you can calculate the p-value associated with the t-test statistic using the t-distribution and the degrees of freedom (n-1).
By comparing the calculated t-test statistic to either the critical value or p-value, you can make a decision about the hypotheses.
If the t-test statistic is greater than the critical value or the p-value is less than the level of significance (α), you can reject the null hypothesis and conclude that there is evidence that the mean shopping time at the local supermarket is different from the claimed value of 22 minutes. Otherwise, if the t-test statistic is less than the critical value or the p-value is greater than the level of significance, you would fail to reject the null hypothesis and conclude that there is no evidence to suggest a difference in mean shopping time.
Note: The calculations for the t-test statistic, critical values, and p-value can be done using statistical software or online calculators specifically designed for hypothesis testing.