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?
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
To determine if there is evidence that the mean shopping time at the local supermarket is different from the claimed value of 22 minutes, we can perform a hypothesis test.
1. State the hypotheses:
- Null hypothesis (H0): The mean shopping time at the local supermarket is equal to 22 minutes.
- Alternative hypothesis (Ha): The mean shopping time at the local supermarket is different from 22 minutes.
2. Set the significance level (alpha): The significance level (also known as alpha) is a predetermined threshold that determines how much evidence is required to reject the null hypothesis. In this case, the significance level is 0.10.
3. Calculate the test statistic:
- The test statistic for this situation is the t-statistic because we are working with a sample and do not know the population standard deviation.
- The formula for the t-statistic is: t = (sample mean - population mean) / (sample standard deviation / √n)
- In this case, the sample mean (x̄) is 25.36 minutes, the population mean (μ) is 22 minutes, the sample standard deviation (s) is 7.24 minutes, and the sample size (n) is 50.
4. Find the degrees of freedom:
- The degrees of freedom for a t-test is equal to the sample size minus 1. In this case, the degrees of freedom is 50 - 1 = 49.
5. Determine the critical value(s):
- Since the alternative hypothesis is two-tailed (stating that the mean could be either greater or less than 22 minutes), we need to find the critical values for both tails of the t-distribution.
- The critical values can be found using a t-table or a statistical software at the chosen significance level of 0.10 and the degrees of freedom of 49.
6. Calculate the p-value:
- The p-value is the probability of observing a test statistic as extreme as or more extreme than the one calculated, assuming that the null hypothesis is true.
- With the calculated t-statistic and the degrees of freedom, we can determine the p-value from the t-distribution.
7. Make a decision:
- If the calculated p-value is less than the significance level (0.10), then we reject the null hypothesis and conclude that there is evidence that the mean shopping time at the local supermarket is different from 22 minutes.
- If the calculated p-value is greater than or equal to the significance level, then we fail to reject the null hypothesis, meaning there is not enough evidence to conclude that the mean shopping time is different from 22 minutes.
Performing the calculations and comparing the p-value to the significance level will give you the final decision on whether there is evidence that the mean shopping time at the local supermarket is different from the claimed value of 22 minutes.