"Working Mom’s Journal" reported that the mean time a mother, with her small children, spends in at the convenience store is 7.3 minutes. A sample of 20 moms is chosen from your neighborhood, and it is found that the mean time they spend in a convenience store was 8.2 minutes with a standard deviation of 1.4 minutes. Using , test the claim that the average amount of time a mom and her children spend in a convenience store is greater than 7.3 minutes.
Determine which test statistic you will use: the standard normal distribution, or the student’s t distribution. Explain why you chose this test statistic.
Establish the null and alternative hypotheses, state the claim.
Test the claim at and discuss your results, should you reject or not reject the null hypothesis, should you reject or except the claim.
Try a t-test since your sample size is rather small.
Formula:
t = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
sample mean = 8.2
population mean = 7.3
standard deviation = 1.4
sample size = 20
Plug in the values and calculate the t-test statistic.
Find the critical value for a one-tailed test using degrees of freedom (df = n - 1). Use a t-table. Compare to your t-test statistic calculated above. If the t-test statistic exceeds the critical value from the table, reject the null. If the t-test statistic does not exceed the critical value from the table, do not reject the null.
I hope this will help get you started.
To test the claim that the average amount of time a mom and her children spend in a convenience store is greater than 7.3 minutes, we need to determine which test statistic to use - the standard normal distribution or the student's t distribution.
In this case, we will use the student's t distribution because the sample size (n = 20) is relatively small, and the population standard deviation is not known. The t distribution is appropriate when working with small samples and unknown population standard deviation.
Now, let's establish the null and alternative hypotheses and state the claim:
Null hypothesis (H0): The average amount of time a mom and her children spend in a convenience store is equal to or less than 7.3 minutes.
Alternative hypothesis (H1): The average amount of time a mom and her children spend in a convenience store is greater than 7.3 minutes.
Claim: The average amount of time a mom and her children spend in a convenience store is greater than 7.3 minutes.
Next, we can perform the hypothesis test. We will calculate the test statistic, compare it with the critical value, and assess whether to reject or fail to reject the null hypothesis based on the significance level (α).
To calculate the test statistic, we will use the formula:
t = (sample mean - population mean) / (sample standard deviation / √n)
Substituting the values:
t = (8.2 - 7.3) / (1.4 / √20)
t = 0.9 / (1.4 / √20)
t ≈ 0.9 / 0.313
Calculating this expression gives us t ≈ 2.875.
Next, we'll determine the critical value for a one-tailed test using a significance level (α). Since the claim states that the average time is "greater than" 7.3 minutes, this is a one-tailed test. Let's assume a significance level of α = 0.05.
Looking up the critical value in the t-distribution table at α = 0.05 for a one-tailed test with degrees of freedom (df = n-1 = 19), we find the critical value to be approximately 1.729.
Now, we can compare the test statistic (t = 2.875) with the critical value (1.729). Since the test statistic is greater than the critical value, it falls in the rejection region.
Based on the calculation and comparison, we can reject the null hypothesis. This means there is sufficient evidence to support the claim that the average amount of time a mom and her children spend in a convenience store is greater than 7.3 minutes.
In conclusion, we reject the null hypothesis and accept the alternative hypothesis, suggesting that the average time spent by moms and their children in a convenience store is indeed greater than 7.3 minutes.