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2 samples were collected to determine whether there was a difference between two weight loss methods targeting men. The average...Asked by kristen
2 samples were collected to determine whether there was a difference between two weight loss methods targeting men. The average weight loss method is provided for the 2 samples along with the respective standard deviations. Is the second method - with an average weight loss of 23 pound better than its competitor (for = .10) Provide the test statistic and critical value.
Chart -
n Avg Stand Dev
27 18 6.4
29 23 8.2
You can probably use an independent groups t-test for your 2 samples to determine your test statistic.
Null hypothesis:
µ1 = µ2 -->meaning no difference between the two groups.
Alternate hypothesis:
µ1 < µ2 -->meaning there is a difference. (The mean of group 1 lost less weight than the mean of group 2.)
Find the critical value using a t-distribution table using .10 (this is a one-tailed test since the alternate hypothesis is showing a specific direction).
I hope this will help.
One other comment:
If you find that the test statistic exceeds the critical value from the table, then the null is rejected in favor of the alternate hypothesis. You can conclude µ1 < µ2. If you find that the test statistic does not exceed the critical value from the table, then the null is NOT rejected and you cannot conclude a difference.
Chart -
n Avg Stand Dev
27 18 6.4
29 23 8.2
You can probably use an independent groups t-test for your 2 samples to determine your test statistic.
Null hypothesis:
µ1 = µ2 -->meaning no difference between the two groups.
Alternate hypothesis:
µ1 < µ2 -->meaning there is a difference. (The mean of group 1 lost less weight than the mean of group 2.)
Find the critical value using a t-distribution table using .10 (this is a one-tailed test since the alternate hypothesis is showing a specific direction).
I hope this will help.
One other comment:
If you find that the test statistic exceeds the critical value from the table, then the null is rejected in favor of the alternate hypothesis. You can conclude µ1 < µ2. If you find that the test statistic does not exceed the critical value from the table, then the null is NOT rejected and you cannot conclude a difference.
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