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.
To determine if the second weight loss method is better than the first method, we can use the independent groups t-test.
First, we need to calculate the test statistic. The formula for the test statistic in an independent groups t-test is:
t = (x1 - x2) / sqrt((s1^2 / n1) + (s2^2 / n2))
where:
x1 and x2 are the average weight losses for the two groups
s1 and s2 are the respective standard deviations
n1 and n2 are the sample sizes
Given the information from the chart, the values for the test statistic can be calculated as follows:
x1 = 18
x2 = 23
s1 = 6.4
s2 = 8.2
n1 = 27
n2 = 29
t = (18 - 23) / sqrt((6.4^2 / 27) + (8.2^2 / 29))
Once we have the calculated test statistic, we can compare it to the critical value to determine if the difference is statistically significant.
To find the critical value, we need to consult a t-distribution table with the appropriate degrees of freedom. In this case, the degrees of freedom can be calculated as:
df = n1 + n2 - 2
df = 27 + 29 - 2
Once we have the degrees of freedom, we can find the critical value by looking up the significance level (0.10) and the degrees of freedom in the t-distribution table.
If the calculated test statistic exceeds the critical value from the table, then we can reject the null hypothesis and conclude that the second weight loss method is better. If the calculated test statistic does not exceed the critical value, then we fail to reject the null hypothesis and cannot conclude a difference.
I hope this helps! Let me know if you have any other questions.