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 a 2-sample t-test on your data. Find the critical value using a t-table. Once you calculate the test statistic, compare the test statistic to the critical value. If the test statistic exceeds the critical value, you will reject the null and conclude a difference. If the test statistic does not exceed the critical value, you cannot reject the null.
I hope these few suggestions will help.
To determine if the second weight loss method is better than its competitor, we can perform a two-sample t-test. The null hypothesis (H₀) is that there is no difference between the two methods, and the alternative hypothesis (H₁) is that the second method has a higher average weight loss.
First, let's calculate the test statistic using the formula:
t = (X₁ - X₂) / √[(s₁²/n₁) + (s₂²/n₂)]
Where:
- X₁ and X₂ are the average weight losses of the first and second methods, respectively.
- s₁ and s₂ are the standard deviations of the first and second samples, respectively.
- n₁ and n₂ are the sample sizes of the first and second samples, respectively.
From the given information, we have:
X₁ = 18
X₂ = 23
s₁ = 6.4
s₂ = 8.2
n₁ = 27
n₂ = 29
Substituting these values into the formula, we get:
t = (18 - 23) / √[(6.4²/27) + (8.2²/29)]
Simplifying further, we have:
t = -5 / √[(41.60/27) + (67.24/29)]
Calculating the values within the square root:
t = -5 / √[1.54 + 2.32]
Adding the values inside the square root:
t = -5 / √3.86
t ≈ -5 / 1.965
t ≈ -2.54
Now, we need to find the critical value from the t-table. Since the significance level (α) is 0.10 and we have a two-tailed test, we need to find the critical values for an alpha of 0.05.
Looking up the t-table with degrees of freedom (df) being the sum of the two sample sizes minus two (df = n₁ + n₂ - 2 = 27 + 29 - 2 = 54), we find the critical value to be approximately ±1.675.
Comparing the test statistic (-2.54) to the critical value (-1.675), we see that the test statistic is more extreme (further into the tails) than the critical value. Therefore, we reject the null hypothesis and conclude that there is evidence to suggest that the second weight loss method with an average weight loss of 23 pounds is better than its competitor.
In summary:
- Test statistic: t = -2.54
- Critical value: ±1.675