Question

Two-Sample T-Test and CI: Weight, Gender

Method

μ₁: mean of weight when Gender = Female

µ₂: mean of weight when Gender = Male

Difference: μ₁ - µ₂

Equal variances are not assumed for this analysis.

Descriptive Statistics: Weight

Gender : N ----   Mean ----- StDev   

Female : 315 -----  170 ------- 50     

Male : 42 ----- 196 -------- 65          

Test

Null hypothesis          H₀: μ₁ - µ₂ = 0

Alternative hypothesis          H₁: μ₁ - µ₂ ≠ 0

T-Value                       P-Value

-----                                 -------

Calculate the p-value associated the test statistic.   b. What is your conclusion based on the above calculated p-value? c.What is the p-value to test if the mean weight of males is less than that of females? d. Use a statistical procedure to determine if the average female weight is less than 165 e. Find the 90th percentile of the female weight assuming that weight is normally distributed. .

Answer 1

(a)

(b)

Since p-value is less than 0.05 so we reject the null hypothesis at 5% level of significance.

(c)

Test is right tailed. The p-value for one tailed test is 1- (0.016 / 2)= 0.992

Since p-value is not less than 0.05 so we fail to reject the null hypothesis at 5% level of significance. That is we cannot conclude that the mean weight of males is less than that of females.

(d)

(e)

We need z-score that has 0.90 area to its left. The z-score 1.28 has 0.90 area to its left. The required weight is