Question

A sample of scores for men and women from an examination in Statistics 201 were:

Men 94 97 66 54 80 59 83 45 Women 57 56 48 98 84 55 70 94

Given that the null hypothesis and the alternative hypothesis are:

  H₀: μ_m - μ_w ≤ -4
  H₁: μ_m - μ_w > -4

and using a 0.05 significance level conduct a t-test about a difference in population means:

a)

What is the correct decision rule? Reject H0 in favour of H1 if the computed value of the statistic is less than 1.761. Reject H0 in favour of H1 if the computed value of the statistic is less than -1.761 or greater than 1.761. Reject H0 in favour of H1 if the computed value of the statistic is greater than 1.761. Reject H0 in favour of H1 if the computed value of the statistic is between -1.761 and 1.761. None of the above.

b) Compute the pooled variance.
For full marks your answer should be accurate to at least four decimal places.

Pooled variance: 0

c) Compute the value of the test statistic.
For full marks your answer should be accurate to at least three decimal places.

Test statistic: 0

d)

What is your decision regarding H0? There is sufficient evidence, at the given significance level, to reject H0, and accept H1 or at least there is not enough evidence to reject H1. There is insufficient evidence, at the given significance level, to reject H0. There is insufficient evidence to reject or not reject the null hypothesis.

Answer 1

a)  

Reject H0 in favour of H1 if the computed value of the statistic is greater than 1.761.

b)

Sample #1   ---->   1              
mean of sample 1,    x̅1=   72.250              
standard deviation of sample 1,   s1 =    19.0919              
size of sample 1,    n1=   8              
                      
Sample #2   ---->   2              
mean of sample 2,    x̅2=   70.250              
standard deviation of sample 2,   s2 =    19.3815              
size of sample 2,    n2=   8              
                      
difference in sample means =    x̅1-x̅2 =    72.2500   -   70.3   =   2.000
                      
pooled Variance , Sp²=   ([(n1 - 1)s1² + (n2 - 1)s2²]/(n1+n2-2)) = 370.0714

c)

std error , SE =    Sp*√(1/n1+1/n2) =    9.6186                  
                          
t-statistic = ((x̅1-x̅2)-µd)/SE = (   2.0000   -   -4   ) /    9.62   =   0.6238

d)

There is insufficient evidence to reject or not reject the null hypothesis.