Question 1150935: A sample of scores on an examination given in Statistics 201 are:
Men : 72 69 98 66 85 76 79 80 77
Women : 81 67 90 78 81 80 76
At the .01 significance level, is the mean grade of the women higher than that of the men?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
mu1 = population mean of the men's test scores
mu2 = population mean of the women's test scores
H0: Null hypothesis
H0: mu1 = mu2
H0: mu1 - mu2 = 0
H1: Alternative hypothesis
H1: mu1 < mu2
H1: mu1 - mu2 < 0
The claim "the mean grade of the women is higher than that of the men" is in the alternative hypothesis.
This is a one-tailed test to the left due to the "less than" sign in the alternative hypothesis. This means when it comes to computing the area under the curve, we'll be shading to the left of the test statistic.
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alpha = 0.01
men's scores = M = {72,69,98,66,85,76,79,80,77}
womens scores = W = {81,67,90,78,81,80,76}
n1 = sample size of men's scores
n1 = 9
n2 = sample size of women's scores
n2 = 7
We're doing a two sample T test. This is not a paired T test because the n1 and n2 values are different. We use a T distribution because n1 and n2 are both smaller than 30. Also, we dont know sigma1 and sigma2 (the population standard deviations of each group), so that is another reason why we are not doing a two sample Z test.
The degrees of freedom (df) is the smaller of n1-1 and n2-1
n1-1 = 9-1 = 8
n2-1 = 7-1 = 6
The df is 6
Use a calculator to find that
xbar1 = mean of men's scores
xbar1 = 78
xbar2 = mean of women's scores
xbar2 = 79
v1 = sample variance of men's scores
v1 = 90
v2 = sample variance of women's scores
v2 = 47.33333
Compute the standard error SE
SE = sqrt(v1/n1 + v2/n2)
SE = sqrt(90/9 + 47.33333/7)
SE = 4.094130467598
Now compute the test statistic
t = ((xbar1-xbar2)-(mu1-mu2))/SE
t = ((78-79)-(0))/4.094130467598
t = -1/4.094130467598
t = -0.24425210869909
t = -0.2443
Use this calculator
https://stattrek.com/online-calculator/t-distribution.aspx
with df = 6, to find that
P(T < -0.2443) = 0.4076
So we can say
p-value = 0.4076
T Distribution Diagram:
Image generated by GeoGebra (free graphing software).
This p-value is very large, much larger than the significance level alpha = 0.01, so we fail to reject the null hypothesis.
In other words, we dont have enough statistically significant evidence to overturn the null hypothesis. We have no choice but to accept the null hypothesis.
Interpretation: The men and women have the same average test scores.
The answer to the question "is the mean grade of the women higher than that of the men?" is "no, the mean grade of the women is not higher than that of the men".
Side notes:
- No assumptions are made that sigma1 = sigma2
- The two samples (men's and women's test scores) are independent and randomly selected through simple random sampling (SRS).
- The sample variance v is equal to the sample standard deviation squared, so v = s^2. This helps simplify notation. Instead of writing (s1)^2, we can write v1.
- I'm not pooling the standard deviations s1 and s2 (instructions dont say to)
- There is a really ugly formula to help generate a more accurate degrees of freedom (df) value, but many stats books use the conservative estimate that the df is the smaller of n1-1 or n2-1.
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