Question

(a) There are 26 F and 13 Males out of a total of 39 Students

The Mens Quizzes scores are as below

 1 M 55 2 M 86 3 M 58 4 M 70 5 M 83 6 M 80 7 M 49 8 M 55 9 M 67 10 M 74 11 M 81 12 M 78 13 M 84

Average = Sum of observations / Total Observations = 920 / 13 = 70.77

The Median. Since n is odd, the median = (n + 1)/ 2 th number = (13 + 1) / 2 = 7th number = 49

For Females

 1 F 61 2 F 59 3 F 54 4 F 62 5 F 49 6 F 73 7 F 66 8 F 66 9 F 73 10 F 75 11 F 69 12 F 66 13 F 87 14 F 74 15 F 78 16 F 91 17 F 67 18 F 66 19 F 89 20 F 77 21 F 63 22 F 87 23 F 82 24 F 81 25 F 81 26 F 92

Average = Sum of observations / Total Observations = 1888 / 26 = 72.62

The Median = Middle value

Since n is even, the median = Average of n/2th and the next number = Average of 26/2 = 13th and 14th numbers

= (87 +74) / 2 = 80.5

 Men Women Mean 70.8 72.62 Median 49 80.5

________________________________________

(a) Test For proportion

The Hypothesis:

p = 0.5

p > 0.5

This is a right tailed test

The Test Statistic: = 26 / 39 = 0.667

The p Value:    The p value (Right tail) for Z = 2.08, is; p value = 0.0188

The Critical Value:   The critical value (Right tail) at = 0.05 (default level) Z critical = +1.645

The Decision Rule: If Z observed is > Z critical Then Reject H0.

Also If the P value is < , Then Reject H0

The Decision:    Since Z observed (2.08) is > Z critical (1.645), We Reject H0.

Also since P value (0.0188) is < (0.05), We Reject H0.

The Conclusion:   There is sufficient evidence at the 95% significance level to conclude that the majority of students in the class are women

____________________________________________

For Grades the sample statistics are as below

 Men Women Total 937 1898 n 13 26 Mean 72.08 73 SD 8.391 8.841

Since s1/s2 = 8.391 / 8.841 = 0.949 (it lies between 0.5 and 2) we used the pooled variance.

The degrees of freedom used is n1 + n2 - 2 = 13 + 26 - 2 = 37 (since pooled variance is used)

The Hypothesis:

This is a Two tailed test.

The Test Statistic:We use the students t test as population standard deviations are unknown.

The p Value:   The p value (2 Tail) for t = -0.32, df = 37, is; p value = 0.7575

The Critical Value:   The critical value (2 tail) at = 0.05 (default level), df = 37, t critical = - 2.03 and + 2.03

The Decision Rule: If t observed is > t critical or If   t observed is < -t critical, Then Reject H0.

Also If the P value is < , Then Reject H0

The Decision: Since t lies in between -2.03 and +2.03, We Fail To Reject H0

Also since P value (0.7575) is > (0.05), We Fail to Reject H0.

The Conclusion: There isn’t sufficient evidence at the 95% significance level to warrant rejection of the claim that Men and Women have the same mean grade.

___________________________________________

Calculation for the mean and standard deviation:

Mean = Sum of observation / Total Observations

Standard deviation = SQRT(Variance)

Variance = Sum Of Squares (SS) / n - 1, where SS = SUM(X - Mean)2.

 Men # X X - Mean (X - Mean)2 1 55 72.08 291.7264 2 72 72.08 0.0064 3 73 72.08 0.8464 4 70 72.08 4.3264 5 64 72.08 65.2864 6 75 72.08 8.5264 7 62 72.08 101.6064 8 68 72.08 16.6464 9 72 72.08 0.0064 10 82 72.08 98.4064 11 82 72.08 98.4064 12 81 72.08 79.5664 13 81 72.08 79.5664 Total 937 844.923
 Women # X X - Mean (X - Mean)2 1 59 73 196 2 63 73 100 3 65 73 64 4 66 73 49 5 61 73 144 6 59 73 196 7 59 73 196 8 67 73 36 9 70 73 9 10 74 73 1 11 72 73 1 12 72 73 1 13 79 73 36 14 75 73 4 15 73 73 0 16 88 73 225 17 76 73 9 18 73 73 0 19 78 73 25 20 83 73 100 21 73 73 0 22 85 73 144 23 80 73 49 24 81 73 64 25 77 73 16 26 90 73 289 Total 1898 1954

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