Question

I would like to know if the sex of a math student is a statistically significant factor in predicting average math exam scores. The following lists are exam scores for a math exam, separated by sex.

male 89 33 104 48 90 80 98 32 98 55 75 74 73 90 105 47 48 67 99 103 63

female 99 80 81 88 94 83 70 42 78 75

Perform a hypothesis test to determine whether the sex of a math student is statistically significant for performance on math tests. In other words, is there a statistically significant difference between the scores of these two groups of students?

(a) State the null and alternative hypotheses. Also, state the meaning of your parameters.

(b) Perform the test. Use α = .05. Show your work. Clearly indicate the value of the test statistic. Be sure to mention the value of df if it is relevant. Also, make sure you clearly state your final answer to the question above.

(c) Compute an appropriate 95% confidence interval that would confirm your final answer from part (b). Explain why it confirms that answer.

Answer 1

(a)

Null Hypothesis(H0):

Alternative Hypothesis(H1):

(two-tailed test).

where, =Population mean math score of males; =Population mean math score of females.

(b)

Male math scores =sample 1

Female math scores =sample 2

For sample 1:

For sample 2:

Standard Error, SE = = = = =7.1607

Test statistic, t =[(​​​​​​) - 0]/SE =[(74.8095 - 79) - 0]/7.1607 = −4.1905/7.1607 = −0.5852

Calculation of degrees of freedom, df:

Numerator, Nr =SE⁴ =7.1607⁴ =2629.1896668394

Denominator, Dr = = =36.175620345943+66.030672153635 =102.20629249957

Degrees of freedom, df =Nr/Dr =2629.1896668394/102.20629249957 =25.72 rounding to the integer, df =26

At df =26, at =0.05, for a two-tailed test, the critical value of t is: t-critical =2.0555

Conclusion:

Absolute t: 0.5852 < Absolute t-critical: 2.0555

Since the absolute value of the test statistic is less than the absolute value of the critical value, we fail to reject the null hypothesis(H0) at 5% significance level. Thus, we do not have enough evidence to claim that there is a statististically significant difference between the scores of the given two groups of students. So, the sex of a math student is not statistically significant for performance on math tests.

(c)

Margin of Error, MoE =t-critical*SE =2.0555*7.1607 =14.7188

95% confidence interval for the population mean difference is:

=(​​​​​​) MoE = -4.1905 14.7188 =(-18.9093, 10.5283)

Since the above confidence interval includes 0, the population mean difference is not significantly different from 0 which means that the difference between the scores of the given two groups of students is not statistically significant. Thus, the sex of a math student is not statistically significant for performance on math tests.

95% confidence interval confirms the answer from part (b) because it lead to the same conclusion as in part (b) and we can check this because 95% confidence level means 5% significance level which we used in part (b).