Question

The SAT is required of most students applying for college admission in the United States. This standardized test has gone through many revisions over the years. In 2005, a new writing section was introduced that includes a direct writing measure in the form of an essay. People argue that female students generally do worse on math tests but better on writing tests. Therefore, the new section may help reduce the usual male lead on the overall average SAT score (The Washington Post, August 30, 2006). Consider the following scores on the writing component of the test of eight male and eight female students.
Males Females
620 660
570 590
540 540
580 560
590 610
580 590
480 610
620 650
570 600
610 620
590 630
570 640
610 590
590 640
570 580
550 560
530 570
560 560
620 600
520 600
560 590
620 590
580 590
610 630
530 560
480 600
590 560
620 560
590 560
580 560

Required:
a. Construct the null and the alternative hypotheses to test if females outscore males on writing tests.
b. Assuming the difference in scores is normally distributed, calculate the value of the test statistic. Do not assume that the population variances are equal.
c. Implement the test at α=0.01 and interpret your results.

Answer 1

Answer:\

a

The null hypothesis is  $H_o : \mu_1 - \mu_2 \le 0$

The alternative hypothesis is  $H_a : \mu_1 - \mu_2 > 0$

b

$t = -0.39$

c

The conclusion

There is sufficient evidence to conclude that the female outscores the male.

Step-by-step explanation:

From the question we are told that

  The  sample size for male  is  $n_1 = 30$

    The sample size for female is  $n_2 = 30$

 The level of significance is  $\alpha = 0.01$

The null hypothesis is  $H_o : \mu_1 - \mu_2 \le 0$

The alternative hypothesis is  $H_a : \mu_1 - \mu_2 > 0$

Generally the sample mean for male is  

     $\= x_1 = \frac{\sum x_i}{n_1}$

=>   $\= x_1 = \frac{620 + 570 +540 + \cdots + 580 }{30}$

=>   $\= x_1 = 574.33$  

Generally the standard deviation of  male is  

     $s_1 = \sqrt{\frac{\sum (x_1 - \= x)^2}{n_1} }$

=>  $s_1 = \sqrt{\frac{ (620 - 574.33)^2 + (570 - 574.33)^2 + (540 - 574.33)^2 + \cdots +(580 - 574.33)^2 }{30} }$

=>  $s_1 =206.24$

Generally the sample mean for female is  

     $\= x_2 = \frac{\sum x_i}{n_2}$

=>   $\= x_2 = \frac{660 + 590 +560 + \cdots + 560 }{30}$

=>   $\= x_2 = 593.33$      

Generally the standard deviation of  male is  

 $s_2 = \sqrt{\frac{\sum (x_1 - \= x)^2}{n_2} }$

=>  $\sigma_1 = \sqrt{\frac{ (660 - 593.33)^2 + (590 - 593.33)^2 + (560 - 593.33 )^2 + \cdots +(560 - 593.33)^2 }{30} }$

=>  $s_2 =169.31$

Generally the degree of freedom for unequal  variance is mathematically represented as  

$df = \frac{[\frac{s_1^2}{n_1} +\frac{s_2^2}{n_2} ]^2}{ \frac{[\frac{s_1^2}{n_1}]^2}{n_1 -1} +\frac{[\frac{s_2^2}{n_2}]^2}{n_2 -1} }$

=> $df = \frac{[\frac{206.24^2}{30} +\frac{169.31^2}{30} ]^2}{ \frac{\frac{206.24^2}{30}}{30 -1} +\frac{\frac{169.31^2}{30}}{30 -1} }$

=>$df = 56$

Generally the test statistics is mathematically represented as

  $t = \frac{\= x _1- \= x_2 }{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} }$

=> $t = \frac{574.33- 593.33 }{\sqrt{\frac{206.24^2}{30} + \frac{169.31^2}{30}} }$

=> $t = -0.39$

From the t distribution table the value of  $P(t < -0.39)$ at a degree of freedom of  $df = 56$ is  

   $P(t < -0.39) = 0.3490080$

Hence the p-value  is   $p-value = 0.3490080$

From the values obtained we see that the p-value  is  >$\alpha$

Hence we fail to reject the null hypothesis.

The conclusion is  

  There is sufficient evidence to conclude that the female outscores the male