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). The data provided is a sample of 60 students and their writing scores on the SAT from last year. The goal of this case study is to analyze the data and make conclusions about the population SAT scores and the difference between male and female scores.

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

Note: Use Excel to solve, including screenshots, and directions on how to solve on excel, construct the Null hypothesis too, Include each answer with its question number (1,2,3)

1.Construct the null and the alternative hypotheses to test if females outscore males on writing tests.  

2.Assuming the difference in scores is normally distributed, calculate the value of the test statistic. Do not assume that the population variances are equal.

3.Implement the test at alpha=0.01 and interpret your results.

Answer 1

(Part 1)

Here, we have to use the independent samples t test for difference between two population means. The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: There is no significant difference in the average scores on writing test by male and female students.

Alternative hypothesis: H_a: Average score on writing test for female is more than average score on writing test for male.

H₀: µ_male = µ_female vs. H_a: µ_male < µ_female

This is a one tailed test.

(Part 2)

We are given

Level of significance = ? = 0.01

From given data, we have

For male,

X1bar = 574.3333333

S1 = 38.2986

N1 = 30

For female,

X2bar = 593.3333333

S2 = 31.4405

N2 = 30

DF = N1+ N2 – 2 = 30 + 30 – 2 = 58

Test statistic = t = (X1bar – X2bar) / sqrt((S1^2/N1)+(S2^2/N2))

t = (574.3333333 - 593.3333333)/sqrt((38.2986^2/30)+( 31.4405^2/30))

t = -19/ sqrt((38.2986^2/30)+( 31.4405^2/30))

t = -2.100211466

Test statistic = -2.1002

(Part 3)

From above test statistic, df, and by using t-table or excel, we have

P-value = 0.02

? = 0.01

P-value > ? = 0.01

So, we do not reject the null hypothesis that there is no significant difference in the average scores on writing test by male and female students.

There is insufficient evidence to conclude that Average score on writing test for female is more than average score on writing test for male.

Excel Procedure:

Open Excel > Click on Data > Click on Data Analysis > Drop down to t-test > select two sample assuming unequal variances > Enter Variable 1 range > Enter Variable 2 range > Enter difference > Check box for Label > Enter Alpha value > Press OK

Excel output:

t-Test: Two-Sample Assuming Unequal Variances
	Males	Females
Mean	574.3333333	593.3333333
Variance	1466.781609	988.5057471
Observations	30	30
Hypothesized Mean Difference	0
df	56
t Stat	-2.10021166
P(T<=t) one-tail	0.020113417
t Critical one-tail	1.672522304
P(T<=t) two-tail	0.040226834
t Critical two-tail	2.003240704