Question

Independent random samples of 17 sophomores and 13 juniors attending a large university yield the following data on grade point averages:

Sophomores			Juniors
3.04	2.92	2.86	2.56	3.47	2.65
1.71	3.60	3.49	2.77	3.26	3.00
3.30	2.28	3.11	2.70	3.20	3.39
2.88	2.82	2.13	3.00	3.19	2.58
2.11	3.03	3.27	2.98
2.60	3.13
??1= 2.84, s1 = 0.52			??2= 2.98, s2 = 0.30

At the 5% significance level, do the data provide sufficient evidence to conclude that the mean GPAs of sophomores and juniors at the university differ?

Answer 1

n1 = 17

n2 = 13

$\bar{x1}=2.84$

$\bar{x2}=2.98$

s1 = 0.52

s2 = 0.30

Null and alternative hypothesis is

H0 :u1 - u2 =0

Vs

H1 :u1 - u2 ? 0

Level of significance = 0.05

Before doing this test we have to check population variances are equal or not.

Null and alternative hypothesis is

$H0:\sigma _{1}^{2}=\sigma _{2}^{2}$

Vs

$H1 : :\sigma _{1}^{2 } \neq \sigma _{2}^{2}$

Test statistic is

F = Larger variance / Smaller variance = 0.2704 / 0.09 = 3.004

Degrees of freedoms

Degrees of freedom for numerator = n1 - 1 = 17 - 1 = 16

Degrees of freedom for denominator = n2 - 1 = 13 - 1 = 12

Critical value = 2.60       ( using f-table )

F test statistic > critical value , reject null hypothesis.

Conclusion: Population variances are unequal .

So we have to use unpooled variance.

Test statistic

Formula

$t=\frac{\bar{x1}-\bar{x2}}{\sqrt{\frac{s1^{2}}{n1}+\frac{s2^{2}}{n2}}}$

$t=\frac{2.84-2.92}{\sqrt{\frac{0.52^{2}}{17}+\frac{0.30^{2}}{13}}}$

$t=-0.927$

$\alpha$ = 0.05      ,d.f = n1 + n2 – 2 = 17 + 13 - 2 = 28

p-value = 0.3626 ( using t table )

p-value > $\alpha$ ,Failed to Reject H0

conclusion : The data provide sufficient evidence to conclude that the mean GPAs of sophomores and juniors at the university differ