A math instructor claims that college women have more credit card debt than college men. She conducts a random sample of 38 college men and 32 college women, determines their average credit card debt, and obtains the following statistics: women n1 =32 x1= 781 s1 = 1489 men n2 = 38 x2 = 435 s2 = 1026
Test the claim that college women have more credit card debt than college men at the a = .05 level of significance. Assume unequal variances.

There is no significant evidence to support the claim that college women have more credit card debt than college men

Step-by-step explanation:

Given :

women n1 =32 x1= 781 s1 = 1489 men n2 = 38 x2 = 435 s2 = 1026

H0 : μ1 = μ2

H0 : μ1 > μ2

Assume unequal variance :

The test statistic :

(x1 - x2) / √(s1²/n1) + (s2²/n2)

T= (781 - 435) / √(1489²/32) + (1026²/38)

T = 346 / 311.42740

Test statistic = 1.111

Degree of freedom, df

(s1²/n1+s2²/n2)²÷1/(n1-1)*(s1²/n1)²+1/(n2-1)*(s2²/n2)²

The Pvalue :

(s1²/n1+s2²/n2)² = ((1489²/32) + (1026²/38))² = 9406484230.6884765625

1/(n1-1)*(s1²/n1)²+1/(n2-1)*(s2²/n2)²:

1/31(1489^2/32)^2 + 1/37(1026^2/38)^2 = 1.755926E8

df = 9406484230.6884765625 / 1.755926E8 = 53.569

df = 54

The Pvalue, from t score ;

Pvalue(1.111, 54) = 0.136

Pvalue > α ; Hence, we fail to reject the null ; There is no significant evidence to support the claim that college women have more credit card debt than college men

x/3-10=8

x/3=18

x=48