Question

Difference of Means Test. A sample of seniors taking the SAT in Connecticut in 2015 revealed the following results for the math portion of the exam by Gender. Males Females Mean 492 520 Std Dev 119 129 N 150 165 The test statistic for a hypothesis test that the mean level of Math SAT scores between males and females is different (assuming equal variances) is:_______. A. 1.982 B. 13.973 C. 1.96 D. 2.004

Answer 1

Answer:

$t=\frac{(520 -492)-(0)}{124.340\sqrt{\frac{1}{150}+\frac{1}{165}}}=1.996$

And the most near value would be:

D. 2.004

Step-by-step explanation:

When we have two independent samples from two normal distributions with equal variances we are assuming that  

$\sigma^2_1 =\sigma^2_2 =\sigma^2$

And the statistic is given by this formula:

$t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}$

Where t follows a t distribution with $n_1+n_2 -2$ degrees of freedom and the pooled variance $S^2_p$ is given by this formula:

$\S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}$

This last one is an unbiased estimator of the common variance $\sigma^2$

The system of hypothesis on this case are:

Null hypothesis: $\mu_1 = \mu_2$

Alternative hypothesis: $\mu_1 \neq \mu_2$

Or equivalently:

Null hypothesis: $\mu_1 - \mu_2 = 0$

Alternative hypothesis: $\mu_1 -\mu_2 \neq 0$

Our notation on this case :

$n_1 =165$ represent the sample size for group female

$n_2 =150$ represent the sample size for group male

$\bar X_1 =520$ represent the sample mean for the group female

$\bar X_2 =492$ represent the sample mean for the group male

$s_1=129$ represent the sample standard deviation for group 1female

$s_2=119$ represent the sample standard deviation for group male

First we can begin finding the pooled variance:

$\S^2_p =\frac{(150-1)(119)^2 +(165 -1)(129)^2}{150 +165 -2}=15460.42$

And the deviation would be just the square root of the variance:

$S_p=124.340$

And now we can calculate the statistic:

$t=\frac{(520 -492)-(0)}{124.340\sqrt{\frac{1}{150}+\frac{1}{165}}}=1.996$

And the most near value would be:

D. 2.004