Question

asked Feb 13, 2021 57.5k views

1 Answer

Manhar Sapra · Answer 1 · 2021-02-20T11:14:57+0000

Answer:

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

And the most near value would be:

D. 2.004

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:

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

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:

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

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:
$\mu_1 \\eq \mu_2$

Or equivalently:

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

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

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:

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

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

S_p=124.340

And now we can calculate the statistic:

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

And the most near value would be:

D. 2.004

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