Question

Given the following information about a hypothesis test of the difference between two means based on independent random samples, what is the calculated value of the test statistic? Assume that the samples are obtained from normally distributed populations having equal variances. HA: μA > μB, = 12, = 9, s1 = 5, s2 = 3, n1 = 13, n2 = 10. t = 1.674 t = 2.823 t = 1.063 t = 1.5 t = 1.96

Answer 1

`S^2_p =((13-1)(5)^2 +(10 -1)(3)^2)/(13 +10 -2)=18.143`

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

`S_p=4.259`

Then the statistic is given by:

`t=\frac{(12 -9)-(0)}{4.259\sqrt{(1)/(13)+(1)/(10)}}=1.674`

And the correct option would be:

t = 1.674

Explanation:

Data given:

`n_1 =13` represent the sample size for group 1

`n_2 =10` represent the sample size for group 2

`\bar X_1 =12` represent the sample mean for the group 1

`\bar X_2 =9` represent the sample mean for the group 2

`s_1=5` represent the sample standard deviation for group 1

`s_2=3` represent the sample standard deviation for group 2

We are assuming 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)`

The system of hypothesis on this case are:

Null hypothesis:
`\mu_1 \leq \mu_2`

Alternative hypothesis:
`\mu_1 > \mu_2`

The pooled variance is given by:

`S^2_p =((13-1)(5)^2 +(10 -1)(3)^2)/(13 +10 -2)=18.143`

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

`S_p=4.259`

Then the statistic is given by:

`t=\frac{(12 -9)-(0)}{4.259\sqrt{(1)/(13)+(1)/(10)}}=1.674`

And the correct option would be:

t = 1.674