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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

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Answer:


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

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