Question

Use the t-distribution and the given sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distributions are relatively normal. Test Upper H Subscript 0 Baseline : mu Subscript 1 Baseline equals mu Subscript 2 vs Upper H Subscript a Baseline : mu Subscript 1 Baseline greater-than mu Subscript 2 using the sample results x Overscript bar EndScripts Subscript 1 Baseline equals 56, s Subscript 1 Baseline equals 8.2 with n Subscript 1 Baseline equals 30 and x Overscript bar EndScripts Subscript 2 Baseline equals 51, s Subscript 2 Baseline equals 6.9 with n Subscript 2 Baseline equals 40.
A. Give the test statistic and the p-value.
B. What is the conclusion of the test? Test at a 10 % level.

Answer 1

Answer:

(A) The value of t test statistics is 2.767 and P-value is 0.0042.

(B) We conclude that the mean of first group is greater than the mean of second group.

Step-by-step explanation:

We are given the following hypothesis below;

Null Hypothesis, $H_0$ : $\mu_1=\mu_2$      {means that the mean of first group is equal to the mean of second group}

Alternate Hypothesis, $H_A$ : $\mu_1>\mu_2$      {means that the mean of first group is greater than the mean of second group}

The test statistics that would be used here Two-sample t-test statistics as we don't know about population standard deviation;

                            T.S. =  $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p\sqrt{\frac{1}{n_1} +\frac{1}{n_2} } }$  ~ $t__n__1-_n__2-2$

where, $\bar X_1$ = sample mean of first group = 56

$\bar X_2$ = sample mean of second group = 51

$s_1$ = sample standard deviation of first group = 8.2

$s_2$ = sample standard deviation of second group = 6.9

$n_1$ = sample of first group = 30

$n_2$ = sample of second group = 40

Also, $s_p=\sqrt{\frac{(n_1-1)s_1^{2} +(n_2-1)s_2^{2} }{n_1+n_2-2} }$ = $\sqrt{\frac{(30-1)\times 8.2^{2} +(40-1)\times 6.9^{2} }{30+40-2} }$  = 7.482

So, the test statistics  =  $\frac{(56-51)-(0)}{7.482 \times \sqrt{\frac{1}{30} +\frac{1}{40} } }$  ~ $t_6_8$

                                     =  2.767

(A) The value of t test statistics is 2.767.

Also, P-value of the test statistics is given by;

              P-value = P( $t_6_8$ > 2.767) = 0.0042

(B) Now, at 10% significance level the t table gives critical value of 1.295 at 68 degree of freedom for right-tailed test.

Since our test statistic is more than the critical values of t as 2.767 > 1.295, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.

Therefore, we conclude that the mean of first group is greater than the mean of second group.