Question

You wish to test the following claim (HaHa) at a significance level of α=0.10α=0.10.

      Ho:μ1=μ2Ho:μ1=μ2
      Ha:μ1>μ2Ha:μ1>μ2

You obtain the following two samples of data.

Sample #1	Sample #2
76.6 68.8 83.7 70.2 82.1 89.6 76.6 83.7 78.4 72.8 79.7 76.6 85.6 80.6 81.2 74.9 77.3 77.8 77.3 85.9 65.4 79.7 73.6 73.6 78.5 79 66.6 76.1 77.7 79 81.1 74.7 83.9 77.7 80.6 86.1 74.7 81.4 77.3 91.8 71.3 81.4 77.3 75.1 82.1 89.6	81.6 56.1 67.4 76.7 84.6 85.9 88.6 81 56.1 64.8 68.8 83.8 67.1 72.1 75.3 82.7 75.3 96 67.8 70 60.8 84.6 74.7 56.1 66.4 66.7 77.3 58.8 77 71.2 54.9 80 61.4 88.6 81 68.4 67.8 96 89.2 101 73.8 76.4 69.4 67.1 97.9 77.6 73.8 88.6 79.4 64.4 88.6 62.5 74.7 67.1 81 70 62

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? For this calculation, use the degrees of freedom reported from the technology you are using. (Report answer accurate to four decimal places.)
p-value =

Also Can it be done on the TI-84 Plus?

Answer 1

Ho:μ1=μ2

Ha:μ1>μ2

From the given data,

n1= 46 n2= 57 We will use t test to compare means of two samples.

Calculating mean and standard deviation for each sample.

x1_bar= 78.5804 Sigma1= 5.6258

x2_bar= 74.7 Sigma2 = 11.1985

Testing for Equality of Variances

A F-test is used to test for the equality of variances. The following F-ratio is obtained:

F = {s_1^2}/ {s_2^2} = 5.6258^2} / { 11.1984^2} = 0.252

The critical values are F_L = 0.47 and F_U = 2.0722, and since F = 0.252, then the null hypothesis of equal variances is rejected.

t

the significance level is α=0.01, and the degrees of freedom are df = 86.032. In fact, the degrees of freedom are computed as follows, assuming that the population variances are unequal:

Hence, it is found that the critical value for this right-tailed test is t_c = 2.37, for α=0.01 and df = 86.032.

The rejection region for this right-tailed test is R={t:t>2.37}.

(78.5804 - 74.7) / sqrt(( 5.6258/46) + (11.1985/ 57 ) ) = 2.283

Since it is observed that t = 2.283 \le t_c = 2.37t=2.283≤tc​=2.37, it is then concluded that the null hypothesis is not rejected.

Using the P-value approach: The p-value is p = 0.0124, and since p=0.0124≥0.01, it is concluded that the null hypothesis is not rejected.

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean M1​ is greater than M2​, at the 0.01 significance level.