Question

You wish to test the following claim (HaHa) at a significance level of α=0.05α=0.05.

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

You obtain the following two samples of data.

Sample #1	Sample #2
46 61.3 37.8 45.3 63.2 52.6 26.3 49.6 49 63.8 60.4 40.6 67.9 55 50.8 49.3 33.9 40.6 55 54.4 53.5 34.7 62.2 33.9 59.2 43.6 51.1 52.6 52.9 23 71.2 46 60.4 41.8 62.2 54.4 41.4 36 32.2 49.3 41.4 48.4	57.9 67.2 35.2 55.3 53.6 75.4 48.9 66.7 57.6 39.9 43.4 28.5 45.6 63.1 57 49.9 41.8 62.3 55 53.6 28.5 71.5 56.7 67.2 68.8 61.7 64.2 46.8 45.2 57.3 41.2 63.4 49.6 70 44.3 51.8 51.2 50.6 58.5 53.9 54.1 48.3 51.8 47.9 56.5

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 =

Answer 1

Since population variance is not known, we should use two sample t-test for testing the given hypotheses.

1) For two sample t-test the test statistic used is given by

  $t = (\overline{x1}-\overline{x2}-(\mu 1-\mu 2))/(s_{p}*\sqrt{1/n1+1/n2})$ ~ t_n1+n2-2

where $s_{p}^{2}$ = ((n1-1)*$s_{1}^{2}$ + (n2-1) * $s_{2}^{2}$)/(n1+n2-2)

Using excel commands average for sample mean, var.s for sample variance and count for n we get,

$s_{1}^{2}$ = 125.175 ,  $\overline{x1}$ = 48.909 and n1 = 42 and $s_{2}^{2}$ = 112.892 , $\overline{x2}$ = 53.753 and n2 = 45.

Hence $s_{p}^{2}$ = (41*125.175 + 44 * 112.892) / (85) = 118.817

Hence t = (48.909 - 53.753)/ ($\sqrt{118.817*(1/42+1/45)}$ = -2.071

2) p-value = 2* P(t_n1+n2-2 > |t| ) = 2* P ( t₈₅ > 2.071 ) = 0.0415 . Here we have used degree of freedom = n1+n2 -2 = 85.