Question

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

      Ho:μ1=μ2
      Ha:μ1≠μ2

You obtain the following two samples of data.

Sample #1	Sample #2
52.2 57 58.8 52.6 43.6 67.8 49.2 80.7 81.6 53.5 76.6 74 50.7 60.2 51.7 64.3 79.2 46 62.6 67.4 67.8 46 61.6 71.3 70.5 39 52.2 37.2 66.7 37.2 67.8 64 76 70.9 64 60.9 81.6 74 70.1 77.8 67.8 67.1 90.3 70.9 68.5	83.4 64.4 74.7 58.1 69.7 47.8 50.5 53.7 70.3 68.6 75.6 34.7 60.5 81.1 60.5 66.1 24.3 61.3 53.7 81.1 71.6 33.5 60.5 50.9 70.3 59.1 24.3 58.4 48.2 60.5 56.7 66.1 70.3 50.9 71.6 63.6 60.5 75.6 75.6 43.3 30.3 68.1 62 76.7 49.8 65.7 55.4 42.3 64.4 76.7

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

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁ u ₂

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s₁²/n₁) + (s₂²/n₂)]
SE = 2.77614
DF = 93
t = [ (x₁ - x₂) - d ] / SE

t = 1.19

where s₁ is the standard deviation of sample 1, s₂ is the standard deviation of sample 2, n₁ is the size of sample 1, n₂ is the size of sample 2, x₁ is the mean of sample 1, x₂ is the mean of sample 2, d is the hypothesized difference between the population means, and SE is the standard error.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 93 degrees of freedom is more extreme than -1.19; that is, less than -1.19 or greater than 1.19.

Thus, the P-value = 0.237.

Interpret results. Since the P-value (0.237) is greater than the significance level (0.10), we have to accept the null hypothesis.