Question

For the new methodology to be efficient, the difference in average scores between two population means (old - new, μ1-μ2) should be negative. Consider that the class with old methodology is the first group. We do not know the population variances, but assume they are NOT equal. We do hypothesis testing to see if the new methodology is efficient. Based on the Excel output, is the new methodology efficient?

Answer 1

Check the explanation

Step-by-step explanation:

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

Null hypothesis: μNew< μOld

Alternative hypothesis: μNew > μOld

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. 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[(s_1^2/n_1) + (s_2^2/n_2)]`

S.E = 4.29

DF = 31

t =
`[ (x_1 - x_2) - d ] / S`E

t = - 0.4997

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 population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of - 0.499. We use the t Distribution Calculator to find P(t < - 0.499) = 0.311

Therefore, the P-value in this analysis is 0.311.

Interpret results. Since the P-value (0.311) is greater than the significance level (0.05), we cannot reject the null hypothesis.

From the above test we do have sufficient evidence in the favor of the claim that new method is efficient than the old method.