Question

A teacher of statistics wants to know if a new teaching methodology that includes IT is efficient in terms of increased average score. He took a class with old methodology and a class with new methodology for samples and gave a same test. Open the file by clicking the file name above. Once you open the file and run Excel, you need not open it again. What is Ha? Find it from Excel output that you generate.

a) 0.62.
b) 0.5.
c) 0.31.
d) -0.5.

Answer 1

The answer is 0.31

Explanation:

Old Method New Method .

Mean 73.5625 Mean 75.70588

Standard Error 3.143736 Standard Error 2.923994

Median 72 Median 75

Mode 90 Mode 64

Standard deviation 12.57494 Standard deviation 12.05594

Sample Variance 158.1292 Sample Variance 145.3456

Kurtosis -1.14544 Kurtosis -0.76646

Skewness 0.171025 Skewness 0.091008

Range 39 Range 41

Minimum 55 Minimum 56

Maximum 94 Maximum 97

Sum 1177 Sum 1287

Count 16 Count 17

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) } \\\\SE=4.29`

DF = 31

`t = ((x_1-x_2)-d)/(SE) \\\\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.