Question

This is the first homework assignment for STA 3112. It is due in class on Monday, January 28. 1. Comparing Independent Samples (Unequal Variances): A new Statistics I textbook is being evaluated in order to improve student success. There are 20 different sections of Statistics I for which 7 are randomly sele the new textbook and 13 will use the old textbook. Both grou Assume that final exam scores follow a bell-shaped curve and each section has the same number of students following data is obtained: cted to use ps of students will take a common final exam. . The New Textbook 7 82.1 3.2 Old Textbook 13 78.3 2.5 Average final exam grade Standard Deviation Conduct a hypothesis test (Ho: μι α-0.05) without assuming that the variances are equal. State the degrees of freedom for your test. Compute a 95% Confidence Interval for the difference in the average between the two textbooks. Estimate the effect size of the new textbook.Illustrate the effect size by sketching the two normal distributions for the class averages (New Textbook and Old Textbook) side by side. a. b. final exam grades (μ-μ)

Answer 1

Solution:-

a)

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

Null hypothesis: u₁ = u ₂
Alternative hypothesis: u₁$\neq$ u ₂

Note that these hypotheses constitute a two-tailed test.

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[(s₁²/n₁) + (s₂²/n₂)]
SE = 1.39414
DF = 6
t = [ (x₁ - x₂) - d ] / SE

t = 2.73

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 6 degrees of freedom is more extreme than -2.73; that is, less than -2.73 or greater than 2.73.

Thus, the P-value = 0.034

Interpret results. Since the P-value (0.034) is less than the significance level (0.05), we cannot accept the null hypothesis.

From the above test we do not have sufficient evidence in the favor of the claim that u₁ = u ₂.

b)

95% confidence interval for the difference in average final exam grades is C.I = ( 0.389, 7.211).

$C.I = \left (\bar{x}_{1}-\bar{x}_{2} \right )\pm t_{\alpha /2}\times \sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}$

C.I = 3.80 + 2.447*1.39414

C.I = 3.80 + 3.41146

C.I = ( 0.389, 7.211)

c) Effect size is 0.9357.

$d = \frac{M_{1}-M_{2}}{S.D}$

$d = \frac{82.1-78.3}{4.061}$

d = 0.9357