Pad 12:16 AM * 56% 7· 0-2 points My Notes Ask Your Teacher Suppose two independe
ID: 3051407 • Letter: P
Question
Pad 12:16 AM * 56% 7· 0-2 points My Notes Ask Your Teacher Suppose two independent samples are associated with the following information: Sample 1: mean 2.06; Sample 2: mean = 157.4, sample size = 26, SD = 2.59, [From Salkind e5, p. 214] 158, sample size = 22, SD = (a) Using the formula, compute the test statistic t (the obtained value) by hand. Keep lots of decimals until the end, and then round to two decimal places. (b) How many degrees of freedom are there? (What is df?) (c) Let's perform a two-tailed tailed test at the alpha = 0.05 level of significance. Find the critical value in Table B2. (If you cannot find an entry with the exact df, then choose the neighboring one that is closest to your actual total sample size, n1+n2.) Give the exact decimal places that you find in the table. (d) Compare the obtained value to the critical value. What is your conclusion? O We fail to reject the null hypothesis, because pExplanation / Answer
Solution:
Given information
x1 = 158, n1 = 22, S.D1 = 2.06
x2 = 157.4, n2 = 26, S.D2 = 2.59
Test hypotheses:
H0 : 1 - 2 = 0
Ha : 1 - 2 0
Standard Error = sqrt[(s1^2/n1) + (s2^2/n2)]
= sqrt[(2.06^2/22) + (2.59^2/26)]
= 0.6715
df = 22+26 -2 = 46
t = [ (x1 - x2) - d ] / SE
= 0.8935
tcritical = 2.013
Thus, the P-value = 0.189 + 0.189 = 0.378
Interpret results. Since the P-value (0.378) is greater than the significance level (0.05), we have to accept the null hypothesis.
We fail to reject the null hypothesis beacuse p > 0.05
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