webassign.net CH 8 HW of 36 data For a random sample of 36 data pairs, the sampl
ID: 3066991 • Letter: W
Question
webassign.net CH 8 HW of 36 data For a random sample of 36 data pairs, the sample mean of the differences was 0.81. The sample standard deviation or the diterences was 2. At the 5% level of significance, test the claim that the population mean of the differences is different from o (a) Is it appropriate to use a Student's t distribution for the sample test statistic? Explain Yes, the standard deviation is larger than the sample mean. No, the sample size is not larger than 30 O No, the standard deviation is not smaller than the sample mean. O Yes, the sample size is larger than 30. What degrees of freedom are used? (b) State the hypotheses. (c) Compute the t value. (Round your answer to three decimal places.) (d) Estimate the P-value of the sample test statistic. P-value>0.500 0.250 P-valueExplanation / Answer
Solution:-
a) The sample size is larger than 30.
The degree of freedom is 35.
b) State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: ?d = 0
Alternative hypothesis: ?d ? 0
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.
Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).
s = sqrt [ (?(di - d)2 / (n - 1) ]
s = 2
SE = s / sqrt(n)
S.E = 0.3333
DF = n - 1 = 36 -1
D.F = 35
c)
t = [ (x1 - x2) - D ] / SE
t = 2.43
where di is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.
Since we have a two-tailed test, the P-value is the probability that a t statistic having 35 degrees of freedom is more extreme than 2.43; that is, less than - 2.43 or greater than 2.43.
d) Thus, the P-value = 0.0204
0.01 < p-value < 0.05
e) Interpret results. Since the P-value (0.0204) is less than the significance level (0.05), we have to reject the null hypothesis.
f) Reject H0. There is sufficient evidence to conclude that population mean of the differences is not zero.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.