Consider the following competing hypotheses and accompanying sample data. H_0: m
ID: 3246022 • Letter: C
Question
Consider the following competing hypotheses and accompanying sample data. H_0: mu_1 - mu_2 = 5 H_A: mu_1 - mu_2 notequalto 5 x_1 = 57 x_2 = 43 s_1 = 21.5 s_2 = 15.2 n_1 = 22 n_2 = 18 Assume that the populations are normally distributed with unknown but equal variances. a. degree Calculate the value of the test statistic. b. degree Using the p-value approach, test the above hypotheses at the 5% significance level. The p-value is H_0. At the 5% significance level, we conclude that the difference between the means differs from 5. c. Repeat the analysis using the critical value approach.Explanation / Answer
Solution:
a)
Formulating the null and alternative hypotheses,
Ho: u1 - u2 = 5
Ha: u1 - u2 =/ 5
At level of significance = 0.05
As we can see, this is a two tailed test.
Calculating the means of each group,
X1 = 57
X2 = 43
Calculating the standard deviations of each group,
s1 = 21.5
s2 = 15.2
Thus, the pooled standard deviation is given by
S = sqrt[((n1 - 1)s1^2 + (n2 - 1)(s2^2))/(n1 + n2 - 2)]
As n1 = 22 , n2 = 18
Then
S = 18.94238494
Thus, the standard error of the difference is
Sd = S sqrt (1/n1 + 1/n2) = 6.020285132
As ud = the hypothesized difference between means = 5 , then
t = [X1 - X2 - ud]/Sd = 1.494945805 [ANSWER, TEST STATISTIC]
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b)
df = n1 + n2 - 2 = 38
Getting the p value using tables,
The p-value is (0.10 < p-value < 0.20).
[(Do not reject)] H0. At the 5% significance level, we cannot conclude that the difference between the means differs from 5.
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c)
Getting the critical value using table/technology,
df = n1 + n2 - 2 = 38
tcrit = +/- 2.024394164 [ANSWER]
As |t| < 2.024, we DO NOT REJECT HO. [ANSWER]
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