A random sample of 8 observations from one population revealed a sample mean of
ID: 3225120 • Letter: A
Question
A random sample of 8 observations from one population revealed a sample mean of 23 and a sample standard deviation of 3.8. A random sample of 9 observations from another population revealed a sample mean of 27 and a sample standard deviation of 3.8.
State the decision rule. (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.)
Compute the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
The null and alternate hypotheses are: H0 : 1 = 2 H1 : 1 2A random sample of 8 observations from one population revealed a sample mean of 23 and a sample standard deviation of 3.8. A random sample of 9 observations from another population revealed a sample mean of 27 and a sample standard deviation of 3.8.
Explanation / Answer
a. Difference in mean = 23 - 27 = -4
Degree of freedom = 15 (Rounded to nearest integer)
(calculated by the formula DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
where s1 = s2 = 3.8 and n1 = 8, n2 = 9)
For 2 tail test, the significance level = 0.05/2 = 0.025
t value for p=0.025 at df = 15 is -2.131
t value for p=0.975 at df = 15 is -2.131
The decision rule is to reject H0 if t < -2.131 or t > 2.131
b. Pooled estimate of the population variance
Variancepooled = [ (n1 -1) * s12) + (n2 -1) * s22) ] / (n1 + n2 - 2)
s1 = s2 = 3.8 and n1 = 8, n2 = 9
Variancepooled = [ (8 -1) * 3.82) + (9 -1) * 3.82) ] / (8 + 9 - 2)
= 14.44
c. Standard error of the 2 samples = 1.846
(calculated from the formula SE = sqrt[ (s12/n1) + (s22/n2) ]
where s1 = s2 = 3.8 and n1 = 8, n2 = 9)
t = -4/1.846 = -2.167
Test statistic = -2.167
d. State your decision about the null hypothesis.
As, t (-2.167) < critical t value (-2.131), we reject the null hypothesis H0.
e. The p-value for t value = -2.131 and df = 15 is 0.02337.
So the correct answer is between 0.02 and 0.05
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.