The null and alternate hypotheses are: H0 : 1 = 2 H1 : 1 2 A random sample of 10
ID: 3232596 • Letter: T
Question
The null and alternate hypotheses are: H0 : 1 = 2 H1 : 1 2
A random sample of 10 observations from one population revealed a sample mean of 23 and a sample standard deviation of 3.5. A random sample of 4 observations from another population revealed a sample mean of 27 and a sample standard deviation of 3.6. At the 0.01 significance level, is there a difference between the population means?
a. State the decision rule. (Negative amounts should be indicated by a minus sign. Round your answer to 3 decimal places.) The decision rule is to reject H0 if t < or t > .
b. Compute the pooled estimate of the population variance. (Round your answer to 3 decimal places.) Pooled estimate of the population variance
c. Compute the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.) Test statistic d. State your decision about the null hypothesis. (Click to select)RejectDo not reject H0 .
e. The p-value is (Click to select)between 0.1 and 0.2between 0.01 and 0.1between 0.001 and 0.01between 0.05 and 0.1less than 0.001.
Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: H0 : 1 = 2
Alternative hypothesis: H1 : 1 2
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 0.01. 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 = sqrt[(3.52/10) + (3.62/4)] = 2.113
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (3.52/10 + 3.62/4)2 / { [ (3.52 / 10)2 / (9) ] + [ (3.62 / 4)2 / (3) ] }
DF = 19.94 / (0.1667 + 3.499) = 5.44
t = [ (x1 - x2) - d ] / SE = [ (23 - 27) - 0 ] / 2.113= -1.893
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 5.44 degrees of freedom is more extreme than -1.893; that is, less than -1.893 or greater than 1.893.
We use the t Distribution Calculator to find P(t < -1.893) = 0.112659, and P(t > 1.893) = 0.112659. Thus, the P-value = 0.112659 + 0.112659 = 0.225318.
The p-value is between 0.1 and 0.2
Interpret results. Since the P-value (0.225318) is greater than the significance level (0.01), we can accept the null hypothesis or we can say we do not reject the null hypothesis.
Pooled estimate of population variance = (df1 / df)(s12) + (df2 / df)(s22)
= (9 / 5.44) * (3.5)2 + (3 / 5.44) * (3.6)2
= 27.414