A real estate company tries to figure out whether the mean time that families li
ID: 3258392 • Letter: A
Question
A real estate company tries to figure out whether the mean time that families live in area A is different from that of families in area B. Assume that the two population variances are equal. What can you conclude from the survey data as below? The absolute value of t statistic from the mean comparison test is between 2 and 4. At the alpha = 0.10 level, the confidence interval for the difference between the 2 sample means include 0. At the alpha = 0.05 level, the null hypothesis that the mean time that families live in area A is not different from that of families in area B is rejected. The estimated standard error of the difference between the 2 sample means is > 5. At the alpha = 0.05 level, the null hypothesis that the mean time that families live in area A is no more than that of families in area B is rejected.Explanation / Answer
Soluion:-
The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below:
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: 1 - 2 = 0, the mean time is same.
Alternative hypothesis: 1 - 2 0, the mean time is different.
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 0.10. 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[(302/100) + (332/150)] = 4.03
DF = (s12/n1 + s22/n2)2 / { [ (s12 / n1)2 / (n1 - 1) ] + [ (s22 / n2)2 / (n2 - 1) ] }
DF = (302/100 + 332/150)2 / { [ (302 / 100)2 / (99) ] + [ (332 / 150)2 / (149) ] }
DF = 264.3876 / (0.8182 + 0.3537) = 225.61
t = [ (x1 - x2) - d ] / SE = [ (43 - 50.5) - 0 ] / 4.03 = -1.86
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 degrees of freedom is more extreme than -1.86; that is, less than -1.86 or greater than 1.86.
We use the t Distribution Calculator to find P(t < -1.86)
The P-Value is 0.064185.
The result is significant at p < 0.10.
Interpret results. Since the P-value (0.064) is less than the significance level (0.10), we cannot accept the null hypothesis.
Statements C and D can be related to the above results.
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.