A rental agent claims that the mean monthly rent, mu for apartments on the east
ID: 3296531 • Letter: A
Question
A rental agent claims that the mean monthly rent, mu for apartments on the east side of town is less than $675. A random sample of 16 monthly rents for apartments on the east side has a mean of $663, with a standard deviation of $21. If we assume that the monthly rents for apartments on the east side are normally distributed, is there enough evidence to conclude, at the 0.05 level of significance, that is less than $675? Perform a one-tailed test. Then fill in the table below. Carry your intermediate computations to at least three decimal places and round your answers as specified in the table. (If necessary, consult a list of formulas.)Explanation / Answer
Solution:-
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: > 675
Alternative hypothesis: < 675
Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the sample mean is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a one-sample t-test.
Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).
SE = s / sqrt(n)
S.E = 5.25
DF = n - 1 = 16 - 1
D.F = 15
t = (x - ) / SE
t = - 2.29
where s is the standard deviation of the sample, x is the sample mean, is the hypothesized population mean, and n is the sample size.
The observed sample mean produced a t statistic test statistic of - 2.29. We use the t Distribution Calculator to find P(t < - 2.29) = 0.0185.
Thus the P-value in this analysis is 0.0185.
Interpret results. Since the P-value (0.0185) is less than the significance level (0.05), we have to reject the null hypothesis.
Yes, From the above test we have sufficient evidence in the favor of the claim that the mean monthly rent for apartments on the east side is less than $675.
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