Question

101T 2019 (2192 Time Remoining 109:45 Qui 6 Chapters & n 3 of 8 A rental agent claims that the mean monthly rent, u, for apartments on the east side of town is less than $700. A random sample of 19 monthly rents for apartments on the east side has a mean of $690, with a standard deviation of $20. 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 μ is less than $700? 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 specifed in the table. (If The alternative hypothesis H (Round to at least three decimal places.) The critical value at the 0.05 level of (Round to at least three Using the 0.05 level of significance, can we conclude that the mean monthly rent for apartments on the east side is less than $700 O No

Where it says “the type of test statistic , choose one) the options are Z, t , Chi square , F . Thank you!

Answer 1

Solution:-

The type of test statistic is t.

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u > 700
Alternative hypothesis: u < 700

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 = 4.5883
DF = n - 1

D.F = 18
t = (x - u) / SE

t = - 2.18

tcritical = -1.734

Rejection region is t < -1.734

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

The observed sample mean produced a t statistic test statistic of - 2.18.

Thus the P-value in this analysis is 0.021.

Interpret results. Since the P-value (0.021) is less than the significance level (0.05), we have to reject the null hypothesis.

Interpret results. Since the t-value (-2.18) lies in the rejection region, hence we have to reject the null hypothesis.

Yes, From the above test we have sufficient evidence in the favor of the claim that u is less than $700.