On their website, a property management company states that the mean monthly rent for apartments on the east side of town is $700. A researcher for a consumer advocacy group believes that, due to the construction of newer apartment complexes in neighboring towns, the mean monthly rent on the east side, , is now lower. A random sample of 12 monthly rents for apartments on the east side has a mean of $687, with a standard deviation of $17. Assume that current monthly rents for apartments on the east side are approximately normally distributed. Based on the sample, is there enough evidence to conclude, at the 0.05 level of significance, that the population mean monthly rent is now less than what is stated on the website?
Perform a one-tailed test. Then complete the parts below. Carry your intermediate computations to three or more decimal places.
(a) State the null hypothesis and the alternative hypothesis . (b) Determine the type of test statistic to use. (c) Find the value of the test statistic. (Round to three or more decimal places.) (d) Find the critical value. (Round to three or more decimal places.) (e) Can we conclude that the mean monthly rent for apartments on the east side is less than what is stated on the website?

The null hypothesis is μ≥700 and alternative hypothesis is μ<700, the type of test statistic is t statistic, the statistical test value is -2.650, the critical value is -1.796 and the null hypothesis is reject.

Given the average monthly rent for apartments on the east side of the cities is $700, and the mean for a random sample of 12-month rents for apartments on the east side is $687, with a standard deviation of $17.

Statistical hypothesis testing is an important technique in inferential statistics. Used to test hypotheses about population parameters. Statistical tests are used to compute test statistics and p-values. Finally, use this test statistic and the p-value to make a decision about rejecting the null hypothesis.

(a) Let μ be the population and monthly rent. the null and alternative hypotheses are given below.

Null Hypothesis: μ≥700

Alternative hypothesis: μ< 700 This is a left-sided hypothesis test.

(b) Here, the sample size is less than 30 and the population mean is unknown. Therefore, a one-sample t-test can be used here. Therefore, the t statistic is the test statistic.

(c) Given [tex]\bar{x}[/tex]=687, s=17 and n=12, the test statistic is computed as

[tex]\begin{aligned}t&=\frac{\bar{x}-\mu_{0}}{s/\sqrt{n}}\\ &=\frac{687-700}{17/\sqrt{12}}\\ &=-2.64901888216\\ &=-2.650\end[/tex]

(d) For n-1=12-1=11 degrees of freedom

The critical value for a significance level of 0.05 with 11 degrees of freedom is

[tex]\begin{aligned}t_{crit}&=-1.7959\\ &=-1.796\end[/tex]

(e) Reject the null hypothesis because the statistical test value (-2.650) is less than the critical value (-1.796). There is ample evidence to support the claim that the average monthly rent for East Side apartments is lower than stated on the website.

Hence, The value of test statistic and critical value has a mean of $687 and a standard deviation of $17, which are -2.650 and -1.796.

Learn more about test statistic from here brainly.com/question/16194574

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