Question

The mean square area of several thousand apartments in a new development is advertised to be 1200 square feet. A tenant group thinks the apartments are smaller, on average, than advertised. The result of sampling 9 apartments yields a sample mean of x=1160 and sample standard deviation s = 120 feet. The histogram shows that the sampling distribution is approximately normal.) Carry out the hypothesis test at significance level 0.01 via the following: State the null and alternative hypotheses, derive the test statistic, and state a conclusion in real world terms. (You can use the fact that the area to the left of -2.8965 is 1%, using the t distribution with 8 df)

Answer 1

Answer:

We conclude that apartments are 1200 square feet, on average, as advertised.

Step-by-step explanation:

We are given the following in the question:  

Population mean, μ = 1200 square feet

Sample mean, $\bar{x}$ = 1160

Sample size, n = 9

Alpha, α = 0.01

Sample standard deviation, s = 120

First, we design the null and the alternate hypothesis

$H_{0}: \mu = 1200\text{ square feet}\\H_A: \mu < 1200\text{ square feet}$

We use one-tailed t test to perform this hypothesis.

Formula:

$t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }$

Putting all the values, we have

$t_{stat} = \displaystyle\frac{1160 - 1200}{\frac{120}{\sqrt{9}} } = -1$

Now,

$t_{critical} \text{ at 0.01 level of significance, 8 degree of freedom } = -2.89$

Since,                    

$t_{stat} > t_{critical}$

We fail to reject the null hypothesis and accept the null hypothesis.

Thus, we conclude that apartments are 1200 square feet, on average, as advertised.