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)

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ChiKesselman ChiKesselman · Answer 1 · 2019-12-30T04:38:04+01:00

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, [tex]\bar{x}[/tex] = 1160

Sample size, n = 9

Alpha, α = 0.01

Sample standard deviation, s = 120

First, we design the null and the alternate hypothesis

[tex]H_{0}: \mu = 1200\text{ square feet}\\H_A: \mu < 1200\text{ square feet}[/tex]

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

Formula:

[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]

Putting all the values, we have

[tex]t_{stat} = \displaystyle\frac{1160 - 1200}{\frac{120}{\sqrt{9}} } = -1[/tex]

Now,

[tex]t_{critical} \text{ at 0.01 level of significance, 8 degree of freedom } = -2.89[/tex]

Since,

[tex]t_{stat} > t_{critical}[/tex]

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.