Question

Tecside real estate, inc. is a research firm that tracks the cost of apartment rentals in southwest virginia. in mid-2002, the regional average apartment rental rate was $895 per month. assume that, based on the historical quarterly surveys, it is reasonable to assume that the population standard deviation is $225. in a current study of apartment rental rates, a sample of 180 apartments in the region provided the apartment rental rates.a. do the sample data enable tecside real estate, inc. to conclude that the population mean apartment rental rate now exceeds the level reported in 2002? the sample mean is $915 and the sample standard deviation is $227.50. make your decision based on α=0.10.b. what is the p-value?

Answer 1

The decision is to fail to reject the null hypothesis

This means that the population mean apartment rental rate now does not exceeds the level reported in 2002

Explanation:

From the question we are told that

The population mean is
`\mu = \$895`

The population standard deviation is
`\sigma = \$225`

The sample size is
`n = 180`

The sample mean is
`\= x = \$ 915`

The sample standard deviation is
`s = \$ 227.50`

The level of significance is
`\alpha = 0.01`

The null hypothesis is
`H_o : \mu \le \$ 895`

The alternative hypothesis is
`H_a : \mu > \$895`

The test statistics is evaluated as

`t = (\= x- \mu)/((\sigma )/(√(n) ) )`

=>
`t = ( 915 - 895)/( (225)/( √(180) ) )`

=>
`t =1.193`

So from z-table the p-value is obtained, the value is

`p-value = P(Z > t) = P(Z > 1.193 ) = 0.11643`

So From the value obtained we see that

`p-value > \alpha` so we fail to reject the null hypothesis