Question

Your local newspaper contains a large number of advertisements for unfurnished one-bedroom apartments in a particular neighborhood. You choose 36 at random and calculate that their mean monthly rent is $530 with a standard deviation of $78. You want to find out if the sample data gives a good reason to believe that the average rent for all advertised one-bedroom apartments is less than $550 per month. Conduct the following hypothesis test at 0.01 level of significance.

Answer 1

Answer:

μ < 550

Step-by-step explanation:

Let us take,

Null Hypothesis(H₀) : μ = 550

Alternative Hypothesis(H₁) : μ < 550

Now, The Z-statistic for mean is given by,

$Z=\frac{\bar{x}-\mu}{\sqrt{\frac{s^{2}}{n}}}$

Here, n = 36, μ = 550, $\bar{x}$ = 530, s = 78

Z = (530 - 550) ÷ √(78 / 36)

Z = -20 ÷ 1.47196

Z = -13.58732

Thus, Z-value = -13.587 with 35 degree of freedom.

and α = 0.01

The value of p is < .00001.

Since, the value of p is less than α.

The result is significant at p < .01.

Thus, we reject the null-hypothesis.

Hence, μ < 550