A housing official in a certain city claims that the mean monthly rent for apartments in the city is more than $1000. To verify this claim, a simple random sample of 40 renters in the city was taken, and the sample mean rent paid was $1100 with a sample standard deviation of $300. Can you conclude that the mean monthly rent in the city is greater than $1000

Answer :

Answer:

Yes, we can conclude that the mean monthly rent in the city is greater than $1000.

Step-by-step explanation:

We are given that a housing official in a certain city claims that the mean monthly rent for apartments in the city is more than $1000.

To verify this claim, a simple random sample of 40 renters in the city was taken, and the sample mean rent paid was $1100 with a sample standard deviation of $300.

Let, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu \leq[/tex] $1000  {means that the mean monthly rent for apartments in the city is less than or equal to $1000}

Alternate Hypothesis, [tex]H_a[/tex] : [tex]\mu[/tex] > $1000  {means that the mean monthly rent for apartments in the city is more than $1000}

The test statistics that will be used here is t-test statistics;

           T.S. = [tex]\frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]

where, [tex]\bar X[/tex] = sample mean rent paid = $1100

             s = sample standard deviation = $300

             n = sample of renters = 40

So, test statistics = [tex]\frac{1100-1000}{\frac{300}{\sqrt{40} } }[/tex]  ~ [tex]t_3_9[/tex]  

                              = 2.1082

Since in the question we are not given the significance level to test this hypothesis, so we assume it to be 5%. At 5% level of significance, t table gives critical value of 1.685 at 39 degree of freedom. Since our test statistics is more the critical value of t so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region.

Therefore, we conclude that the mean monthly rent in the city is greater than $1000.

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