A random sample of 16 apartments in NYC showed that the average rent is $2,850 with standard deviation of $50. Assume normal distribution. Test the claim that the average rent has increased at α= 0.05

At  α= 0.05, there is enough evidence to support the claim that the average rent has increased from $2800.

P-value = 0.001.

Step-by-step explanation:

The question is incomplete:

The average monthly rent for a one bedroom apartment in NYC was $2,800 in 2013. A random sample of 16 apartments in NYC showed that the average rent is $2,850 with standard deviation of $50. . Test the claim that the average rent has increased at α= 0.05.

This is a hypothesis test for the population mean.

The claim is that the average rent has increased from $2800.

Then, the null and alternative hypothesis are:

The significance level is 0.05.

The sample has a size n=16.

The sample mean is M=2850.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=50.

The estimated standard error of the mean is computed using the formula:

Then, we can calculate the t-statistic as:

The degrees of freedom for this sample size are:

This test is a right-tailed test, with 15 degrees of freedom and t=4, so the P-value for this test is calculated as (using a t-table):

As the P-value (0.001) is smaller than the significance level (0.05), the effect is significant.

The null hypothesis is rejected.

There is enough evidence to support the claim that the average rent has increased from $2800.

for this pair of functions g(x)=x-2 and f(x)=|x| is (g o f)(a)= |a| -2

step-by-step explanation:

g(x)=x-2

f(x)=|x|

(g o f)(a)=?

(g o f)(x)=g(f(x))

(g o f)(x)=g(|x|)

(g o f)(x)=|x|-2

(g o f)(a)=|a|-2

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step-by-step explanation: