Question

In testing the hypothesis H 0 colon mu subscript D less or equal than 5 space v s. space H 1 colon mu subscript D greater than 5 space, two random samples from two normal populations produced the following statistics: n subscript D equals 20 comma space X with bar on top subscript D equals 9 space a n d space s subscript D equals 7.5 . What conclusion can we draw at the 1% significance level?

Answer 1

The null hypothesis failed to be rejected (P-value=0.014).

Explanation:

This is a hypothesis test for the population mean.

The null and alternative hypothesis are:

`H_0: \mu=5\\\\H_a:\mu> 5`

The significance level is α=0.01.

The sample has a size n=20.

The sample mean is M=9.

As the standard deviation of the population is not known, we estimate it with the sample standard deviation, that has a value of s=7.5.

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

`s_M=(s)/(√(n))=(7.5)/(√(20))=1.677`

Then, we can calculate the t-statistic as:

`t=(M-\mu)/(s/√(n))=(9-5)/(1.677)=(4)/(1.677)=2.385`

The degrees of freedom for this sample size are:

`df=n-1=20-1=19`

This test is a right-tailed test, with 19 degrees of freedom and t=2.385, so the P-value for this test is calculated as (using a t-table):

`P-value=P(t>2.385)=0.014`

As the P-value (0.014) is bigger than the significance level (0.01), the effect is not significant.

The null hypothesis failed to be rejected.