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

Answer:

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

Step-by-step 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:

$s_M=\dfrac{s}{\sqrt{n}}=\dfrac{7.5}{\sqrt{20}}=1.677$

Then, we can calculate the t-statistic as:

$t=\dfrac{M-\mu}{s/\sqrt{n}}=\dfrac{9-5}{1.677}=\dfrac{4}{1.677}=2.385$

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.