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?

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:

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:

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

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.

center= (-2,0)

radius= sqrt2/2

step-by-step explanation:

42%

step-by-step explanation:

we know,

total people = 60

total males = 25

percentage of males

= 25/60 × 100

= 25/6 × 10

= 25/3 × 5

= 125 / 3

= 41.

rounding of = 42%

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