Question

Assume that µ = 500 and s = 100. Your study shows a sample of size 22 with a mean of 530 and standard deviation of 113. a. What is the most powerful test to use to test the hypothesis that the mean of the sample was drawn from the above Null Hypothesis Population? b. What is the value of the test statistic? c. What do you conclude using a = 0.052 tail?

Answer 1

Answer:   using t - test value is 1.37<2.08( At 0.05 level of significance)

        The mean of the sample was drawn from the population.

Step-by-step explanation:

Here sample size n=22

Given sample mean =530

Given population mean   =500

$t=\frac{sample mean-µ}{\frac{S}{\sqrt{n-1} } }$

given sample standard deviation s=100

a) null hypothesis:-  The mean of the sample was drawn from the population   µ =500

Alternative hypothesis:-

The most powerful test you can use is t - distribution.

The test statistic is  

$t=\frac{sample mean- µ}{\frac{S}{\sqrt{n-1} } } here S is the standard deviation of the sampleb) The value of the test statistic [tex]t=\frac{sample mean- µ}{\frac{S}{\sqrt{n-1} } } substitute given values sample size n=22sample mean =530sample standard deviation s =100mean of the population µ =500the calculated value of t =[tex]\frac{530-500}{\frac{100}{\sqrt{22} } }$

                      the calculated value=1.3748

c) The degrees of freedom =n-1 = 22-1 = 21

     The table value of t are 0.05 level of significance with 21 degrees of freedom is 2.08

       The calculated value 1.37<2.08

∴ we accept null hypothesis at 0.05 level of significance

conclusion:-

The mean of the sample was drawn from the  population.