Question

You are given the following information obtained from a random sample of 5 observations. 20 18 17 22 18 At 90% confidence, you want to determine whether or not the mean of the population from which this sample was taken is significantly less than 21. (Assume the population is normally distributed.) a) State the null and the alternative hypotheses. b) Compute the standard error of the mean. c) Determine the test statistic. d) Test to determine whether or not the mean of the population is significantly less than 21.

Answer 1

Answer:

a

  The null hypothesis is  

         $H_o : \mu = 21$

The Alternative  hypothesis is  

           $H_a : \mu< 21$

b

     $\sigma_{\= x} = 0.8944$

c

   $t = -2.236$

d

  Yes the  mean population is  significantly less than 21.

Step-by-step explanation:

From the question we are given

           a set of  data  

                               20  18  17  22  18

       The confidence level is 90%

       The  sample  size  is  n =  5  

Generally the mean of the sample  is  mathematically evaluated as

        $\= x = \frac{20 + 18 + 17 + 22 + 18}{5}$

       $\= x = 19$

The standard deviation is evaluated as

        $\sigma = \sqrt{ \frac{\sum (x_i - \= x)^2}{n} }$

         $\sigma = \sqrt{ \frac{ ( 20- 19 )^2 + ( 18- 19 )^2 +( 17- 19 )^2 +( 22- 19 )^2 +( 18- 19 )^2 }{5} }$

         $\sigma = 2$

Now the confidence level is given as  90 %  hence the level of significance can be evaluated as

         $\alpha = 100 - 90$

        $\alpha = 10$%

         $\alpha =0.10$

Now the null hypothesis is  

         $H_o : \mu = 21$

the Alternative  hypothesis is  

           $H_a : \mu< 21$

The  standard error of mean is mathematically evaluated as

         $\sigma_{\= x} = \frac{\sigma}{ \sqrt{n} }$

substituting values

         $\sigma_{\= x} = \frac{2}{ \sqrt{5 } }$

        $\sigma_{\= x} = 0.8944$

The test statistic is  evaluated as  

              $t = \frac{\= x - \mu }{ \frac{\sigma }{\sqrt{n} } }$

substituting values

              $t = \frac{ 19 - 21 }{ 0.8944 }$

              $t = -2.236$

The  critical value of the level of significance is  obtained from the critical value table for z values as  

                   $z_{0.10} = 1.28$

Looking at the obtained value we see that $z_{0.10}$ is greater than the test statistics value so the null hypothesis is rejected