Question

Consider the following hypothesis test:

Answer 1

Answer:

a

 $z = - 2.12$

b

 $p- value = 0.017003$

c

There is sufficient evidence to show that the population mean is less than 20

d

 $z_{0.05} =- 1.645$

e

   The decision rule is  

Reject the null hypothesis

f

The conclusion is  

  There is sufficient evidence to show that the population mean is less than 20.

Step-by-step explanation:

From the question we are told that

   The null hypothesis  is  $H_o: \mu = 20$

    The alternative hypothesis is  Ha: µ < 20

    The sample size is  n  =  50  

    The sample mean is  $\= x = 19.4$

    The level of significance is  $\alpha = 0.05$

      The population standard deviation is  $\sigma = 2$

Generally the test statistics is mathematically represented as

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

=>    $z = \frac{19.4 - 20 }{ \frac{2}{ \sqrt{50 } } }$

=>    $z = - 2.12$

From the z table  the area under the normal curve to the left corresponding to  -2.12  is  

       $(P< -2.12) = 0.017003$

Generally the p-value is  

        $p- value = 0.017003$

From values obtained we see that $p-value < \alpha$ hence  

   The decision rule is  

Reject the null hypothesis

   The conclusion is  

  There is sufficient evidence to show that the population mean is less than 20.

Generally form the normal distribution table the critical value at a level of significance of  $\alpha = 0.05$ is

        $z_{0.05} =- 1.645$

Generally given that  $z < z_{0.05}$

   The decision rule is  

Reject the null hypothesis

   The conclusion is  

  There is sufficient evidence to show that the population mean is less than 20.