Question

The time needed for all college students to complete a certain paper-and-pencil maze follows a normal distribution with a population mean of 30 seconds and a population standard deviation of 3 seconds.
You wish to see if the mean time μ (where μ stands for population mean) is changed by vigorous exercise, so you have a group of nine college students exercise vigorously for 30 minutes and then complete the maze. It takes them an sample average of 31.2 seconds to complete the maze. Use this information to test the hypotheses:

H0: μ = 30,

Ha: ≠30

at the significance level of 0.01.

You conclude:


a) There is not enough evidence to support the claim.

b) There is enough evidence to support the claim.

Answer 1

Answer:

The correct option is a

Step-by-step explanation:

From the question we are told that

   The population mean is  $\mu = 30\ seconds$

    The standard deviation is  $\sigma = 3 \ seconds$

     The sample size is  n  = 9

    The null hypothesis is $H_o: \mu = 30$

     The alternative hypothesis is  $H_a : \mu \ne 30$

    The level of significance is  $\alpha = 0.01$

    The sample mean is  $\= x = 31.2$

Generally the test statistics is mathematically represented as

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

=>   $z = \frac{ 31.2 - 30}{ \frac{3}{\sqrt{9} } }$

=>   $z = 1.2$

From the z table  the area under the normal curve to the right corresponding to  1.2  is    

         $P(Z > 1.2) = 0.11507$

Generally the p-value is mathematically represented as

         $p-value = 2 * P(Z > 1.2)$

=>       $p-value = 2 *0.11507$

=>       $p-value = 0.230$

From the value obtained w can see that

         $p-value > \alpha$ hence

The decision rule is  

Fail to reject the null hypothesis

The conclusion is there is not enough evidence to support the claim