Question

A consumer group is testing camp stoves. To test the heating capacity of a stove, they measure the time required to bring 2 quarts of water from 50 degrees to boiling.
Two competing models are under consideration. Thirty-six stoves of each model were tested and the following results were obtained.
Model 1: mean time is 11.4 and standard deviation is2.5
Model 2: mean time is 9.9 and standard deviation is 3.0
Is there any difference between the performances of these two models? {use a .05 level of significance}. Find the p-value of the sample statistic and do a significance test.
Find a 95% confidence interval for the difference of the means.

Answer 1

Answer:

a

The decision rule  is  

Reject the null hypothesis

  The conclusion is  

There is sufficient evidence to show that there is a difference between the performances of these two models

b

The  95% confidence interval is  $0.224 < \mu_1 - \mu_2 < 2.776$

Step-by-step explanation:

From the question we are told that

    The sample size is  n  =  36

    The first sample mean is  $\= x_1 = 11.4$

    The first standard deviation is  $s_1 = 2.5$

    The second sample mean is   $\= x_2 = 9.9$

     The second standard deviation is  $s_2 = 3.0$

      The level of significance is  $\alpha = 0.05$

The null hypothesis is  $H_o : \mu_1 - \mu_2 = 0$

The alternative hypothesis is $H_a : \mu_1 - \mu_2 \ne 0$

Generally the test statistics is mathematically represented as

      $z = \frac{ (\= x_1 - \= x_2 ) - (\mu_1 - \mu_2 ) }{ \sqrt{ \frac{s_1^2 }{n} + \frac{s_2^2 }{ n} } }$

=>    $z = \frac{ ( 11.4 - 9.9) - 0 }{ \sqrt{ \frac{2.5^2 }{36} + \frac{ 3^2 }{36 } } }$

=>     $z = 2.3$

From the z table  the area under the normal curve to the left corresponding to  2.3 is  

       $P( Z > 2.3 ) = 0.010724$

Generally the p-value is mathematically represented as

      $p-value = 2 * P( Z > 2.3 )$

=>    $p-value = 2 * 0.010724$

=>    $p-value = 0.02$

From the value 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 there is a difference between the performances of these two models

Considering question b

From the question we are told the confidence level is  95% , hence the level of significance is    

      $\alpha = (100 - 95 ) \%$

=>   $\alpha = 0.05$

Generally from the normal distribution table the critical value  of  $\frac{\alpha }{2}$ is  

   $Z_{\frac{\alpha }{2} } = 1.96$

Generally the margin of error is mathematically represented as  

      $E = Z_{\frac{\alpha }{2} } * \sqrt{ \frac{s_1^2 }{n } + \frac{s_2^2}{n}}$

 => $E = 1.96 * \sqrt{ \frac{2.5^2 }{ 36 } + \frac{ 3^2}{36}}$

  => $E = 1.276$

Generally 95% confidence interval is mathematically represented as  

      $( \= x_1 - \= x_2) -E < \mu_1 - \mu_2 < ( \= x_1 - \= x_2) + E$

=>  $( 11.4 - 9.9 ) -1.276 < \mu_1 - \mu_2 < ( 11.4 - 9.9 ) + 1.276$

=>  $0.224 < \mu_1 - \mu_2 < 2.776$