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:

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  [tex] 0.224   <  \mu_1 - \mu_2  < 2.776  [/tex]

Step-by-step explanation:

From the question we are told that

    The sample size is  n  =  36

    The first sample mean is  [tex]\= x_1 = 11.4[/tex]

    The first standard deviation is  [tex]s_1 = 2.5[/tex]

    The second sample mean is   [tex]\= x_2 = 9.9[/tex]

     The second standard deviation is  [tex]s_2 = 3.0[/tex]

      The level of significance is  [tex]\alpha = 0.05[/tex]

The null hypothesis is  [tex]H_o : \mu_1 - \mu_2 = 0[/tex]

The alternative hypothesis is [tex]H_a : \mu_1 - \mu_2 \ne 0[/tex]

Generally the test statistics is mathematically represented as

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

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

=>     [tex]z = 2.3[/tex]

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

       [tex]P( Z > 2.3 ) = 0.010724[/tex]

Generally the p-value is mathematically represented as

      [tex]p-value = 2 * P( Z > 2.3 )[/tex]

=>    [tex]p-value = 2 * 0.010724[/tex]

=>    [tex]p-value = 0.02[/tex]

From the value obtained we see that  [tex]p-value < \alpha[/tex] 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    

      [tex]\alpha = (100 - 95 ) \%[/tex]

=>   [tex]\alpha = 0.05[/tex]

Generally from the normal distribution table the critical value  of  [tex]\frac{\alpha }{2}[/tex] is  

   [tex]Z_{\frac{\alpha }{2} } =  1.96[/tex]

Generally the margin of error is mathematically represented as  

      [tex]E = Z_{\frac{\alpha }{2} } *  \sqrt{ \frac{s_1^2 }{n } + \frac{s_2^2}{n}} [/tex]

 => [tex]E = 1.96  *    \sqrt{ \frac{2.5^2 }{ 36 } + \frac{ 3^2}{36}}  [/tex]

  => [tex]E = 1.276   [/tex]

Generally 95% confidence interval is mathematically represented as  

      [tex]( \= x_1 - \= x_2) -E <  \mu_1 - \mu_2  < ( \= x_1 - \= x_2) + E [/tex]

=>  [tex]( 11.4 - 9.9 ) -1.276  <  \mu_1 - \mu_2 < ( 11.4 - 9.9 ) + 1.276   [/tex]

=>  [tex] 0.224   <  \mu_1 - \mu_2  < 2.776  [/tex]

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