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

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`

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 \\e 0`

Generally the test statistics is mathematically represented as

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

=>
`z = \frac{ ( 11.4 - 9.9) - 0 }{ \sqrt{ (2.5^2 )/(36) + ( 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
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.96`

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *  \sqrt{ (s_1^2 )/(n ) + (s_2^2)/(n)}`

=>
`E = 1.96  *    \sqrt{ (2.5^2 )/( 36 ) + ( 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`