Question

A randomized trial tested the effectiveness of diets on adults. Among 36 subjects using Diet 1, the mean weight loss after a year was 3.5 pounds with a standard deviation of 5.9 pounds. Among 36 subjects using Diet 2, the mean weight loss after a year was 0.6 pounds with a standard deviation of 4.4 pounds. Construct a 95% confidence interval estimate of the difference between the population means, assuming the population standard deviations are equal.

Answer 1

The 95% confidence interval is

`0.45 <  \mu_1 - \mu_2  < 5.35`

Explanation:

From the question we are told that

The first sample size is
`n_1 = 36`

The first sample mean is
`\= x_1 = 3.5`

The first standard deviation is
`\sigma_1 = 5.9 \ pounds`

The second sample size is
`n_2 = 36`

The second sample mean is
`\= x_2 = 0.6`

The second standard deviation is
`\sigma = 4.4`

Generally the degree of freedom is mathematically represented as

`df = ( [ (s_1^2 )/(n_1 ) + (s_2^2 )/(n_2) ]^2 )/( (1)/((n_1 - 1 )) [ (s_1^2)/(n_1) ]^2 + (1)/((n_2 - 1 )) [ (s_2^2)/(n_2) ]^2 )`

=>
`df = ( [ (5.9^2 )/(34 ) + (4.4^2 )/(34) ]^2 )/( (1)/((34 - 1 )) [ (5.9^2)/(34) ]^2 + (1)/((34- 1 )) [ (4.4^2)/( 34) ]^2 )`

=>
`df =63`

Generally the standard error is mathematically represented as

`SE = \sqrt{ (s_1 ^2 )/(n_1) + (s_2^2 )/( n_2 ) }`

=>
`SE = \sqrt{ ( 5.9 ^2 )/( 36 ) + ( 4.4^2 )/(36) }`

=>
`SE = 1.227`

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 t distribution table the critical value of at a degree of freedom of is

`t_{(\alpha )/(2), 63  } =  1.998`

Generally the margin of error is mathematically represented as

`E = t_{(\alpha )/(2), 63 } * SE`

=>
`E = 1.998 * 1.227`

=>
`E = 2.45`

Generally 95% confidence interval is mathematically represented as

`(\= x_1 - \x_2) -E <  \mu <(\= x_1 - \x_2)  + E`

=>
`(3.5  - 0.6) - 2.45 <  \mu_1 - \mu_2  < (  3.5  - 0.6)  + 2.45`

=>
`0.45 <  \mu_1 - \mu_2  < 5.35`