Question

You test calories for a food item. The brand name has a mean of 158.706 and a sample standard deviation = 25.236, when seventeen are tested. The generic item has a mean of 122.471 and a sample standard deviation = 25.183, when seventeen are tested. Which is a confdence interval of 95%?

Multiple Choice: (Show work)

A) 17.21 to 55.26
B) 18.12 to 54.35
C) 17.79 to 54.67
D) 18.622 to 53.848

Answer 1

option C

Explanation:

given,

`\bar{x_1} = 158.706, \sigma_1 = 25.36 , n_1= 17`

`\bar{x_1} = 122.471, \sigma_1 = 25.183 , n_2= 17`

α = 1 - 0.95 = 0.05

degree of freedom (df) = 17 -1 = 16

critical value
`= t_(\alpha/2),df = t_(0.025),16 =2.120` (from t-table)

margin of error =
`t_(\alpha/2)\sqrt{(s_1^2)/(n_1)+(s_2^2)/(n_2)}`

=2.120\times \sqrt{\dfrac{25.36^2}{17}+\dfrac{25.183^2}{17}}[/tex]

= 2.120 x 8.6467

= 18.33

Margin of error = 18.33

Point estimation of difference =
`\bar{x_1} - \bar{x_2}`

= 36.235

lower limit = 36.235 - 18.33 = 17.91

upper limit = 36.235 + 18.33 = 54.57

hence, the nearest option near to answer is option C