Question

Listed below are the amounts of mercury? (in parts per? million, or? ppm) found in tuna sushi sampled at different stores. The sample mean is 0.967 ppm and the sample standard deviation is 0.356 ppm. Use technology to construct a 90?% confidence interval estimate of the mean amount of mercury in the population. 0.91 0.61 1.49 1.02 0.64 1.39 0.71 What is the confidence interval estimate of the mean amount of mercury in the? population? nothing ppmless thanmuless than nothing ppm ?(Round to three decimal places as? needed.)

Answer 1

here the sample mean ( ) = 0.967 and sample standard deviation ( s ) = 0.356

the population standard deviation are not known hence use t statistic

90% confidence interval formuis :

= ( - t (_{alpha / 2 ,n -1 )} * ( s / ) , + t (_{alpha / 2 ,n -1 )} * ( s / ) )

use excel to calculate t value , the excel formula is ' = tinv ( 0.1, 6 ) ' and enter . the answer is  1.94

here alpha = 0.1 , n = sample size = 7 , degree of freedom = n-1 = 7-1 = 6 hence t value = 1.94

= ( 0.967 - (1.94) * ( 0.356 / 2.6457) ,  0.967 + (1.94) * ( 0.356 / 2.6457) )

=( 0.967 - 0.2610 ,   0.967 + 0.2610 )

=( 0.706 , 1.228 )

90% confidence interval estimate of the mean amount of mercury in the population is :

( 0.706 , 1.228 )