Question

A random sample of 17 brand energy bars of chocolate has, on average, 230 calories per bar. The standard deviation of the calorie content of this energy bar brand is 11.2 calories. Calculate the margin of error for a 98% confidence interval for the true average calorie content of this brand of chocolate energy bars. Suppose the distribution of caloric content is approximately normal.

Important note: Use the critical value rounded to 3 decimal places and your answer to 2 decimal places.

Write the result with up to 4 decimals.

Answer 1

Solution :

Given that,

n = 17

  $url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1596159762&ymreqid=69d777f6-ff7f-1f97-1cee-62000201a800&sig=29kDB7ASptGJw0WBPhx86w--~D$ = 230

s = 11.2  

Note that, Population standard deviation($\sigma$) is unknown..So we use t distribution.

Our aim is to construct 98% confidence interval.   

$\therefore$ c = 0.98

$\therefore$$\alpha$ = 1 - c = 1 - 0.98 = 0.02

$\therefore$  $\alpha$/2 = 0.02 $\slash$ 2 = 0.01

Also, d.f = n - 1 = 17 - 1 = 16

$\therefore$  
ta/2.d.f-
  =  
ta/2.n-1
  =  $url=https://latex.codecogs.com/gif.latex?t&t=1596159762&ymreqid=69d777f6-ff7f-1f97-1cee-62000201a800&sig=eX.DLtNSzxt3lEtTnO_y.g--~D$_0.01,16 = 2.583

( use t table or t calculator to find this value..)

The margin of error is given by

E =  $url=https://latex.codecogs.com/gif.latex?t&t=1596159762&ymreqid=69d777f6-ff7f-1f97-1cee-62000201a800&sig=eX.DLtNSzxt3lEtTnO_y.g--~D$$\alpha$/2,d.f. * ($url=https://latex.codecogs.com/gif.latex?s&t=1596159762&ymreqid=69d777f6-ff7f-1f97-1cee-62000201a800&sig=bVI2NcbH778X0zIQKD99_g--~D$ / $\sqrt{}$ n )

= 2.583* ( 11.2/ $\sqrt{}$ 17 )

= 7.018

= 7.02

Answer :  The margin of error is 7.02