Question

Use technology and the given confidence level and sample data to find the confidence interval for the population mean u. Assume that the population does not exhibit a normal distribution. 95% confidence Weight lost on a diet **3.0 kg n=51 s5.4 kg nd Master- What is the confidence interval for the population mean u? Okg u kg (Round to one decimal place as needed.)

Answer 1

Solution:

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

Our aim is to construct 95% confidence interval.

$\therefore$ c = 0.95

$\therefore$$\alpha$ = 1- c = 1- 0.95 = 0.05

$\therefore$  $\alpha$/2 = 0.05 $\slash$ 2 = 0.025

Also, d.f = n - 1 = 51 - 1 = 50

$\therefore$  
ta/2.0.f.
  =  
ta/2,1-1
  =  $url=https://latex.codecogs.com/gif.latex?t&t=1576272452&ymreqid=50870fb5-0ab3-84d7-1c68-64003101e900&sig=mSHboqsQxXCG3QTQXH2JZg--~C$_0.025,6 = 2.009

( 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=1576272452&ymreqid=50870fb5-0ab3-84d7-1c68-64003101e900&sig=mSHboqsQxXCG3QTQXH2JZg--~C$$\alpha$/2,d.f. * ($url=https://latex.codecogs.com/gif.latex?s&t=1576272452&ymreqid=50870fb5-0ab3-84d7-1c68-64003101e900&sig=m2gJe0wbNFXvGYcwcm2WYw--~C$ / $\sqrt{}$ n )

= 2.009 * (5.4 / $\sqrt{}$ 51 )

= 1.5

Now , confidence interval for mean($\mu$) is given by:

($url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1576272452&ymreqid=50870fb5-0ab3-84d7-1c68-64003101e900&sig=_9jD8nSDFMvG2f0w5We8rg--~C$ - E ) <  $\mu$ <  ($url=http://latex.codecogs.com/gif.latex?%5Cbar%20x&t=1576272452&ymreqid=50870fb5-0ab3-84d7-1c68-64003101e900&sig=_9jD8nSDFMvG2f0w5We8rg--~C$ + E)

(3.0 - 1.5)   <  $\mu$ <  (3.0 + 1.5)

1.5 <  $\mu$ < 4.5

Answer :  1.5 <  $\mu$ < 4.5

Is the confidence interval affected by the fact that the data appear to be from a population that is not normally distributed?

Answer : No . because the sample size is large enough