Question

(3 points) A news report states that the 99% confidence interval for the mean number of daily calories consumed by participants in a medical study is (1750, 1880). Assume the population distribution for daily calories consumed is normally distributed and that the confidence interval was based on a simple random sample of 20 observations. Calculate the sample mean, the margin of error, and the sample standard deviation based on the stated confidence interval and the given sample size. Use the t distribution in any calculations and round non-integer results to 4 decimal places. 1. What is the sample mean? 1815 2. What is the margin of error? 65 3. What is the sample standard deviation? 101.6039

Answer 1

Solution:

A 99% confidence interval for a population mean is (1750, 1880)

Upper limit = 85

Lower limit = 77

Since population SD is unknown , this interval is constructed using t distribution.

n = 20

c = 0.99

$\therefore$$\alpha$ = 1- c = 1- 0.99 = 0.01

$\therefore$  $\alpha$/2 =  0.005

Also, d.f = n - 1 = 20 - 1 = 19

$\therefore$  
ta/2.0.f.
  =  
ta/2,1-1
  =  $url=https://latex.codecogs.com/gif.latex?t&t=1573429298&ymreqid=50870fb5-0ab3-84d7-1ca9-bb005d01e500&sig=LDzLQzLb_38IgLICVzGsyw--~C$_0.005,19 =  2.861   ..use t table

1) Sample mean = (Upper limit + Lower limit)/2

= (1880 + 1750)/2

=1815

Sample mean = 1815

2)Margin of error = (Upper limit - Lower limit)/2

= (1880 - 1750)/2

= 65

Margin of error = 65

3)Find Sample standard deviation s

We know , margin of error =   $url=https://latex.codecogs.com/gif.latex?t&t=1573429298&ymreqid=50870fb5-0ab3-84d7-1ca9-bb005d01e500&sig=LDzLQzLb_38IgLICVzGsyw--~C$$\alpha$/2,d.f. * ($url=https://latex.codecogs.com/gif.latex?s&t=1573429298&ymreqid=50870fb5-0ab3-84d7-1ca9-bb005d01e500&sig=PbAj1n0jdgG03fbPTVFRzw--~C$ / $\sqrt{}$ n)

Bur from part 2, margin of error = 65

$url=https://latex.codecogs.com/gif.latex?%5Ctherefore&t=1573429298&ymreqid=50870fb5-0ab3-84d7-1ca9-bb005d01e500&sig=OwJjwxMnJQBF7eabbPKOcQ--~C$ 65 = $url=https://latex.codecogs.com/gif.latex?t&t=1573429298&ymreqid=50870fb5-0ab3-84d7-1ca9-bb005d01e500&sig=LDzLQzLb_38IgLICVzGsyw--~C$ $\alpha$/2,d.f. * ($url=https://latex.codecogs.com/gif.latex?s&t=1573429298&ymreqid=50870fb5-0ab3-84d7-1ca9-bb005d01e500&sig=PbAj1n0jdgG03fbPTVFRzw--~C$ / $\sqrt{}$ n)

$url=https://latex.codecogs.com/gif.latex?%5Ctherefore&t=1573429298&ymreqid=50870fb5-0ab3-84d7-1ca9-bb005d01e500&sig=OwJjwxMnJQBF7eabbPKOcQ--~C$ 65 = 2.861 * ($url=https://latex.codecogs.com/gif.latex?s&t=1573429298&ymreqid=50870fb5-0ab3-84d7-1ca9-bb005d01e500&sig=PbAj1n0jdgG03fbPTVFRzw--~C$ / $\sqrt{}$ 20)

$url=https://latex.codecogs.com/gif.latex?%5Ctherefore&t=1573429298&ymreqid=50870fb5-0ab3-84d7-1ca9-bb005d01e500&sig=OwJjwxMnJQBF7eabbPKOcQ--~C$ s = (65 * $\sqrt{}$ 20) /2.861

=  101.6039

Sample standard deviation = 101.6039