Question

Twelve different video games showing substance use were observed and the duration times of game play (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. Use the data to construct a 90% confidence interval estimate of u, the mean duration of game play. 4050 3875 3850 4022 4323 4812 4650 4040 5010 4816 4327 4325 Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. What is the confidence interval estimate of the population mean u? (Round to one decimal place as needed.)

Answer 1

Solution:

x	x2
4050	16402500
3875	15015625
3850	14822500
4022	16176484
4323	18688329
4812	23155344
4650	21622500
4040	16321600
5010	25100100
4816	23193856
4327	18722929
4325	18705625
∑x=52100	∑x2=227927392

Mean ˉx=∑xn

=4050+3875+3850+4022+4323+4812+4650+4040+5010+4816+4327+4325/12

=52100/12

=4341.6667

Sample Standard deviation S=√∑x2-(∑x)2nn-1

=√227927392-(52100)212/11

=√227927392-226200833.3333/11

=√1726558.6667/11

=√156959.8788

=396.1816

Degrees of freedom = df = n - 1 = 12 - 1 = 11

At 90% confidence level the z is ,

$\alpha$ = 1 - 90% = 1 - 0.90 = 0.10

$\alpha$ / 2 = 0.10 / 2 = 0.05

t_{$\alpha$ /2,df} = t_0.05,11=1.796

Margin of error = E = t_{$\alpha$/2,df} * (s /$\sqrt$n)

= 1.796 * (396.18 / $\sqrt$ 12)

= 205.39

Margin of error = 205.39

The 90% confidence interval estimate of the population mean is,

$\bar x$ - E <$\mu$  < $\bar x$ + E

4341.67 - 205.39< $\mu$ < 4341.67 + 205.39

4136.28 < $\mu$ < 4547.06

(4136.3 , 4547.1)