The Nielsen Company reported that U.S. residents aged 18 to 24 years spend an av

Question

The Nielsen Company reported that U.S. residents aged 18 to 24 years spend an average of 41.9 hours per month using the Internet on a computer. You think this is quite low compared with the amount of time that students at your university spend using the Internet on a computer, and you decide to do a survey to verify this. You collect an SRS of n = 50 students and obtain x^bar = 40.2 hours with s = 28.3 hours a) Report the 95% confidence interval for mu, the average number of hours per month that students at your university use the Internet on a computer. Use software to calculation. 95% confidence interval (plusminus 0.001) is from hours to hours (b) Use this interval to test whether the average time for students at your university is different from the average reported by Nielsen. Use the 5% significance level. We reject the average reported by Nielsen We cannot reject the average reported by Nielsen

Explanation / Answer

PART A.
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x)=40.2
Standard deviation( sd )=28.3
Sample Size(n)=50
Confidence Interval = [ 40.2 ± t a/2 ( 28.3/ Sqrt ( 50) ) ]
= [ 40.2 - 2.01 * (4.002) , 40.2 + 2.01 * (4.002) ]
= [ 32.156,48.244 ]

PART B.
Given that,
population mean(u)=41.9
sample mean, x =40.2
standard deviation, s =28.3
number (n)=50
null, Ho: =41.9
alternate, H1: !=41.9
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.01
since our test is two-tailed
reject Ho, if to < -2.01 OR if to > 2.01
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =40.2-41.9/(28.3/sqrt(50))
to =-0.425
| to | =0.425
critical value
the value of |t | with n-1 = 49 d.f is 2.01
we got |to| =0.425 & | t | =2.01
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.4248 ) = 0.6729
hence value of p0.05 < 0.6729,here we do not reject Ho
ANSWERS
---------------
null, Ho: =41.9
alternate, H1: !=41.9
test statistic: -0.425
critical value: -2.01 , 2.01
decision: do not reject Ho
p-value: 0.6729
we don't have evidence that it is diffrent

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