Question

For this activity, you will be creating a confidence interval for the average number of hours of TV watched. You will also run a hypothesis test to test the claim that the average number of hours of TV watched is more than 7 hours. Students from a former MAT 152 class asked the question, “How many hours of TV do you watch in a typical week?” Please use the data they collected to answer the following questions.

Data: 15, 11, 11, 12, 6, 4.5, 6, 15.5, 10, 10, 5, 5, 2, 15, 2, 10, 15.5, 4, 5, 12

7.     Test the claim that the mean hours of TV watched is more than 7 using a 5% significance level. Show the steps you use to do this. Show your hypotheses, calculated numbers, and the critical values or p-values that you use to make your decision.

8.    What are your results from the hypothesis test? Do you reject or fail to reject your null hypothesis? Why?

9.     Suppose you knew the standard deviation for all Americans for watching TV was 4 hours. If you wanted your results to be within a margin of error of 0.25 hours, how many people would you have to survey (use a 95% confidence level)?

10.Name at least 2 ways this data could be biased and hence the confidence interval would not be a good estimate of the population mean.

Thank you in advance!

Answer 1

Given that,
population mean(u)=7
sample mean, x =8.825
standard deviation, s =4.439
number (n)=20
null, Ho: μ=7
alternate, H1: μ>7
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.729
since our test is right-tailed
reject Ho, if to > 1.729
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =8.825-7/(4.439/sqrt(20))
to =1.8386
| to | =1.8386
critical value
the value of |t α| with n-1 = 19 d.f is 1.729
we got |to| =1.8386 & | t α | =1.729
make decision
hence value of | to | > | t α| and here we reject Ho
p-value :right tail - Ha : ( p > 1.8386 ) = 0.04083
hence value of p0.05 > 0.04083,here we reject Ho
ANSWERS
---------------
7.
null, Ho: μ=7
alternate, H1: μ>7
test statistic: 1.8386
critical value: 1.729
decision: reject Ho
p-value: 0.04083
8.
we have enough evidence to support the claim that the average number of hours of TV watched is more than 7 hours.
9.
Given data,
margin of error = 0.25
standard deviation =4
confidence level is 95%
Compute Sample Size
n = (Z a/2 * S.D / ME ) ^2
Zα/2 at 0.05% LOS is = 1.96 ( From Standard Normal Table )
Standard Deviation ( S.D) = 4
ME =0.25
n = ( 1.96*4/0.25) ^2
= (7.84/0.25 ) ^2
= 983.45 ~ 984          
10.
Bias is taken to mean interference in the outcomes of research by predetermined ideas, prejudice or influence in a certain direction.
Data can be biased but so can the people who analyse the data. When data is biased, we mean that the sample is not representative of the entire population.
TRADITIONAL METHOD
given that,
sample mean, x =8.825
standard deviation, s =4.439
sample size, n =20
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 4.439/ sqrt ( 20) )
= 0.993
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.093
margin of error = 2.093 * 0.993
= 2.077
III.
CI = x ± margin of error
confidence interval = [ 8.825 ± 2.077 ]
= [ 6.748 , 10.902 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =8.825
standard deviation, s =4.439
sample size, n =20
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 19 d.f is 2.093
we use 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
confidence interval = [ 8.825 ± t a/2 ( 4.439/ Sqrt ( 20) ]
= [ 8.825-(2.093 * 0.993) , 8.825+(2.093 * 0.993) ]
= [ 6.748 , 10.902 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 6.748 , 10.902 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean