In: Math
A survey of 1057 parents who have a child under the age of 18 living at home asked about their opinions regarding violent video games. A report describing the results of the survey stated that 89 % of parents say that violence in today's video games is a problem. (a) What number of survey respondents reported that they thought that violence in today’s video games is a problem? You will need to round your answer. X = (b) Find a 95% confidence interval ( ±0.001) for the proportion of parents who think that violence in today's video games is a problem. 95% confidence interval is from to (c) Convert the estimate of your confidence interval to percents ( ±0.1) % to %
n = total number of parents in a survey = 1057
p = proportion of the parents who say that violence in today's video games is a problem = 0.89
(a) What number of survey respondents reported that they thought that violence in today’s video games is a problem?
X = n*p = 1057*0.89 = 940.73 = 941
Find a 95% confidence interval ( ±0.001) for the proportion of parents who think that violence in today's video games is a problem.
b) Formula of confidence interval for population proportion is
Where E is called margin of error
Formula of E is
......( 1 )
It is given that ; c = confidence level = 0.95
so that level of significance = = 1 - c = 1 - 0.95 = 0.05
this implies that \alpha/2 = 0.05/2 = 0.025
So we want to find Z_{/2} such that
P(Z >Z /2) =
0.025
Therefore ,
P(Z < Z /2) = 1 - 0.025 = 0.975
The general excel command to find critical z value is
"=NORMSDIST(probability)"
Here probability = 0.975
So that critical Z /2 is = "=NORMSINV(0.975)" = 1.96
also n = 1057 and = 0.89
Put this value in equation (1) so we get
So E = 0.019
Lower limit = - E = 0.89 - 0.019 = 0.871
Upper Limit = + E = 0.89 + 0.019 = 0.909
So that 95% confidence interval for population proportion P is (0.871 , 0.909)
c) Convert the estimate of your confidence interval to percents ( ±0.1) % to %
The confidence interval to percents ( ±0.1) 87.1% to 90.9%