QUESTION 2 (a) Suppose that your task is to estimate the mean of a normally dist
ID: 2949612 • Letter: Q
Question
QUESTION 2 (a) Suppose that your task is to estimate the mean of a normally distributed population to within 10 units with 95% confidence and that the population standard deviation is known to be 70. What sample size should be used if you wish to estimate the population mean to within 5 units? (b) Determine the minimum sample size required for estimating the population proportion of number of people who drive to work, to within 0.003, with 90% confidence. As a manufacturer of golfclubs, a major corporation wants to estimate the proportion of golfers who are right-handed. How many golfers must be surveyed if they want to be within 0.02 with a 95% confidence level: (c) assuming that there is no information available that could be used as an estimate of p? assuming that the manufacturer has an estimate of p obtained from a previous study that suggests that 75% of golfers are right-handed? (d)The population proportion of voters in favour of a particular political candidate is being estimated with a confidence interval. A random sample of 55 voters is taken, and 28 are found to be in favour. Find and interpret a 90% confidence interval for the population proportion of people in favour of this political candidate. (e) Arandom sample of 10 university stuents was surveyed to determine the amount of time they spent weekly using a personal computer. The times (in hours) were 13 14 10 12 15 If the times are normally distributed with a standard deviation of 5.2 hours, find and interpret the 90% confidence interval for the mean weekly time spent using a personal computer for all university students (f) Suppose that the amount of time teenagers spend on the Internet is normally distributed with a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample mean is computed as 6.5 hours. Find and interpret the 95% confidence interval estimate of the population mean.Explanation / Answer
a)
Given CI Level = 95%
Margin of Error(ME) = 5
std. dev. = 70
z-value of 95% CI = 1.9600
n = (z*sigma/ME)^2
n = (1.96*70/5)^2
= 753
b)
Given CI Level = 90%
As proportion of number of people who drive to work is not given, we assume p = 0.5
ME = 0.003
z-value of 90% CI = 1.6449
n = (z/ME)^2*p*(1-p)
= (1.6449/0.003)^2*0.5*0.5
= 75154
c)
i)
Given CI Level = 95%
Assume p = 0.5
ME = 0.02
z-value of 95% CI = 1.9600
n = (z/ME)^2*p*(1-p)
= (1.96/0.02)^2*0.5*0.5
= 2401
ii)
Given CI Level = 95%
p = 0.75
ME = 0.02
z-value of 95% CI = 1.9600
n = (z/ME)^2*p*(1-p)
= (1.96/0.02)^2*0.75*0.25
= 1801
d)
p = 28/55 = 0.5091 , n = 55
z value at 90% = 1.645
CI = p +/- z *sqrt(p*(1-p)/n)
= 0.5091 +/- 1.645 *sqrt(0.5091 *(1-0.5091)/55)
= (0.3982, 0.62)
A 90% Confidence interval for the population proportion of people in favour of this political cancdidate is (0.3982,0.62 )
f)
z value at 95% = 1.96
mean = 6.5 , std.dev = 1.5 , n =100
CI = mean +/- z *(s/sqrt(n))
= 6.5 +/- 1.96 *(1.5/sqrt(100))
= (6.206 , 6.794 )
A 95% conficdence interval of the population mean includes (6.206 , 6.794 )