1. Do one of the following, as appropriate. (a) Find the critical value Z /2 , (
ID: 3219656 • Letter: 1
Question
1.
Do one of the following, as appropriate. (a) Find the critical value Z/2, (b) find the critical value t/2, (c) state that neither the normal nor the t distribution applies.
Confidence level 99%; n=25; is known; population appears to be very skewed.
2.
Use technology and the given confidence level and sample data to find the confidence interval for the population mean . Assume that the population does not exhibit a normal distribution.
Weight lost on a diet: 90 % confidence; n=41; x=2.0 kg; s=3.8 kg
What is the confidence interval for the population mean ?
? Kg < < ? Kg (Round to one decimal place as needed.)
Is the confidence interval affected by the fact that the data appear to be from a population that is not normally distributed?
A. No, because the population resembles a normal distribution.
B. Yes, because the sample size is not large enough.
C. Yes, because the population does not exhibit a normal distribution.
D. No, because the sample size is large enough.
3.
In order to estimate the mean amount of time computer users spend on the internet each month, how many computer users must be surveyed in order to be 95% confident that your sample mean is within 13 minutes of the population mean? Assume that the standard deviation of the population of monthly time spent on the internet is 212 min. What is a major obstacle to getting a good estimate of the populationmean? Use technology to find the estimated minimum required sample size
The minimum sample size required is____computer users. (Round up to the nearest whole number.)
What is a major obstacle to getting a good estimate of the population mean?
A. There may not be 1,022 computer users to survey.
B. The data does not provide information on what the computer users did while on the internet.
C. It is difficult to precisely measure the amount of time spent on the internet, invalidating some data values.
D. There are no obstacles to getting a good estimate of the population mean.
Explanation / Answer
Solution:
1) (c) state that neither the normal nor the t distribution applies.
2) Given a=1-0.99=0.01, Z(0.01) =2.575 (from standard normal table)
So the confidence interval is
x ± Z*s/n =2 ± 2.575*3.8/sqrt(41)
= (0.4718, 3.5282)
0.4718<?<3.5282
(D) No, because the sample size is large enough.
3) 95% confident
13 = 1.96.(212/n)
n = (1.96(212)/13)^2
= 1021.63
(A) The minimum sample size required is 1022 computer users.
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