The 2008 General Social Survey asked, “What do you think is the ideal number of
ID: 3359222 • Letter: T
Question
The 2008 General Social Survey asked, “What do you think is the ideal number of children for a ... The 2008 General Social Survey asked, “What do you think is the ideal number of children for a family to have?” The 678 females who responded had a median of 2, mean of 3.22, and standard deviation of 1.99. Determine whether the following statements are true or false, and explain your reasoning. (a) From the data summary, the distribution of the sample is not normal. (b) This confidence interval is not valid since the distribution of the sample is not normal (c) We are 99% confident that the average of the ideal number of children in a family answered by the 678 females in the sample is between 3.02 and 3.42. (d) 99% the women in the samples thought that the ideal number of children is between 3.02 and 3.42. (e) If a new sample was taken, we are 99% confident that the sample mean of the new sample would lie between 3.02 and 3.42. (f) The margin of error of the 99% confidence interval (3.02, 3.42) is 0.2. (g) In order to decrease the margin of error of a 99% confidence interval by half, we would need to use a sample twice as large. (h) A 95% confidence interval would be narrower than the 99% confidence interval since we don’t need to be as sure about our estimate.
Explanation / Answer
a) From the data summary, the distribution of the sample is not normal. ;
yes not normal since mean is not equal median
TRUE
(B)
This confidence interval is not valid since the distribution of the sample is not normal
We construct confidence interval it he distribution is normal
and given sample is SRS
TRUE
Solutionc:
l (c) We are 99% confident that the average of the ideal number of children in a family answered by the 678 females in the sample is between 3.02 and 3.42.
let us construct 99% CI
mean=3.22
sd=1.99
n=678
z crit for 99%=2,576
99% CI for the mean is
sampl mean-margin of error and sample mean +margin of error
margin of error=zcrit*samples(n)
=2.576*1.99/sqrt(678)
= 0.1968721
=0.2
99% CI is
sampl mean-margin of error and sample mean +margin of error
3.22-0.2 and 3.22+0.2
3.02 and 3.42
TRUE
If a new sample was taken, we are 99% confident that the sample mean of the new sample would lie between 3.02 and 3.42
FALSE
SINCE CONFIDENCE INTERVAL GIVEN FOR POPULATION MEAN NOT SAMPLE MEAN.
99% CI means
we takerandom samples and in 99% ot times we get the population mean will be in between 3.02 and 3.42
(f) The margin of error of the 99% confidence interval (3.02, 3.42) is 0.2.
TRUE
(g)
n order to decrease the margin of error of a 99% confidence interval by half, we would need to use a sample twice as large
TRUE
needed margin of error=0.1
doube sample szie=n+n=678+678=1356
margin of error=z crit*samplesd/sqrt(n)
=2.576*1.99/sqrt(1356)
=0.139
=0.1
TRUE
A 95% confidence interval would be narrower than the 99% confidence interval since we don’t need to be as sure about our estimate.
TRUE
as llevel of significance incraeses ,Z crit incraeses
incraesing confidence interval range
99% will be wider than 95%
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