Chapter 8 Review

Which of the following sources of error is included in the margin of error?

A. Voluntary response
B. undercoverage
C. some of the subjects did not understand the question
D. chance variation in choosing a random sample
E. all of the above

D. chance variation in choosing a random sample

margin of error = (critical value)(standard error)
Since the standard error is included in the margin of error, this is true.
Remember, the standard error describes how close the sample statistic will be, on average, to the population parameter.

A 95% confidence interval for the mean (mu) of a population is computed from a random random sample and found to be 9 + or – 3. We may conclude that

A. there is a 95% probability that (mu) is between 6 and 12.
B. 95% of values samples are between 6 and 12.
C. If we took many, many additional random samples and from each computed a 95% confidence interval for (mu), approximately 95% of these intervals would contain (mu).
D. there is a 95% probability that the true mean is 9 and a 95% chance that the true margin of error is 3.
E. all of the above.

C. If we took many, many additional random samples and from each computed a 95% confidence interval for µ, approximately 95% of these intervals would contain µ.
What is the critical value t* that satisfies the condition that the t distribution with 7 degrees of freedom has probability 5% to the right of t*?
invT(.95, 7) = 1.89

This represents a t-distribution curve with 5% in the tail to the right of the area.

So really this would be a 90% confidence level because there would be 5% on each tail!

Other things being equal, the margin of error of a confidence interval increases as

A. the sample size increases
B. the population standard deviation increases
C. the sample mean increases
D. the confidence level decreases
E. none of the above

B. the population standard deviation increases

margin of error = (critical value)(standard error)

So for estimating a mean and knowing the population standard deviation, the formula is

ME = z*(σ)
σ being the population standard deviation

If σ increases, ME will increase as well.

An SRS of 35 students at a BPHS were asked “Do you like to go to the beach? Twenty of the students responded “yes.” If p-hat = the proportion of students who answered “yes,” what is the standard error of p-hat?

A. 0.084
B. 0.057
C. 0.043
D. 0.005
E. The standard error cannot be calculated unless we know the standard deviation.

A. 0.084

p-hat = 20/35 = .57
SE = √((.57)(.43)/35)

A marketing consultant wants to estimate the proportion of people who buy cigarettes at a certain store. He stands in front of a liquor store one afternoon and uses a random number generator to determine if he should ask them if they bought cigarettes. He randomly generates a number between 1 and ten and if it is odd, he asks the person the question. If it’s even, he does not ask them. After about three hours, he has 85 responses, 12 of whom made a purchase. Which condition for constructing a confidence interval for a proportion has the consultant failed to satisfy?

A. n(p-hat) greater than equal to 10
B. n(1 – (p-hat)) greater than or equal to 10
C. n is greater than or equal to 30
D. The data is a random sample from the population of interest
E. The sample is less than 10% of the population.

D. The data is a random sample from the population of interest

There is no way this is random. He’s there for only on afternoon and only for 3 hours. The sample would be biased.

The critical value used to construct a confidence interval for a proportion depends upon
A. The confidence level being used and the size of the sample.
B. The size of the sample and the sample proportion
C. The confidence level, the size of the sample, and the standard deviation of the population
D. Only on the sample size
E. Only on the confidence level
E. Only on the confidence level

Since it’s for a proportion, we use z* and z* is calculated from a normal distribution and only use areas under the curve and not degrees of freedom like means do.

An agricultural researcher plants 25 plots with a new variety of corn. A 90% confidence interval for the average yield for these plots is found to be (158.25, 167.19) bushels per acre. What is the statistic and the margin of error for this example?
Statistic = (158.25+167.19)/2
= 162.72

Margin of Error = 167.19 – 162.72
= 4.47

An agricultural researcher plants 25 plots with a new variety of corn. A 90% confidence interval for the average yield for these plots is found to be (158.25, 167.19) bushels per acre. Which of the following would produce a confidence interval with a smaller margin of error than this one?

A. Using a 95% confidence level
B. Reducing bias in the study design
C. Planting 100 plots, rather than 25
D. Using 25 control plots with an old variety of corn
E. None of the above

C. Planting 100 plots, rather than 25

Margin of Error formula for x-bar is

(critical value)(standard error)

However, if we look at just the standard error = Sx/(√(n)), we can see that as n gets bigger, the standard error get smaller.

You are told that the proportion p of those who answered “yes” to a poll about internet use is p = 0.70, and the standard error SE of the proportion is 0.0459. What is the sample size?
The sample size is 100.

Since the standard error for a population is
SE = √((p-hat)(1-p-hat)/n),
we just solve for n.
0.0459 = √((.70)(.30)/n)
n = (.21)/(0.0459)²
n = 99.68

Some scientists believe that a new drug would benefit about 1/4 of all people with a certain blood disorder. To estimate the proportion of patients who would benefit from taking the drug, the scientists will administer it to a random sample of patients who have the blood disorder. What sample size is needed so that the 90% confidence interval will have a margin of error of no more than 2%?
n = 1261

We use the margin of error formula:
z*√((p-hat)(1 – p-hat)/n) = ME
1.64√((.25)(.75)/n) = .02
n = (1.64/.02)²(.1875)
n = 1260.75

To assess the accuracy of a laboratory scale, a standard weight that is known to weigh 2 grams is repeatedly weighed a total of n times and the mean of the weighing is computed. Suppose the scale readings are Normally distributed with unknown mean x-bar and standard deviation σ = 0.03 g. How large should n be so that a 95% confidence interval for µ has a margin of error of ± 0.0004?
n ≥ 21,609

We use the margin of error formula:
z*(σ/√n) ≤ ME
1.96(.03/√n) ≤ .0004 ((1.96)(.03)/(.0004))² ≤ n
21,609 ≤ n