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

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. 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.

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!

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

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.

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.

p-hat = 20/35 = .57

SE = √((.57)(.43)/35)

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.

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

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

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.

= 162.72

Margin of Error = 167.19 – 162.72

= 4.47

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

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.

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

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

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