Review Problems for Quiz 4 with Solutions

Why do we generally expect some error when estimating a parameter (such
as a population mean) by a statistic (such as a sample mean)?
Sampling variability (error)

Explain why increasing the sample size results in a tendency for smaller
sampling error when using a sample mean to estimate a population mean.
The standard deviation of the mean will decrease as the sample size
increases.

Define sampling error (also called sampling variability).
Sampling error is the variability that occurs because the value
of a sample statistic will vary from sample to sample.

What exactly is a confidence interval and why is it better to report a
confidence interval instead of a single number for estimating the population
mean, ??
A confidence interval is simply an estimate +/ a margin of error.
It is better to report a confidence interval because it allows us to include
some estimate of the sampling variability present when estimating a population
parameter by a sample statistic.

Suppose that we have obtained data by taking a simple random sample from
a population. What should we do (e.g., what assumptions should we verify)
before we construct a confidence interval from the sample?
Verify that the sample mean has a normal distribution.

Suppose that we have obtained data by taking a
simple random sample from a population and we intend to find a confidence
interval for the population mean. We will either use a 95% confidence interval
or a 99% confidence interval. Which confidence level will give us a narrower
interval?
The 95% confidence interval will give a narrower interval for estimating
the mean.

The Gallup Organization conducts annual national surveys on home gardening.
Results are published by the national Association for Gardening. A random
sample is taken of 25 households with vegetable gardens. The size of vegetable
gardens is normally distributed. The mean size of the vegetable gardens
from the sample was 643 sq ft.

Find a 90% confidence interval for the mean size of all household vegetable
gardens in the United States. Assume that
sigma=247 sq ft.
xbar +/ z^{*}(sigma / sqrt(n))
643 +/ (1.645)(247 / sqrt(25))
643 +/ 81.263

Explain in words what the confidence interval from part (a) means.
We are 90% confident that the true average size of vegetable gardens
in the U.S. is between 561.74 and 724.26 square feet.

A qualitycontrol engineer in a bakery goods
plant needs to estimate the mean weight of bags of potato chips that are
packed by a machine. He knows from experience that sigma=0.1 oz
for this machine. Weights of bags are normally distributed. A random
sample of 12 bags has a mean weight of 16.01 oz.

Find a 99% confidence interval for the mean weight bags of potato chips.
xbar +/ z^{*}(sigma / sqrt(n))
16.01 +/ (2.576)(0.1 / sqrt(12))
16.01 +/ 0.074

Explain in words what the confidence interval from part (a) means.
We are 99% confident that the true mean weight of the bags of potato
chips is between 15.94 and 16.08 ounces.

Suppose that we are considering a hypothesis test for a population mean.
In each part below, express the alternative hypothesis symbolically and
give the formula for the pvalue that would be used for the test:

We want to decide whether the population mean
is different from a specified value m_{o}
.
H_{a}: m ¹m_{o}
pvalue = 2*P(Z ³ z)

We want to decide whether the population mean
is less than a specified value m_{o}
H_{a}: m < m_{o}
pvalue = P(Z £ z)

We want to decide whether the population mean
is greater than a specified value m_{o}
.
H_{a}: m > m_{o}
pvalue = P(Z ³ z)

The Radio Advertising Bureau of New York reports in Radio Facts that in
1994 the mean number of radios per U.S. household was 5.6. A random sample
of 45 U.S. households taken this year showed that the average number of
radios owned is 5.9. Do the data provide sufficient evidence to conclude
that this year’s mean number of radios per U.S. household has changed from
the 1994 mean of 5.6? Assume that the standard deviation of this year’s
number of radios per U.S. household is 1.9. Use the following steps to
answer the question.

State the null and alternative hypotheses.
H_{o}: m = 5.6
H_{a}: m ¹
5.6

Discuss the logic of conducting the hypothesis test (e.g., how will you
determine whether you have enough evidence to reject the null hypothesis).
We will calculate the test statistic and then calculate the pvalue.
The pvalue will tell us whether or not the test statistic is extreme enough
for us to reject the null hypothesis.

Identify the distribution of the variable xbar; that is, the sampling
distribution of the mean for samples of size 45.
Since n>30, the Central Limit Theorem (CLT) tells us that the sampling
distribution of the mean is NORMAL.

Obtain a precise criterion for deciding whether to reject the null hypothesis
in favor of the alternative hypothesis (e.g., pick out a significance level
that you will use for conducting the test. This can be any value that you
would like to use, but most of the time, a 5% significance level is used).
We will use 0.05 to conduct the test.

Calculate your pvalue.
To do this, we must first calculate the test statistic:
z = (xbar  m_{o} ) / {s
/ sqrt(n)} = (5.9  5.6)/{1.9/sqrt(45)} = 1.07
pvalue = 2*P(Z ³ 1.07) = 2{1
– P(Z £ 1.07)} = 2*(1 – 0.8577) = 0.2846

Apply the criterion in part (d) to the problem and state your conclusion.
Since the pvalue (0.2846) is greater than a
(0.05), fail to reject the null hypothesis and conclude that the average
number of radios per household has not changed since 1994.