Stastics exam

MATH 105 - Worksheet Week 7 Names:

1. In January 2007 Consumer Reports published their study of bacterial contamination of chicken sold

in the United States. They purchased 525 broiler chickens from various kinds of food stores in 23

states and tested them for types of bacteria that cause food-borne illnesses. Laboratory results

indicated that 86% of these chickens were infected with Campylobacter.

(a) State the parameter we are trying to estimate.

p is the proportion of chickens that are contaminated with bacteria that cause food-borne ill-

nesses.

(b) Is 86% a parameter or a statistic? Explain

86% is a statistic because it comes from the sample.

(c) Calculate a 95% confidence interval for the proportion of contaminated chicken sold in the US.

.86± 1.96× √

.86∗.14 525

= .86± .030 = (.83, .89)

(d) Interpret your confidence interval.

We can be 95% confident that between 84% and 89% of chickens are contaminated.

(e) Suppose it is known that a few years ago the proportion of contaminated chicken was 80%.

Based only on the confidence interval you have calculated, does your sample provide evidence

that the contamination problem is getting worse? Explain your answer.

Yes, the confidence limits are both higher than 80%.

2. Find the margin of error, E, that corresponds to the given statistics and confidence level.

(a) n=500, x=200, 95% confidence

E = 1.96

√ .4× .6 500

= .043

(b) n=1200, x=800, 99% confidence

E = 2.576

√ .667× .333

1200 = .0350

(c)

E = 1.645 ∗ 20√ 70

= 3.932

3. Recall that on the first day survey a question was asked “On a typical day, how many hours do you

spend talking/texting on the phone”? For 70 students the mean was 3.9 hours and the standard

deviation, s, was 3.7 hours. Assume that the population is Chico State University students and that

we are trying to estimate µ, the mean amount of hours students spend talking/texting on the phone

per day.

(a) Compute the margin of error for a 90% confidence interval.

E = 1.671 3.7√ 70

= .739

(b) Calculate a 90% confidence interval for µ.

3.9± .74 = (3.16, 4.64)

(c) Interpret the confidence interval in context of the problem. Make sure you use the word “mean”

or “average” in your interpretation.

We can be 90% confident that Chico State students will talk/text, on average, between 3.16

and 4.64 hours per day.

4. A medical researcher measured the pulse rates (beats per minute) of a sample of 100 randomly

selected adults and found the following 95% confidence interval

70.89< µ <74.50

For each of the following statements, indicate whether it is a correct interpretation of a confidence

interval. If it is not a correct interpretation, identify what is wrong with the statement.

(a) We are 95% confident that the mean pulse rate for the 100 randomly selected adults is between

70.89 and 74.50 beats per minute.

This is incorrect. The confidence interval is for the population mean, not the sample mean.

(b) We are 95% confident that the mean pulse rate for all adults is between 70.89 and 74.50 beats

per minute.

This is a correct interpretation for the confidence interval.

(c) We are 95% confident that all adults will have a pulse rate between 70.89 and 74.50 beats per

minute.

This is incorrect because there is no mention of mean pulse rate.