Log inSign up for FREE
Library

Ch 16 Confidence Intervals, Critical Values & Sample Size Practice

Last updated over 4 years ago
17 questions
4

Confidence Interval Review:
A National Nutrition Study by the Center for Disease Control reported that 649 of 1546 African American women tested had vitamin D deficiency.
What is the sample proportion?
Round to three places after the decimal, be careful.

4

Confidence Interval Review:
A National Nutrition Study by the Center for Disease Control reported that 649 of 1546 African American women tested had vitamin D deficiency.
Estimate the variability in such sample proportions by finding the standard error:

Round to three places after the decimal.

4

Confidence Interval Review:
A National Nutrition Study by the Center for Disease Control reported that 649 of 1546 African American women tested had vitamin D deficiency.
Is it appropriate to use the Normal Model?
Check all answers that apply when checking the independence and sample size condition/assumptions.

4

Confidence Interval Review:
A National Nutrition Study by the Center for Disease Control reported that 649 of 1546 African American women tested had vitamin D deficiency.
Find the Margin of Error for a 95% Confidence Interval:

Round to three places after the decimal.

4

Confidence Interval Review:
A National Nutrition Study by the Center for Disease Control reported that 649 of 1546 African American women tested had vitamin D deficiency.
Give the 95% Confidence Interval for the proportion of African American women that are Vitamin D deficient.

Use your calculator to find the 95% Confidence Interval: Stat, Test, A: 1-PropZInt
Enter x, n and C- Level (Confidence Level)

Round to three places past the decimal.
Use the notation from class: (_____, _____)

4

Confidence Interval Review:
A National Nutrition Study by the Center for Disease Control reported that 649 of 1546 African American women tested had vitamin D deficiency.
Interpret the Margin of Error and the Confidence Interval:

4

The Mars Candy Co. makes M&M's. Suppose that the candies are packaged at random in small bags containing about 50 M&M's. A class of elementary school students learning about percents opens up a bag, counts the various colors of the candies and find that 7 are green.
For this sample, what is the proportion that are green?

4

The Mars Candy Co. makes M&M's. Suppose that the candies are packaged at random in small bags containing about 50 M&M's. An elementary class opens a bag of M&M's and find 7 that are green.
For this sample, what is the Margin of Error for a 95% Confidence Interval for the proportion that are green?
1st: Calculate the Std Error (see #2 for the formula)
2nd: Calculate the Margin of Error (Use the formula from #4 if needed)

4

Calculate the 95% Confidence Interval for the bag of 50 M&M's for the proprotion that are green.
Use your calculator: Stat, Test, A: 1-PropZInt.
Enter x, n, C-Level
Round to 3 places past the decimal.

4

Let's see the difference in Confidence Intervals with a larger size bag.
Suppose the class of elementary school students buys bigger bags of candy, with 500 M&M's each.
Assume the proportion that you calculated stays the same.
Using the new # of trials, you need to calculate:
See #2 & 4 for formulas.

What is the Margin of Error for a 95% Confidence Interval for a bag of 500 M&M's for the proportion that will be green?
Round both SE and ME to 3 places past the decimal.

4

Continue with the M&M's that are packaged at random in bags containing about 500 M&M's.
Use your TI-84 calculator to calculate the 95% Confidence Interval for the proportion of green candies in a bag of 500 M&M's.
Stat, Test, A:1-PropZInt,
x= you will need to calculate the # green M&M's using your sample proportion from #7 of 0.14
n= C-Level=
Round to 2 places past the decimal.

4

By increasing the sample size from 50 to 500, what was the effect on the 95% Confidence Interval?
Compare the two 95% Confidence Intervals from #10 & 12.

4

Continue with the M&M's that are packaged at random in bags containing about 50 M&M's.
Remember, the Margin of Error was calculated as .098.
If you wanted to cut that Margin of Error to 1/4 the size, how big would your sample need to be?
Remember from class:

4

In December of 2011, Consumer Reports published their study of labeling of seafood sold in New York, New Jersey, and Connecticut. They purchased 190 pieces of seafood from various kinds of food stores and restaurants in the three states and genetically compared the pieces sampled to standard gene fragments. Laboratory results indicated that 22% of these packages of seafood were mislabeled, incompletely labeled or misidentified by store or restaurant employees.
Construct a 95% Confidence Interval for the proportion of all seafood packages in those three states that are mislabeled or misidentified.
You will need to: First calculate the # of mislabeled seafood products using the proportion, round to a whole number.
Second calculate the 95% Confidence Interval using your TI-84 Calculator.
Use your calculator: Stat, Test, A: 1-PropZInt.
Enter x, n, C-Level
Round to three places past the decimal.

4

What is the percent chance(how confident are you?) that your 95% Confidence Interval contains the true proportion of mislabeled seafood from the three states?

4

Continue the study of mislabeling of seafood sold in New York, New Jersey, and Connecticut. Originaly 190 pieces of seafood were purchased from various kinds of food stores and restaurants in the three states. Suppose you wanted to create a 95% Confidence Interval with a Margin of Error that was 1/3 the size as that from the original sample of 190.
How big does your sample size need to be to shrink the Margin of Error by 1/3?
See # 13 for a hint.

4

In describing the findings of a 2010 Public Religion Research Institute poll, the researchers reported a 3% margin of error at a 95% confidence level. If instead they had used a 99% confidence level, their margin of error would have been: