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Math 225N Statistics Final Exam With 100% Correct Answers

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MATH 225N Statistics Final EXAM WITH 100% CORRECT ANSWERS

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FINAL EXAM Week 8 Exam Chamberlain College of
Nursing MATH 225N Statistics Final EXAM WITH
100% CORRECT ANSWERS


A catalog sales company promises to deliver orders placed on the internet within 3
days. Follow-up calls to a few randomly selected customers show that a 95%
confidence interval for the proportion of all orders that arrive on time is 88% +/- 6%.
What does this mean? Is this conclusion correct? Explain.

Between 82% and 94% of all orders arrive on time. - ANSWER Not correct because
this implies certainty.

A catalog sales company promises to deliver orders placed on the internet within 3
days. Follow-up calls to a few randomly selected customers show that a 95%
confidence interval for the proportion of all orders that arrive on time is 88% +/- 6%.
What does this mean? Is this conclusion correct? Explain.

95% of all random samples of customers will show that 88% of orders arrive on time.
- ANSWER Not correct because different samples will give different results.

A catalog sales company promises to deliver orders placed on the internet within 3
days. Follow-up calls to a few randomly selected customers show that a 95%
confidence interval for the proportion of all orders that arrive on time is 88% +/- 6%.
What does this mean? Is this conclusion correct? Explain.

95% of all random samples of customers will show that 82% to 94% of orders arrive
on time. - ANSWER Not correct because this is about the population proportion, not
the sample proportion of different samples.

A catalog sales company promises to deliver orders placed on the internet within 3
days. Follow-up calls to a few randomly selected customers show that a 95%
confidence interval for the proportion of all orders that arrive on time is 88% +/- 6%.
What does this mean? Is this conclusion correct? Explain.

We are 95% sure that between 82% and 94% of the orders placed by the sampled
customers arrive on time. - ANSWER Not correct because we know that 88% of the
orders arrive on time with a margin of error of 6%.

A catalog sales company promises to deliver orders placed on the internet within 3
days. Follow-up calls to a few randomly selected customers show that a 95%
confidence interval for the proportion of all orders that arrive on time is 88% +/- 6%.
What does this mean? Is this conclusion correct? Explain.

,On 95% of the days, between 82% and 94% of the orders will arrive on time. -
ANSWER Not correct because the parameter is talking about the interval, not the
days.

Several factors are involved in the creation of a confidence interval. Among them are
the sample size, the level of confidence, and the margin of error. Is this statement
true?

For a given sample size, higher confidence means smaller margin of error. -
ANSWER False.

Several factors are involved in the creation of a confidence interval. Among them are
the sample size, the level of confidence, and the margin of error. Is this statement
true?

For a specified confidence level, larger samples provide smaller margins of error. -
ANSWER True.

Several factors are involved in the creation of a confidence interval. Among them are
the sample size, the level of confidence, and the margin of error. Is this statement
true?

For a fixed margin of error, larger samples provide greater confidence. - ANSWER
True.

Several factors are involved in the creation of a confidence interval. Among them are
the sample size, the level of confidence, and the margin of error. Is this statement
true?

For a given confidence level, halving the margin of error requires a sample twice as
large. - ANSWER False.

What fraction of cars made in Japan? The computer output below summarizes the
results of a random sample of 50 autos. Explain carefully what it tells you.

z-interval for proportion
With 90.00% confidence
0.29938661< P(Japan) < 0.46984416 - ANSWER On the basis of this sample, we
are 90 confident that the proportion of Japanese cars is between 29.9% and 47.0%.

Vitamin D, whether ingested as a dietary supplement or produced naturally when
sunlight falls on the skin, is essential for strong, healthy bones. The bone disease
rickets was largely eliminated in England during the 1950s, but now there is concern
that a generation of children more likely to watch TV or play computer games than
spend time outdoors is at increased risk. A recent study of 2700 children randomly
selected from all parts of England found 20% of them deficient in Vitamin D.

Find a 98% confidence interval. - ANSWER Z*98= 2.326 (t-table bottom row)
(18.2%, 21.8%)

, Vitamin D, whether ingested as a dietary supplement or produced naturally when
sunlight falls on the skin, is essential for strong, healthy bones. The bone disease
rickets was largely eliminated in England during the 1950s, but now there is concern
that a generation of children more likely to watch TV or play computer games than
spend time outdoors is at increased risk. A recent study of 2700 children randomly
selected from all parts of England found 20% of them deficient in Vitamin D.

What does your interval mean? - ANSWER I am 98% confident that the interval from
18.2% to 21.8% captures the true percentage of English children who are Vitamin D
deficient.

Vitamin D, whether ingested as a dietary supplement or produced naturally when
sunlight falls on the skin, is essential for strong, healthy bones. The bone disease
rickets was largely eliminated in England during the 1950s, but now there is concern
that a generation of children more likely to watch TV or play computer games than
spend time outdoors is at increased risk. A recent study of 2700 children randomly
selected from all parts of England found 20% of them deficient in Vitamin D.

What does the "98% confidence" mean? - ANSWER If we repeat this process over
and over again with a sample size of 2700, we expect that 98% of all intervals to
capture the true population parameter (English children deficient in Vitamin D).

It is believe that as many as 25% of adults over 50 never graduated from high
school. We wish to see if this percentage is the same among the 25 to 30 age group.

How many of this younger age group must we survey in order to estimate the
proportion of non-grads to within 6% with 90% confidence. - ANSWER (p-hat x q-hat
x Z*90^2) / n^2 = .25 x .75 x 1.645^.06^2 = 141

It is believe that as many as 25% of adults over 50 never graduated from high
school. We wish to see if this percentage is the same among the 25 to 30 age group.

Suppose we want to cut the margin of error to 4%. What's the necessary sample
size? - ANSWER (p-hat x q-hat x Z*90^2) / n^2 = .25 x .75 x 1.645^.04^2 =318

It is believe that as many as 25% of adults over 50 never graduated from high
school. We wish to see if this percentage is the same among the 25 to 30 age group.

What sample size would produce a 3% margin of error? - ANSWER (p-hat x q-hat x
Z*90^2) / n^2 = .25 x .75 x 1.645^.03^2 = 564

Formula for z score - ANSWER observation-mean/standard deviation

Statement for a confidence interval - ANSWER "I am C% confident that the interval
from X% to Y% captures the true percentage of who/what that do this/that"

Conditions that need to be met for sampling distribution models and confidence
intervals for PROPORTIONS - ANSWER 1. Independence assumption
2. Randomization condition
3. 10% condition

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