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90%, 95% and 99% confidence intervals using normal distribution

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Mimi was the 5th seed in 2014 UMUC Tennis open that took place in August. In this tournament, she won 80 of her 100 serving games. Based on UMUC sports Network, she wins 75% of ;the serving games in her 5 year tennis career.

1. Find a 90 % confidence interval estimate of the proportion of serving games Mimi won. (Show work and round the answer to three decimal places).

2. The SAT scores are normally distributed. A simple random sample of 225 SAT scores has a sample mean of 1500 and a sample standard deviation of 300.

(a) What distribution will you use to determine the critical value? Why?

(b) Construct a 95% confidence interval estimate of the mean SAT score. (Show work and round the answer to tow decimal places).

(c ) Is a 99% confidence interval estimate of the mean SAT score wider than 95% confidence interval estimate you got from part (b)? Why? [You don't have to construct the 99% confidence interval]

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1. Find a 90 % confidence interval estimate of the proportion of serving games Mimi won. (show work and round the answer to three decimal places)
At 90% level, critical value is 1.645 from standard normal table.
So a 90% confidence interval estimate of the proportion of serving games Mimi won = [0.80-1.645*sqrt[0.75*0.25/100], 0.80+1.645*sqrt[0.75*0.25/100]]=[0.729, 0.871]

2. ...

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The solution gives detailed steps on determining the 90%, 95% and 99% confidence intervals using normal distribution in real example.

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See Also This Related BrainMass Solution

10 Problems on Descriptive Statistics, Point Estimates and Confidence Interval

1. Use the confidence interval to find the estimated margin error. Then find the sample mean.
A biologist reports a confidence interval of (1.8, 3.0) when estimating the mean height (in centimeters) of a sample of seedlings.

The estimated margin of error is______.

The sample mean is _________.

2. You work for a consumer advocate agency and want to find the mean repair cost of a washing machine. As part of your study, you randomly select 60 repair costs and find the mean to be $122.00. The sample standard deviation is $18.10. Complete parts (a) and (b).

(a) Construct a 95% confidence interval for the population mean repair cost.
The 95% confidence interval is (_____,________). ( Round to two decimal places as needed.)
(b) Change the sample size to n=120. Construct a 95% confidence interval for the population mean repair cost.
The 95% confidence interval is (______,______). (Round to two decimal places as needed.)

Which confidence interval is wider? Explain. Choose the correct answer below.

( ) The n=60 confidence interval is wider because a smaller sample is taken, giving less information about the population.
( ) The n=120 confidence interval is wider because a larger sample is taken, giving more information about the population.
( ) The two intervals are the same size because the confidence interval is based on the level of confidence and sample standard deviation.

3. A machine cuts plastic into sheets that are 30 feet (360 inches) long. Assume that the population of lengths is normally distributed. Complete parts (a) and (b).

(a) The company wants to estimate the mean length the machine is cutting the plastic within 0.25 inch. Determine the minimum sample size required to construct a 95% confidence interval for the population mean. Assume the population standard deviation is 0.5 inch.
n=_______. (Round up to the nearest whole number as needed.)

(b) Repeat part (a) using an error tolerance of 0.125 inch.
n= ________. (Round up to the nearest whole number as needed.)

Which error tolerance requires a larger sample size? Explain.

( ) The tolerance E=0.125 inch requires a larger sample size. As error size decreases, a larger sample must be taken to ensure the desired accuracy.
( ) The tolerance E= 0.125 inch requires a larger sample size. As error size increases, a larger sample must be taken to ensure the desired accuracy.
( ) The tolerance E=0.25 inch requires a larger sample size. As error size increases, a larger sample must be taken to ensure the desired accuracy.
( ) The tolerance E= 0.25 inch requires a larger sample size. As error decreases, a larger sample must be taken to ensure the desired accuracy.

4. The grade point averages (GPA) for 12 randomly selected college students are
2. 2 3.3 2.7
1.7 0.6 4.0
2.1 1.4 3.8
0.4 2.2 3.4

Assume the population is normally distributed.
(a) Find the sample mean.

Type equation here.
Sample mean=_________(Round to two decimal places as needed.)

(b) Find the sample deviation.
s=__________. (Round to two decimal places as needed.)

(c ) Construct a 90% confidence interval for the population mean.
A 90% confidence interval for the population mean is (______,______). (Round to two decimal places as needed.

5. In a random sample of 36 bolts, the mean length was 2.14 inches and the standard deviation was 0.08 inch. Use a normal distribution or a t-distribution to construct the 90% confidence interval for the mean.
Which distribution should be used to construct the 90% confidence interval?

( ) Use a normal distribution because the lengths are normally distributed and the standard deviation is known.
( ) Use a normal distribution because n > 30.
( ) Use a t-distribution because n > 30.
( ) Use a t-distribution because the lengths are normally distributed and the standard deviation is known.
The 905% confidence interval is (______,________). ( Round to two decimal places as needed.)

6. Let p be the population proportion for the following condition. Find the point estimates for p and q.

A study of 4436 adults from country A found that 2545 were obese or overweight.

The point estimate for p, ^p, is _______. (Round to three decimal places as needed.)

The point estimate for q, ^q, is _______. (Round to three decimal
Places as needed.)

7. The table below shows the results of a survey in which 400 adults from the East, 400 adults from the South, 400 adults from the Midwest, and 400 adults from the West were asked if traffic congestion is a serious problem. Complete parts (a) and (b).

Adults who say traffic
Congestion is a serious problem
East 37%
South 32 %
Midwest 25 %
West 57 %

(a) Construct a 99% confidence interval for proportion of adults from the Midwest who say traffic congestion is a serious problem.
The 99% confidence interval for the population of adults from the Midwest who say traffic congestion is a serious problem is (______,_______). ( Round to three decimal places as needed).

(b) Construct a 99% confidence interval for the proportion of adults from the West who say traffic congestion is a serious problem. Is it possible that these two proportions are equal? Explain your reasoning.

The 99% confidence interval for the proportion of adults from the West who say traffic congestion is a serious problem is (_______,________).
(Round to three decimal places as needed.)

Is it possible that these two proportions are equal?
( ) Yes, because 99% confidence interval for the Midwest overlaps with 99% confidence interval for the west.

( ) No, because the 99% confidence interval for the Midwest does not overlap with the 99% confidence interval for the west.

8. A researcher wishes to estimate, with 95% confidence, the percentage of adults who support abolishing the penny. His estimate must be accurate within 4% of the true proportion.

(a) Find the minimum sample size needed, using a prior study th(a)at found that 34% of respondents said they support abolishing the penny?

n=______(Round up to the nearest whole number as needed.)

(b) No preliminary estimate is available. Find the minimum sample size needed.

n= ______(Round up to the nearest whole number as needed.)

9. Find the critical values x²L and X²R for the given confidences level c and sample size n.
C=0.99 , n=20
X²L = __ (Round to three decimal places as needed.)
X²R- __ (Round to three decimal places as needed.)

10. You randomly select and measure the contents of 10 bottles of cough syrup. The results (in fluid ounces) are shown below
4.215 4.297 4.252 4. 244 4.189
4.274 4. 263 4.427 4.221 4.231
Assume the sample is taken from a normally distributed population. Construct 80% confidence intervals for (a) the population variance and (b) the population standard deviation.

(a) The confidence interval for the population variance is (______,_______). (Round to six decimal places as needed.)
(b) The confidence interval for the population standard deviation is (____,____). (Round to four decimal places as needed.).

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