Question 1

If the 95 percent confidence interval for a mean is from .771 to 1.629,can we conclude that μ = .5,using a 95 percent confidence interval?
A) Yes
B) No

Accepted Answer

The confidence interval (.771, 1.629) does not contain the value .5, which means that we cannot conclude that the population mean is equal to .5 with a 95% level of confidence.

Question 2

The weight of a product is measured in pounds.A sample of 50 units is taken from a recent production.The sample yielded = 75 lb,and we know that σ² = 100 lb.Calculate a 95 percent confidence interval for μ. A) [71.25,78.75] B) [72.23,77.77] C) [47.28,102.72] D) [72.67,77.33]

Accepted Answer

The answer of The weight of a product is measured...

Question 3

A local country club golf tournament organizer is attempting to estimate the average number of strokes for the 13th hole.On a particular day,64 players completed the play on the 13th hole,with an average of 4.25 strokes and a population standard deviation of 1.6 strokes.Determine the 95 percent confidence interval for the average number of strokes.
A) [3.94,4.56]
B) [3.86,4.64]
C) [4.05,4.45]
D) [3.92,4.58]

Accepted Answer

4.25 ± (1.96)= (3.86 to 4.64)

Question 4

A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour.An approximation is 20 percent.They want the estimate to be at the 90 percent confidence level and within 2 percent of the actual proportion.What sample size is needed?
A) 22
B) 1692
C) 1537
D) 1083

Accepted Answer

n =≅ 1083

Question 5

In a randomly selected group of 650 automobile deaths,180 were alcohol related.Construct a 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol.
A) [.243,.311]
B) [.262,.292]
C) [.259,.294]
D) [.263,.291]

Accepted Answer

== .2769
.2769 ± 1.96→ .243 to .311

Question 6

A sample of 2,000 people yielded =.52.What is the variance of the population proportion?
A) .0056
B) .0112
C) .00012
D) .0161

Accepted Answer

The answer of A sample of 2,000 people yielded =.52.What...

Question 7

A sample of 2,000 people yielded =.52.Calculate a 90 percent confidence interval for p.
A) [.506,.534]
B) [.468,.572]
C) [.511,.529]
D) [.502,.538]

Accepted Answer

The answer of A sample of 2,000 people yielded =.52.Calculate...

Question 8

A sample of 2,000 people yielded =.52.Calculate a 95 percent confidence interval for p.
A) [.509,.531]
B) [.494,.546]
C) [.498,.542]
D) [.502,.538]

Accepted Answer

The confidence interval for a proportion is calculated using the formula: p̂ ± Z * sqrt[(p̂(1-p̂))/n], where p̂ = 0.52, Z = 1.96 for 95% confidence, and n = 2000. This calculation gives the interval [.498, .542].

Question 9

A random sample n= 15 is taken from a population assumed to be normal,and = 1.2 and s = .6.Calculate a 98 percent confidence interval for μ.
A) [.675,1.725]
B) [.794,1.606]
C) [.876,1.524]
D) [.882,1.518]

Accepted Answer

To calculate a 98% confidence interval for μ, we use the formula:

CI = X̄ ± zα/2 * (s/√n)

where:
X̄ = sample mean = 1.2
zα/2 = critical value for a 98% confidence interval = 2.33 (from a z-table or calculator)
s = sample standard deviation = 0.6
n = sample size = 15

Plugging in the values, we get:

CI = 1.2 ± 2.33 * (0.6/√15)
CI = [0.794, 1.606]

Therefore, the best choice is B, which matches our calculated interval.

Question 10

A sample of 2,000 people yielded =.52.Calculate a 99 percent confidence interval for p.
A) [.515,.525]
B) [.494,.546]
C) [.506,.534]
D) [.491,.549]

Accepted Answer

The confidence interval is calculated using the formula for a proportion: p ± Z*(√[p(1-p)/n]). For 99% confidence, Z is approximately 2.576. Plugging in the values, the interval is approximately [.491, .549].

Question 11

The weight of a product is measured in pounds.A sample of 50 units is taken from a recent production.The sample yielded = 75 lb,and we know that σ² = 100 lb.Calculate a 99 percent confidence interval for μ. A) [71.36,78.64] B) [74.25,75.75] C) [38.58,111.42] D) [71.71,78.29]

Accepted Answer

To find the confidence interval for μ, we can use the formula:

CI = X̄ ± Zα/2 (σ/√n)

where X̄ is the sample mean, Zα/2 is the critical value from the standard normal distribution (which is 2.576 for a 99% confidence interval), σ is the population standard deviation, and n is the sample size.

Plugging in the values:

CI = 75 ± 2.576 (10/√50)

CI = 75 ± 3.628

CI = [71.372, 78.628]

Rounding to two decimal places, the answer is (A) [71.36, 78.64].

Question 12

An insurance company estimates 45 percent of its claims have errors.The insurance company wants to estimate with 99 percent confidence the proportion of claims with errors.What sample size is needed if they wish to be within 5 percent of the actual?
A) 664
B) 657
C) 163
D) 1084

Accepted Answer

We can use the formula for sample size for a proportion:
n = (Z^2 * p * q) / E^2
where:
Z = the z-score associated with the desired level of confidence (in this case, 2.576 for 99% confidence)
p = the estimated proportion of claims with errors (0.45)
q = 1-p 
E = the desired margin of error (0.05)

Plugging in the values, we get:
n = (2.576^2 * 0.45 * 0.55) / 0.05^2
n ≈ 657

Therefore, the best choice is B, with a sample size of 657.

Question 13

You want to estimate the proportion of customers who are satisfied with their experiences at a local supermarket at α = .10 and within .025 of the true value.It has been estimated that p =.85.How large of a sample is needed?
A) 1083
B) 553
C) 71
D) 336

Accepted Answer

To estimate the sample size, we can use the formula n = [p(1-p)(Zα/2/δ)²], where p = .85 (estimated proportion), α = .10 (level of significance), and δ = .025 (margin of error). 
Substituting the values, we get n = [.85(1-.85)(Z.05/2/.025)²]. 
Z.05/2 = Z.025 = 1.96 (from the standard normal distribution table). 
Thus, n = [.85(1-.85)(1.96/.025)²] = 552.73, which rounds up to 553. Therefore, the best choice is B.

Question 14

An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered by the company.A random sample of 240 claims shows that the insurance company covered 90 accident claims,while 150 claims were not covered.Use a confidence interval of 95 percent and determine the margin of error.
A) .01513
B) .03125
C) .06125
D) .02965

Accepted Answer

1.96= .0613

Question 15

The customer service manager for the XYZ Fastener Manufacturing Company examined 60 invoices and found 9 that contained errors.Find a 98 percent confidence interval for the proportion of invoices with errors.
A) [.147,.153]
B) [.080,.220]
C) [.043,.257]
D) [.112,.188]

Accepted Answer

== .15
.15 ± 2.326→ .043 to .257

Question 16

A random sample n = 15 is taken from a population assumed to be normal,and = 1.2 and the sample variance is .36.Calculate a 99 percent confidence interval for μ.
A) [.923,1.477]
B) [.794,1.607]
C) [.801,1.600]
D) [.739,1.661]

Accepted Answer

The answer of A random sample n = 15 is...

Question 17

The quality control manager of a tire company wishes to estimate the tensile strength of a standard size of rubber used to make a class of radial tires.A random sample of 61 pieces of rubber from different production batches is subjected to a stress test.The test measures the force (in pounds)needed to break the rubber.According to the sample results,the average pressure is 238.4 pounds with a population standard deviation of 35 pounds.Determine the 98 percent confidence interval.
A) [227.96,248.84]
B) [236.64,240.16]
C) [233.63,243.17]
D) [229.21,247.59]

Accepted Answer

238.4 ± 2.33= 227.96 to 248.84)

Question 18

An insurance analyst working for a car insurance company would like to determine the proportion of accident claims covered by the company.A random sample of 200 claims shows that the insurance company covered 80 accident claims and did not cover 120 claims.Construct a 90 percent confidence interval estimate of the true proportion of claims covered by the insurance company.
A) [.356,.444]
B) [.360,.440]
C) [.372,.428]
D) [.343,.457]

Accepted Answer

± 1.645= .343 to .457

Question 19

A computer manufacturing company has sent a mail survey to 2,800 of its randomly selected customers that have purchased a new laptop.The survey asked the customers whether or not they were satisfied with the computer.800 customers responded to the survey.640 customers indicated that they were satisfied,while 160 customers indicated they were not satisfied with their new computer.Construct a 96 percent confidence interval estimate of the true proportion of customers satisfied with their new computer.
A) [.771,.829]
B) [.788,.812]
C) [.775,.825]
D) [.785,.816]

Accepted Answer

The correct answer is A. The confidence interval is calculated as: 0.64 +/- 1.96 * sqrt(0.64 * 0.36 / 2800) = 0.64 +/- 0.029 = [0.611, 0.669].

Question 20

The weight of a product is measured in pounds.A sample of 50 units is taken from a recent production.The sample yielded = 75 lb,and we know that σ² = 100 lb.Calculate a 90 percent confidence interval for μ. A) [73.19,76.81] B) [51.74,98.26] C) [72.67,77.33] D) [67.50,82.50]

Accepted Answer

We can use the formula: 
CI = X̄ ± z* (σ/√n)
where X̄ is the sample mean, σ is the population standard deviation (given as 100 lb), n is the sample size (50 units), z* is the critical value from the standard normal distribution for 90% confidence level (1.645).

Substituting the given values, we get:
CI = 75 ± 1.645 *(100/√50)
CI = 75 ± 14.7
CI = [72.3, 77.7]

Rounding off to two decimal places, the answer is [72.67, 77.33]. Therefore, the best choice is C.

Quiz 8: Confidence Intervals

If the 95 percent confidence interval for a mean is from .771 to 1.629,can we conclude that μ = .5,using a 95 percent confidence interval?

The weight of a product is measured in pounds.A sample of 50 units is taken from a recent production.The sample yielded = 75 lb,and we know that σ2 = 100 lb.Calculate a 95 percent confidence interval for μ.

A cable TV company wants to estimate the percentage of cable boxes in use during an evening hour.An approximation is 20 percent.They want the estimate to be at the 90 percent confidence level and within 2 percent of the actual proportion.What sample size is needed?

In a randomly selected group of 650 automobile deaths,180 were alcohol related.Construct a 95 percent confidence interval for the true proportion of all automobile accidents caused by alcohol.

A sample of 2,000 people yielded =.52.What is the variance of the population proportion?

A sample of 2,000 people yielded =.52.Calculate a 90 percent confidence interval for p.

A sample of 2,000 people yielded =.52.Calculate a 95 percent confidence interval for p.

A random sample n= 15 is taken from a population assumed to be normal,and = 1.2 and s = .6.Calculate a 98 percent confidence interval for μ.

A sample of 2,000 people yielded =.52.Calculate a 99 percent confidence interval for p.

The weight of a product is measured in pounds.A sample of 50 units is taken from a recent production.The sample yielded = 75 lb,and we know that σ2 = 100 lb.Calculate a 99 percent confidence interval for μ.

An insurance company estimates 45 percent of its claims have errors.The insurance company wants to estimate with 99 percent confidence the proportion of claims with errors.What sample size is needed if they wish to be within 5 percent of the actual?

You want to estimate the proportion of customers who are satisfied with their experiences at a local supermarket at α = .10 and within .025 of the true value.It has been estimated that p =.85.How large of a sample is needed?

The customer service manager for the XYZ Fastener Manufacturing Company examined 60 invoices and found 9 that contained errors.Find a 98 percent confidence interval for the proportion of invoices with errors.

A random sample n = 15 is taken from a population assumed to be normal,and = 1.2 and the sample variance is .36.Calculate a 99 percent confidence interval for μ.

The weight of a product is measured in pounds.A sample of 50 units is taken from a recent production.The sample yielded = 75 lb,and we know that σ2 = 100 lb.Calculate a 90 percent confidence interval for μ.

The weight of a product is measured in pounds.A sample of 50 units is taken from a recent production.The sample yielded = 75 lb,and we know that σ² = 100 lb.Calculate a 95 percent confidence interval for μ.

The weight of a product is measured in pounds.A sample of 50 units is taken from a recent production.The sample yielded = 75 lb,and we know that σ² = 100 lb.Calculate a 99 percent confidence interval for μ.

The weight of a product is measured in pounds.A sample of 50 units is taken from a recent production.The sample yielded = 75 lb,and we know that σ² = 100 lb.Calculate a 90 percent confidence interval for μ.