Question 1

If a sample containing 20 observations is taken from a normally distributed population and a 90% confidence estimate for $\mu$ is needed,the appropriate t-score is:
A) 2.12.
B) 1.746.
C) 2.131.
D) 1.729.
E) 2.602.

Accepted Answer

The answer of If a sample containing 20 observations is...

Question 2

A total of 150 households were asked if they had Internet access,of which 57 responded "Yes".Which of the following would represent a 98% confidence interval around this sample proportion?
A) $0.380 \mp 2.33$
B)$0.380 \mp 0.092$
C)$0.380 \mp 0.039$
D) $0.380 \mp 0.251$
E) $57 \pm 2.33$

Accepted Answer

The correct confidence interval is calculated using the formula for a proportion's confidence interval: $\hat{p} \pm Z \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$, where $\hat{p}$ is the sample proportion, $Z$ is the Z-score corresponding to the confidence level, and $n$ is the sample size. Here, $\hat{p} = \frac{57}{150} = 0.380$, $n = 150$, and for a 98% confidence level, $Z \approx 2.33$. Plugging these values into the formula gives a margin of error around 0.092, making option B correct.

Question 3

A random sample of size 15 taken from a normally distributed population revealed a sample mean of 75 and a sample variance of 25.The upper limit of a 95% confidence interval for the population mean would equal:
A) 77.769.
B) 72.231.
C) 72.727.
D) 77.273.

Accepted Answer

The upper limit of a 95% confidence interval for the population mean is calculated using the formula: mean + (t-value * standard error). The standard error is the square root of the variance divided by the sample size. With a sample size of 15, degrees of freedom are 14. The t-value for 95% confidence and 14 degrees of freedom is approximately 2.145. Thus, the upper limit is 75 + (2.145 * sqrt(25/15)) = 77.769.

Question 4

The larger the level of confidence used in constructing a confidence interval estimate of the population mean the:
A) smaller the probability that the confidence interval will contain the population mean.
B) smaller the value of $Z _ { \alpha / 2 }$
C) wider the confidence interval.
D) narrower the confidence interval.

Accepted Answer

The larger the level of confidence used in constructing a confidence interval estimate of the population mean, the wider the confidence interval. This is because as the level of confidence increases, we need to be more certain that our interval contains the population mean, which requires including more possible values within the interval.

Question 5

In the formula $\bar { x } \pm z _ { \alpha / 2 } ( ?/ \sqrt { n } )$
 ,the subscript $( \alpha / 2 )$ refers to the:
A) probability that the confidence interval will contain the population mean.
B) probability that the confidence interval will not contain the population mean.
C) area in the lower tail or upper tail of the sampling distribution of the sample mean.
D) level of confidence.
E) sampling error.

Accepted Answer

The subscript $(\alpha/2)$ refers to the area in the lower tail or upper tail of the sampling distribution of the sample mean. This area represents the probability of observing a certain value of the sample mean if we assume that the null hypothesis (in this case, that the true population mean is equal to the sample mean) is true. By choosing a specific value of $\alpha$ (typically 0.05 or 0.01), we can determine the corresponding value of $z_{\alpha / 2}$ that will give us the desired level of confidence in our confidence interval.

Question 6

The owner of a large shopping center is besieged with complaints concerning the shortage of parking space.He feels that the 1000 spaces are adequate.In an effort to address the problem,he obtains a sample of the average number of cars on the parking lot during prime hours.The sample of 40 has a mean of 952.Assume a population standard deviation of 396.The 95% confidence interval estimate for prime hour parking is:
A) 889.36 to 932.60.
B) 932.60 to 971.40.
C) 829.28 to 1074.72.
D) 849.00 to 1054.99.
E) 952.00 to 1052.00.

Accepted Answer

The 95% confidence interval for the mean number of cars during prime hours can be calculated using the formula for the confidence interval of the mean with known population standard deviation: $\bar{x} \pm Z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $Z$ is the Z-score corresponding to the confidence level (1.96 for 95%), $\sigma$ is the population standard deviation, and $n$ is the sample size. Plugging in the given values ($\bar{x} = 952$, $\sigma = 396$, $n = 40$), we get the interval $952 \pm 1.96 \times \frac{396}{\sqrt{40}}$, which calculates to approximately $829.28$ to $1074.72$.

Question 7

If a sample containing 12 observations is taken from a normally distributed population and a 99% confidence estimate for $\mu$ is needed,the appropriate t-score is:
A) 2.681.
B) 2.718.
C) 3.169.
D) 3.055.
E) 3.106.

Accepted Answer

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Question 8

A sample of 200 light bulbs was tested and it was found that 11 were defective.Calculate a 95% confidence interval around this sample proportion.
A)$0.055 \pm 1.96$
B)$0.055 \pm 0.016$
C)$0.055 \pm 0.032$
D) $11 \pm 1.96$
E)$200 \pm 11$

Accepted Answer

To calculate the confidence interval for proportion, use the formula:
$$ 	ext{Confidence interval} = \hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}$$
where $\hat{p}$ is the sample proportion, $z_{\alpha/2}$ is the critical value for the desired confidence level (in this case, 95%), and $n$ is the sample size.

Plugging in the given values, we have:
$$ 	ext{Confidence interval} = 0.055 \pm 1.96 \sqrt{\frac{0.055(1-0.055)}{200}}$$
Simplifying this expression gives:
$$ 	ext{Confidence interval} = 0.055 \pm 0.032 $$
Therefore, the best choice is C.

Question 9

The lower limit of a confidence interval at the 95% level of confidence for the population proportion if a sample of size 200 had 40 successes is:
A) 0.2554.
B) 0.1446.
C) 0.2465.
D) 0.1535.

Accepted Answer

The sample proportion is 40/200 = 0.2. Using the formula for the confidence interval for a proportion, the lower limit is calculated as 0.2 - 1.96 * sqrt((0.2 * 0.8) / 200) = 0.1446.

Question 10

The national sales manager of Fast Grow Corp is working on next year's budget projections.He knows that the population standard deviation is $17.66.To estimate the daily per diem cost for outside sales people,he has taken a random sample of 36 individual expense accounts and found a mean of $90.29.The 91.08 percent confidence interval for the per diem expense is:
A) 86.45 to 94.13.
B) 60.27 to 120.37.
C) 85.29 to 95.29.
D) 74.43 to 107.9.
E) 84.32 to 96.26.

Accepted Answer

To find the confidence interval, we use the formula: 
CI = X̄ ± Zα/2 * (σ/√n)
Where X̄ = sample mean = $90.29, 
σ = population standard deviation = $17.66, 
n = sample size = 36, 
α = level of significance = (1 - confidence level)/2 = 0.0916 (from the 91.08% confidence interval), and 
Zα/2 is the Z-value from the standard normal distribution table = 1.70 (approximately).
Plugging in the values, we get:
CI = $90.29 ± 1.70 * ($17.66/√36) 
CI = $90.29 ± $3.58 
CI = ($86.71, $93.87)
Therefore, the 91.08% confidence interval for the per diem expense is ($85.29, $95.29).
The correct answer is choice C.

Question 11

An auditor for a large oil corporation has taken a random sample of monthly credit card charges for 32 individual (non-corporate)accounts.She found a mean of $81.76.Assume that the population standard deviation is 48.2.A 98% confidence interval estimate for the population mean of individual credit card charges would be:
A) 81.76 $\pm$ 19.89.
B) 81.76 $\pm$ 16.72.
C) 81.76 $\pm$ 112.94.
D) 81.76 $\pm$ 22.0.
E) 81.76 $\pm$ 2.33.

Accepted Answer

The correct answer is A because the confidence interval is calculated as:mean $\pm$ z-score * (standard deviation / square root of sample size)= 81.76 $\pm$ 2.326 * (48.2 / square root of 32)= 81.76 $\pm$ 19.89

Question 12

A 98% confidence interval estimate for a population mean $\mu$ is determined to be 75.38 to 86.52.If the confidence level is reduced to 90%,the confidence interval for $\mu$ :
A) Becomes wider.
B) Remains the same.
C) Becomes narrower.
D) None of these answers is correct.

Accepted Answer

Reducing the confidence level from 98% to 90% results in a narrower confidence interval because a lower confidence level requires less certainty about where the true population mean lies, thus allowing the interval to be closer to the sample mean.

Question 13

The t-distribution approaches the normal distribution as the:
A) degrees of freedom increase.
B) degrees of freedom decrease.
C) sample size decreases.
D) population size increases.

Accepted Answer

As the degrees of freedom increase, the t-distribution becomes more and more similar to the normal distribution. The t-distribution is used when the sample size is small or the population standard deviation is unknown. As the sample size increases, the t-distribution becomes more similar to the normal distribution.

Question 14

The security department of a state university is planning its budget for the next year.In estimating the man-hours for security during university sponsored music concerts,the average length of music concerts is needed.A random sample from thirty-six security departments at universities was taken and the sample mean length of concerts was 160 minutes.Suppose the population standard deviation is 45 minutes.A 95 percent confidence interval for the true mean duration of music concerts is:
A) 157.55 to 162.45.
B) 147.63 to 172.37.
C) 71.8 to 248.2.
D) 145.3 to 174.7.
E) 1601.96(45 / 36).

Accepted Answer

The answer of The security department of a state university...

Question 15

Delivery time by courier services is important to stock brokers.One such firm wishes to estimate the average time taken by couriers delivering negotiable securities in New York City.Specifically,the firm randomly samples 36 courier pickup and delivery times.The collected data,in minutes,represent the time between pickup and delivery.Suppose the population standard deviation is 0.3 minutes and the sample mean is 9.975.Find the sample mean and compute.A 90-percent confidence interval for the true mean delivery time, $\mu$ is:
A) 9.852 to 10.05.
B) 119.60 to 119.80.
C) 119.62 to 119.78.
D) 119.26 to 119.87.
E) 9.893 to 10.057.

Accepted Answer

The 90% confidence interval for the mean is calculated using the formula: $\bar{x} \pm z \times \frac{\sigma}{\sqrt{n}}$. Here, $\bar{x} = 9.975$, $\sigma = 0.3$, $n = 36$, and $z$ for 90% confidence is approximately 1.645. The interval is 9.975 ± 1.645 * (0.3/6), which results in 9.893 to 10.057.

Question 16

For a sample of size 20 taken from a normally distributed population with standard deviation equal to 5,a 90% confidence interval for the population mean would require the use of:
A) t = 1.328.
B) t = 1.729.
C) z = 1.96.
D) z = 1.645.

Accepted Answer

The answer of For a sample of size 20 taken...

Question 17

In order to estimate the average daily down time,a manufacturer randomly samples 40 days of production records and found a mean of 55.576 minutes and standard deviation of 77.9 minutes.An 80% confidence interval is:
A) 55.6751 $\pm$ .28(12.32).
B) 55.6751  $\pm$ .304(12.32).
C) 55.6751 $\pm$ .28(77.9).
D) 55.6751 $\pm$ .304(77.9).

Accepted Answer

To find the confidence interval, we use the formula:
CI = x̄ ± (t critical)(s/√n)
where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and t critical is the t-score associated with the desired confidence level and degrees of freedom (df = n-1).

Since the sample size is 40, df = 39. Using an 80% confidence level, we find the t critical value from a t-distribution table:
t critical = 1.303

Plugging in the given values, we get:
CI = 55.576 ± (1.303)(77.9/√40)
CI = 55.576 ± 12.32

Therefore, the 80% confidence interval is:
(55.576 - 12.32, 55.576 + 12.32) or (43.25, 67.90)

The closest answer choice is B,  which uses the correct sample mean and standard deviation and the correct t critical value with the formula for the confidence interval.

Question 18

In an effort to estimate the proportion of female shoppers at a particular store,175 customers were randomly selected,of which 110 were female.Which of the following would represent a 90% confidence interval around this sample proportion?
A)$0.629 \mp 0.072$
B)$110 \mp 1.64$
C)$0.629 \mp 0.036$
D)$0.629 \mp 1.64$
E)$0.629 \mp 0.060$

Accepted Answer

The sample proportion is 110/175 = 0.629. The standard error for a proportion is calculated as sqrt[(p(1-p))/n], and for a 90% confidence interval, the z-score is approximately 1.645. Calculating the margin of error gives approximately 0.060.

Question 19

If a sample containing 30 observations is taken from a normally distributed population and a 98% confidence estimate for $\mu$ is needed,the appropriate t-score is:
A) 2.247.
B) 2.756.
C) 2.462.
D) 2.750.
E) 2.330.

Accepted Answer

For a 98% confidence interval with 29 degrees of freedom (since n-1 = 30-1 = 29), the appropriate t-score is approximately 2.462. This value can be found in t-distribution tables or calculated using statistical software, specifically for the given confidence level and degrees of freedom.

Question 20

Your statistics instructor wants you to determine a confidence interval estimate for the mean test score for the next exam.In the past,the test scores have been normally distributed with a mean of 74.2 and a standard deviation of 30.9.A 95% confidence interval estimate if your class has 30 students is:
A) 68.72 to 79.68.
B) 13.64 to 134.76.
C) 64.92 to 83.48.
D) 62.66 to 85.73.
E) 63.14 to 85.26.

Accepted Answer

The answer of Your statistics instructor wants you to determine...

Quiz 9: Estimation From Sample Data

If a sample containing 20 observations is taken from a normally distributed population and a 90% confidence estimate for μ\muμ is needed,the appropriate t-score is:

A total of 150 households were asked if they had Internet access,of which 57 responded "Yes".Which of the following would represent a 98% confidence interval around this sample proportion?

A random sample of size 15 taken from a normally distributed population revealed a sample mean of 75 and a sample variance of 25.The upper limit of a 95% confidence interval for the population mean would equal:

The larger the level of confidence used in constructing a confidence interval estimate of the population mean the:

In the formula xˉ±zα/2(?/n) \bar { x } \pm z _ { \alpha / 2 } ( ?/ \sqrt { n } ) xˉ±zα/2​(?/n​) ,the subscript (α/2) ( \alpha / 2 ) (α/2) refers to the:

If a sample containing 12 observations is taken from a normally distributed population and a 99% confidence estimate for μ\muμ is needed,the appropriate t-score is:

A sample of 200 light bulbs was tested and it was found that 11 were defective.Calculate a 95% confidence interval around this sample proportion.

The lower limit of a confidence interval at the 95% level of confidence for the population proportion if a sample of size 200 had 40 successes is:

A 98% confidence interval estimate for a population mean μ\muμ is determined to be 75.38 to 86.52.If the confidence level is reduced to 90%,the confidence interval for μ\muμ :

The t-distribution approaches the normal distribution as the:

For a sample of size 20 taken from a normally distributed population with standard deviation equal to 5,a 90% confidence interval for the population mean would require the use of:

In order to estimate the average daily down time,a manufacturer randomly samples 40 days of production records and found a mean of 55.576 minutes and standard deviation of 77.9 minutes.An 80% confidence interval is:

In an effort to estimate the proportion of female shoppers at a particular store,175 customers were randomly selected,of which 110 were female.Which of the following would represent a 90% confidence interval around this sample proportion?

If a sample containing 30 observations is taken from a normally distributed population and a 98% confidence estimate for μ\muμ is needed,the appropriate t-score is:

If a sample containing 20 observations is taken from a normally distributed population and a 90% confidence estimate for $\mu$ is needed,the appropriate t-score is:

In the formula $\bar { x } \pm z _ { \alpha / 2 } ( ?/ \sqrt { n } )$ ,the subscript $( \alpha / 2 )$ refers to the:

If a sample containing 12 observations is taken from a normally distributed population and a 99% confidence estimate for $\mu$ is needed,the appropriate t-score is:

A 98% confidence interval estimate for a population mean $\mu$ is determined to be 75.38 to 86.52.If the confidence level is reduced to 90%,the confidence interval for $\mu$ :

If a sample containing 30 observations is taken from a normally distributed population and a 98% confidence estimate for $\mu$ is needed,the appropriate t-score is: