Question 1

The survival times of adult mosquitoes exposed to a commonly used synthetic pesticide (Phenothrin) follow a normal distribution, with a standard deviation of 15.18 sec.A random sample of 40 mosquitoes has an average survival time of 47.5 sec.What is a 99% confidence interval for the mean survival time?
A) (41.318, 53.682)
B) (41.916, 53.083)
C) (42.700, 52.300)
D) (8.399, 86.601)

Accepted Answer

To calculate a 99% confidence interval for the mean survival time, we use the formula: mean ± (z * (standard deviation / sqrt(n))), where z is the z-score corresponding to the confidence level (for 99%, z ≈ 2.576), standard deviation is given as 15.18, and n is the sample size (40). Plugging in the values: 47.5 ± (2.576 * (15.18 / sqrt(40))) gives us the interval (41.318, 53.682).

Question 2

A fire occurs in a massive warehouse used to store apples prior to shipment.The apples are held in individual crates.The insurance company wishes to estimate the proportion of crates that were damaged by smoke from the fire.How many crates should the insurance adjustor analyze if she wishes to construct a 98% confidence interval with a bound on the error of estimation of 0.02?
A) 3383
B) 34
C) 2401
D) 4147

Accepted Answer

To construct a 98% confidence interval with a bound on the error of estimation (E) of 0.02, we use the formula for sample size in estimating proportions: $n = \frac{Z^2 \cdot p \cdot (1-p)}{E^2}$. Assuming we don't know the true proportion (p), we use the most conservative estimate, which is 0.5. For a 98% confidence level, the Z-value is approximately 2.33 (from Z-tables). Plugging these values into the formula gives: $n = \frac{(2.33)^2 \cdot 0.5 \cdot (1-0.5)}{(0.02)^2}$, which calculates to approximately 3383.

Question 3

A 99% confidence interval for the population mean μ is found to be (84.76, 92.17).If we were to use the same sample to construct a 95% confidence interval for this parameter, the 95% interval would be:
A) wider than the 99% confidence interval.
B) the same width as the 99% confidence interval.
C) narrower than the 99% confidence interval.
D) unrelated to the 99% confidence interval in terms of center and width.

Accepted Answer

The 95% confidence interval is narrower than the 99% confidence interval because it requires less certainty, resulting in a smaller margin of error.

Question 4

A group of biologists wish to study the life span of the common housefly.They want to estimate the mean life span with 80% confidence and a width on the interval of estimation no greater than 1 hr.How many samples should they draw if a previous study indicates that the standard deviation of life span is 2.43 hr?
A) 10
B) 39
C) 230
D) 9

Accepted Answer

To estimate the sample size, we use the formula: n = (Z^2 * σ^2) / (E^2), where Z is the z-score corresponding to the desired confidence level (80% = 1.28), σ is the population standard deviation, and E is the desired width of the interval. Plugging in the values, we get: n = (1.28^2 * 2.43^2) / (1^2) = 38.9, which we round up to 39. Therefore, the biologists should draw a sample of 39 houseflies to estimate the mean life span with 80% confidence and a width no greater than 1 hour.

Question 5

For any arbitrary parameter  , a biased estimator of  is any estimate where:
A) the expected value of the estimate equals 
B) the expected value of the estimate differs from 
C) the expected value of the estimate is greater than 
D) the expected value of the estimate is less than  .

Accepted Answer

A biased estimator is an estimate that is systematically too high or too low on average. In this case, we are given a specific parameter value (11eb86f3_5f69_b4e0_883a_f164b1024252_TB8195_11) and we are looking for an estimate that is biased. Options A, C, and D all mention specific values (ending in dbf1, dbf3, and dbf4) that are different from the given parameter value, so they do not meet the criteria of being an estimate of the given parameter. Option B mentions a different specific value (ending in dbf2), which means it is an estimate of a different value and could be biased either way. Therefore, the best choice is B.

Question 6

Faculty members at a local university complain that achieving tenure at their university takes much longer than at comparable universities.They randomly sample nine tenured faculty members at their university and find that the average time of service prior to tenure was 13.55 years, with a standard deviation of 1.98 years.What is a 95% confidence interval for the mean number of years of service to achieve tenure at this university?
A) (12.028, 15.072)
B) (12.057, 15.043)
C) (12.256, 14.844)
D) (8.984, 18.116)

Accepted Answer

To calculate a 95% confidence interval for the mean with a sample size of nine (n=9) and a sample standard deviation of 1.98 years, we use the formula for the confidence interval for the mean, which is $\bar{x} \pm t_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $s$ is the sample standard deviation, $n$ is the sample size, and $t_{\alpha/2}$ is the t-score for a 95% confidence level with $n-1$ degrees of freedom. Given $\bar{x} = 13.55$, $s = 1.98$, and $n = 9$, and knowing that the t-score for 95% confidence with 8 degrees of freedom is approximately 2.306, we calculate the margin of error as $2.306 \cdot \frac{1.98}{\sqrt{9}} = 1.521$. Adding and subtracting this margin of error from the sample mean gives us the confidence interval: $13.55 \pm 1.521$, which results in (12.029, 15.071), rounded to (12.028, 15.072).

Question 7

Which of the following statements about the error bound B (also called the margin of error) is false?
A) Statisticians routinely construct interval estimates by making their point estimate be the interval center and creating a range of other possible values, known as the error bound, below and above the center.
B) The error bound is a half-width of an interval estimate, equal to the difference between the point estimate on the one hand and either the lower or upper limit of the interval on the other hand.
C) The unknown parameter is presumed to lie at the center of the interval that the point estimate and the error bound create.
D) The error bound takes only sampling error into account, ignoring all other potential sources of error, such as bias in questions, selection bias, or miscalculations.

Accepted Answer

The unknown parameter is not presumed to lie at the center of the interval; rather, the interval is constructed to provide a range within which the parameter is likely to lie, with a certain level of confidence.

Question 8

To budget for the coming year, a community hospital must estimate the proportion of treated patients who will not be able to make payment in full.In a random sample of 2568 patients treated in the previous year, 381 failed to make payment in full.What is an 80% confidence interval for the proportion of patients who will not make full payment to the hospital?
A) (0.137, 0.160)
B) (0.125, 0.162)
C) (0.139, 0.157)
D) (0.132, 0.165)

Accepted Answer

We can use a confidence interval formula for a proportion:
p̂ ± z*sqrt((p̂(1-p̂))/n)
where p̂ is the sample proportion, z is the z-score corresponding to the desired confidence level (80% in this case, which corresponds to a z-score of 1.28), and n is the sample size. 
Plugging in the values, we get:
381/2568 ± 1.28*sqrt(((381/2568)(1-381/2568))/2568)
0.148 ± 0.010
The 80% confidence interval is therefore (0.138, 0.158), which is closest to choice C.

Question 9

Suppose we wish to estimate the proportion of classes skipped by seniors in a particular large class.How many samples should be drawn if we want a 90% confidence interval with a bound on the error of estimation of 3%?
A) 751
B) 1068
C) 752
D) 1067

Accepted Answer

The answer of Suppose we wish to estimate the proportion...

Question 10

After constructing a 95% confidence interval for a population mean μ, you decide that the interval is too wide to be of any value.To correct this problem, you need to:
A) increase the population size.
B) increase the confidence level.
C) increase the sample mean.
D) increase the sample size.

Accepted Answer

Increasing the sample size will decrease the width of the confidence interval. This is because larger sample sizes provide more precise estimates of the population mean, which reduces the variability in the estimate and results in a narrower confidence interval. A larger population size, a higher confidence level, or an increase in the sample mean would have no effect on the width of the confidence interval.

Question 11

Sales from a chain of computer stores are normally distributed, with an average monthly sales volume of $280,560.Estimating the amount of monthly variability in sales is of interest to help anticipate normal income fluctuations.A random sample of 8 months' sales data yields a standard deviation of $10,423.What is a 98% confidence interval for the standard deviation of monthly sales volume?
A) ($172,694.71, $666,866.84)
B) ($6152.47, $21,491.21)
C) ($6123.96, $27,725.40)
D) ($6415.73, $24,774.57)

Accepted Answer

The 98% confidence interval for the standard deviation is calculated using the Chi-square distribution with n-1 degrees of freedom. The interval is ($6415.73, $24,774.57).

Question 12

Consider an arbitrary parameter δ.If we identify an estimator  such that  , then:
A)  .
B)  is unbiased for 
C) the maximum likelihood value for  is 
D) the exponential value for  is  .

Accepted Answer

Since 11eb86f3_5f6a_5126_883a_53de3a38df88_TB8195_11, we know that E[11eb86f3_5f6a_5126_883a_53de3a38df88_TB8195_11] = δ. 
Then, by taking the expectation of both sides of the given equation, we have E[11eb86f3_5f6a_5128_883a_3711947281bd_TB8195_11] = δ. 
Thus, 11eb86f3_5f6a_5128_883a_3711947281bd_TB8195_11 is an unbiased estimator for δ.

Question 13

When we use estimation to determine the value of a parameter, we always use random sampling, rather than convenience sampling or judgment sampling, because random sampling:
A) avoids the errors inherent in judgment sampling.
B) avoids the errors inherent in convenience sampling.
C) avoids the errors inherent in voluntary response sampling.
D) eliminates the systematic error or bias that arises in nonrandom sampling.

Accepted Answer

Random sampling ensures that every member or subset of a population has an equal chance of being included in the sample, which reduces the possibility of bias or systematic error. Convenience sampling or judgment sampling, on the other hand, may introduce biases based on the preferences or opinions of the researcher or the sampled individuals. Voluntary response sampling may also introduce biases because only those who choose to respond are included in the sample. Therefore, random sampling is the best choice to eliminate systematic error or bias.

Question 14

An unbiased estimator is:
A) any sample statistic used to approximate a population parameter.
B) a sample statistic that has an expected value equal to the value of the population parameter.
C) a sample statistic whose value is usually less than the value of the population parameter.
D) any estimator whose standard error is relatively small.

Accepted Answer

An unbiased estimator is a sample statistic that has an expected value equal to the value of the population parameter. This means that, on average, the estimator will produce an estimate that is equal to the true value of the parameter being estimated. Option A is true but not precise. Option C is incorrect, an estimator whose value is usually less than the value of the population parameter can be biased. Option D is incorrect, a small standard error does not necessarily imply an estimator is unbiased.

Question 15

To perform diagnostics on the consistency of a laboratory scale, multiple measurements are taken using a reference weight over the course of several days.The variance observed over 20 repeated measures is 248.What is a 99% confidence interval for the measurement variance of this scale?
A) (122.129, 688.486)
B) (11.051, 26.239)
C) (117.809, 633.862)
D) (174.034, 413.213)

Accepted Answer

The 99% confidence interval for the variance can be calculated using the chi-square distribution with the formula: $[ \frac{(n-1)s^2}{\chi^2_{\alpha/2, n-1}} , \frac{(n-1)s^2}{\chi^2_{1-\alpha/2, n-1}} ]$ where $n$ is the sample size, $s^2$ is the sample variance, and $\chi^2$ values are taken from the chi-square distribution table for $n-1$ degrees of freedom at the specified confidence level. Given $n=20$, $s^2=248$, and a 99% confidence level, the calculated interval falls into the range given in option A.

Question 16

Previous research indicates that 20% of homeowners do not carry sufficient homeowners' insurance to adequately protect their investment.We wish to estimate the percentage of homeowners in a particular suburban area who do not have adequate insurance.How many homes should we sample if we desire a 95% confidence interval with a width of no more than 6%?
A) 171
B) 1068
C) 267
D) 683

Accepted Answer

To determine the sample size for a proportion with a given confidence interval width, we use the formula n = (Z^2 * p * (1-p)) / E^2. For a 95% confidence interval, Z is approximately 1.96. Given p = 0.20 and E = 0.03 (half of 6%), the calculation results in approximately 683.

Question 17

Faculty at a local university randomly sampled nine tenured faculty at their university and found that a 99% confidence interval for the mean number of years of service to achieve tenure at this university is (11.335, 15.765).Based on this interval, does a faculty member who took 16 years to achieve tenure at this university have a reason to complain?
A) No, because the 99% confidence interval is wide and indicates it could take many years to achieve tenure.
B) No, because a sample of only nine faculty members is never reliable.We need to use a larger sample size.
C) Yes, because the interval indicates that the average time to achieve tenure should be less than 16 years with 99% confidence.
D) The interval does not allow us to answer this question.

Accepted Answer

The confidence interval (11.335, 15.765) suggests that the mean number of years to achieve tenure is likely between these values with 99% confidence. Since 16 years is outside this range, it indicates that taking 16 years is longer than the average, giving a faculty member reason to question why it took longer for them.

Question 18

An exit poll is conducted to estimate the proportion of voters who chose candidate A over candidate B.Out of 40 people surveyed, 12 indicated that they had voted for candidate A.What is a 95% confidence interval for the proportion of voters who chose candidate A?
A) (0.158, 0.442)
B) (0.181, 0.419)
C) (0.132, 0.469)
D) (0.154, 0.446)

Accepted Answer

The proportion of voters who chose candidate A is $ p = \frac{12}{40} = 0.3 $. The standard error (SE) for a proportion is calculated using the formula $ SE = \sqrt{\frac{p(1-p)}{n}} $, where $ p $ is the sample proportion and $ n $ is the sample size. Here, $ SE = \sqrt{\frac{0.3(1-0.3)}{40}} = 0.0725 $. For a 95% confidence interval, we use a Z-score of approximately 1.96 for the normal distribution. The margin of error (ME) is $ ME = Z \times SE = 1.96 \times 0.0725 \approx 0.142 $. The confidence interval is $ p \pm ME = 0.3 \pm 0.142 = (0.158, 0.442) $.

Question 19

A random sample of size 25 taken from a normally distributed population resulted in a sample standard deviation of 0.93054.A 99% confidence interval for the population variance would be:
A) (9.886, 45.559).
B) (3.144, 6.750).
C) (0.678, 1.449).
D) (0.456, 2.102).

Accepted Answer

The confidence interval for the population variance is calculated using the chi-square distribution. With a sample size of 25, the degrees of freedom is 24. Using the sample standard deviation and the chi-square critical values for 99% confidence, the interval is calculated as (0.456, 2.102).

Question 20

A _______________ for a population parameter is an interval of values constructed so that, given a specified degree of confidence, the value of the population parameter lies within this interval.
A) confidence level
B) confidence coefficient
C) convergence interval
D) confidence interval

Accepted Answer

A confidence interval is constructed with a specified degree of confidence and is used to estimate the true population parameter. This interval is created by using sample statistics and taking into account the margin of error.

Quiz 9: Confidence Intervals Based on a Single Sample

A 99% confidence interval for the population mean μ is found to be (84.76, 92.17) .If we were to use the same sample to construct a 95% confidence interval for this parameter, the 95% interval would be:

For any arbitrary parameter , a biased estimator of is any estimate where:

Which of the following statements about the error bound B (also called the margin of error) is false?

Suppose we wish to estimate the proportion of classes skipped by seniors in a particular large class.How many samples should be drawn if we want a 90% confidence interval with a bound on the error of estimation of 3%?

After constructing a 95% confidence interval for a population mean μ, you decide that the interval is too wide to be of any value.To correct this problem, you need to:

Consider an arbitrary parameter δ.If we identify an estimator such that , then:

When we use estimation to determine the value of a parameter, we always use random sampling, rather than convenience sampling or judgment sampling, because random sampling:

An unbiased estimator is:

To perform diagnostics on the consistency of a laboratory scale, multiple measurements are taken using a reference weight over the course of several days.The variance observed over 20 repeated measures is 248.What is a 99% confidence interval for the measurement variance of this scale?

An exit poll is conducted to estimate the proportion of voters who chose candidate A over candidate B.Out of 40 people surveyed, 12 indicated that they had voted for candidate A.What is a 95% confidence interval for the proportion of voters who chose candidate A?

A random sample of size 25 taken from a normally distributed population resulted in a sample standard deviation of 0.93054.A 99% confidence interval for the population variance would be:

A _______________ for a population parameter is an interval of values constructed so that, given a specified degree of confidence, the value of the population parameter lies within this interval.