Question 1

The formula for the maximum error of estimation is&#10;A) $E=z_{\alpha / 2} \cdot \frac{\sigma}{\sqrt{2 n}}$.&#10;B) $E=z_{\alpha} \cdot \frac{\sigma}{\sqrt{n}}$.&#10;C) $E=z_{\alpha / 2} \cdot \frac{\sigma}{\sqrt{n}}$.&#10;D) $E=z_{\alpha / 2} \cdot \frac{\sigma}{n}$.

Accepted Answer

The correct formula for the maximum error of estimation is $E = z_{\alpha / 2} \cdot \frac{\sigma}{\sqrt{n}}$, where $z_{\alpha / 2}$ is the z-score corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. This formula is used in constructing confidence intervals for estimating population parameters.

Question 2

If $n=325$ and $\sigma=8.3$, what is the maximum error of estimate with a probability of 0.95 ?&#10;A) 0.05&#10;B) 1.19&#10;C) 0.90&#10;D) 0.75

Accepted Answer

0.90&#10;

Question 3

Susan wants fo find a $99 \%$ confidence interval for the time it takes her to drive home from school. She kept records for 15 days and found her average time to drive was 17.5 minutes with a standard deviation of 4.7 minutes. Calculate Susan's maximum error of estimate.

A) 2.38
B) 3.12
C) 2.11
D) 1.68

3.12

Accepted Answer

The maximum error of estimate (E) can be calculated using the formula $E = z \times \frac{\sigma}{\sqrt{n}}$, where $z$ is the z-score corresponding to the confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size. Since the population standard deviation is not given, we use the sample standard deviation as an estimate. For a 99% confidence interval and a sample size of 15, the z-score is approximately 2.576 (from z-tables for 99% confidence). Plugging in the values: $E = 2.576 \times \frac{4.7}{\sqrt{15}} \approx 3.12$.

Question 4

A professor wants to use the mean of a random sample to estimate the average amount of time students take to do their homework. He wants to be able to assert with a probability of 0.99 that his error will be at most 15 minutes. If $\sigma=29$ minutes, how large a sample will he need?

A) 15
B) 14
C) 25
D) 4

Accepted Answer

To estimate the sample size needed, the formula $n = \left(\frac{Z\sigma}{E}\right)^2$ is used, where $Z$ is the Z-score corresponding to the desired confidence level (0.99 in this case, which corresponds to a Z-score of approximately 2.576), $\sigma$ is the population standard deviation (29 minutes), and $E$ is the margin of error (15 minutes). Plugging in the values, $n = \left(\frac{2.576 \times 29}{15}\right)^2 \approx 25.1$. Since the sample size must be a whole number, rounding up gives 25.

Question 5

$\bar{x}-z_{\alpha / 2} \cdot \frac{\sigma}{\sqrt{n}}<\mu<\bar{x}+z_{\alpha / 2} \cdot \frac{\sigma}{\sqrt{n}}$ is an example of $\mathrm{a}(\mathrm{n})$&#10;A) confidence interval.&#10;B) interval estimate.&#10;C) distribution interval.&#10;D) confidence limit.

Accepted Answer

This expression represents a confidence interval, which is used to estimate the range within which a population parameter, such as the mean ($\mu$), is expected to lie with a certain level of confidence.

Question 6

Construct a $99 \%$ confidence interval for $n=42, \bar{x}=18.2$, and $\sigma=3.96$.&#10;A) $16.63<\mu<19.77$&#10;B) $17.00<\mu<19.40$&#10;C) $17.00<\mu<19.77$&#10;D) $16.63<\mu<19.40$

Accepted Answer

To construct a 99% confidence interval for the mean when the population standard deviation ($\sigma$) is known, we use the formula: $\bar{x} \pm Z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $n$ is the sample size, and $Z$ is the Z-score corresponding to the desired level of confidence. For a 99% confidence level, the Z-score is approximately 2.576.Given $n=42$, $\bar{x}=18.2$, and $\sigma=3.96$, we calculate the margin of error (ME) as follows:$$ME = Z\frac{\sigma}{\sqrt{n}} = 2.576 \times \frac{3.96}{\sqrt{42}} \approx 1.57$$Thus, the confidence interval is:$$\bar{x} \pm ME = 18.2 \pm 1.57 = (16.63, 19.77)$$Therefore, the correct choice is $16.63<\mu<19.77$.

Question 7

A survey randomly selected 250 top executives. The average height of these executives was 66.9 inches with a standard deviation of 6.2 inches. What is a 95\% confidence interval for the mean height, $\mu$ , of all top executives?

A)

66.1<\mu<67.7

B)

65.3<\mu<68.5

C)

62.8<\mu<66.8

D)

63.5<\mu<66.1

Accepted Answer

To calculate a 95% confidence interval for the mean height, we use the formula: $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the confidence level (for 95%, $z = 1.96$), $\sigma$ is the standard deviation, and $n$ is the sample size. Plugging in the given values: $66.9 \pm 1.96\frac{6.2}{\sqrt{250}}$, we get the interval $66.1<\mu<67.7$.

Question 8

A quality control engineer intends to use the mean of a random sample of $n=85$ to estimate the average time it takes to manufacture an item. If, based on experience, the engineer can assume that $\sigma=4.8$ hours for such data, what can he assert with probability 0.99 about the maximum error of the estimate?

A)

E \approx 1.34

hours
B)

E \approx 0.13

hours
C)

E \approx 0.15

hours
D)

E \approx 1.21

hours

Accepted Answer

The maximum error of the estimate (E) can be calculated using the formula for the margin of error in estimating a population mean: $E = Z \cdot \frac{\sigma}{\sqrt{n}}$, where $Z$ is the Z-score corresponding to the desired confidence level (0.99 in this case), $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 0.99 confidence level, the Z-score is approximately 2.576. Plugging in the given values: $E = 2.576 \cdot \frac{4.8}{\sqrt{85}} \approx 1.34$ hours.

Question 9

The exact shape of the $t$ distribution depends on a parameter called the __________&#10;A) degrees of freedom&#10;B) confidence level&#10;C) Student's $t$ distribution&#10;D) $t$ statistic

Accepted Answer

The shape of the $t$ distribution is determined by the degrees of freedom, which typically depend on the sample size.

Question 10

When constructing confidence intervals for $\sigma$ based on $s$, it requires that the population we are sampling has roughly the shape of __________.&#10;A) a normal distribution&#10;B) a straight line&#10;C) a uniform distribution&#10;D) a chi-square distribution

Accepted Answer

When constructing confidence intervals for the population standard deviation ($\sigma$) based on the sample standard deviation ($s$), it is required that the population from which we are sampling has roughly the shape of a normal distribution. This is because the method relies on the assumption that the sample comes from a normally distributed population for the chi-square distribution to be applicable in calculating the confidence interval for $\sigma$.

Question 11

In general, we make __________ about future values of random variables and __________ once the data have been obtained.&#10;A) confidence statements; probability statements&#10;B) estimation statements; confidence statements&#10;C) probability statements; confidence statements&#10;D) probability statements; estimation statements

Accepted Answer

The answer of In general, we make __________ about future...

Question 12

Suppose that $\bar{x}=81$ is being used as an estimate of the true average score for students' first exam in psychology. What can be said with $95 \%$ confidence about the maximum error if $n=11$ and $s=5$ ?

A)

E \approx 2.73

B)

E \approx 3.32

C)

E \approx 3.36

D)

E \approx 2.71

Accepted Answer

The answer of Suppose that $\bar{x}=81$ is being used as...

Question 13

Suppose that a highway patrol person wants to estimate what proportion of all drivers drive over the speed limit, and she wants to be able to assert with probability of at least 0.95 that its error will not exceed 0.07 . How large a sample will be needed if she knows that the true proportion lies somewhere on the interval from 0.6 or 0.7 ?

A)

n=165

B)

n=196

C)

n=189

D)

n=188.2

Accepted Answer

The answer of Suppose that a highway patrol person wants...

Deck 11: Problems of Estimation