Question 1

Which statement(s) is/are correct about the t distribution?

Accepted Answer

The t distribution is characterized by having a mean of 0, being symmetric around the mean, and its shape depends on the degrees of freedom.

Question 2

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard error of $20. What is the 95% confidence interval to estimate the true cost of the DVD?

Accepted Answer

The answer of A survey of 144 retail stores revealed...

Question 3

i. The t distribution is based on the assumption that the population of interest is normal or nearly normal. ii. The t distribution is a continuous distribution.
iii. There is not one t distribution, but rather a "family" of t distributions.

Accepted Answer

The first statement is true, as the t distribution assumes normality. The second statement is also true, as the t distribution is a continuous distribution. The third statement is true as the t distribution is a family of distributions, each distinguished by the degrees of freedom parameter.

Question 4

A survey of 25 grocery stores revealed that the average price of a 4-litre bag of milk was $2.98 with a standard error of $0.10. What is the 98% confidence interval to estimate the true cost of a 4-litre bag of milk?

Accepted Answer

The 98% confidence interval is calculated as:mean +/- 2.326 * standard error= 2.98 +/- 2.326 * 0.10= 2.98 +/- 0.2326= (2.75, 3.21)

Question 5

Dr. Patton is a professor of English. Recently she counted the number of misspelled works in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a standard deviation of 2.44 words per essay. For her Tuesday class of 50 students, the mean number of misspelled words per essay was 6.05. Construct a 90% confidence interval for the mean number of misspelled words in the population of student essays.

Accepted Answer

The answer of Dr. Patton is a professor of English....

Question 6

i. The test statistic t has n-1 degrees of freedom. ii. All t distributions have the same mean of zero and a standard deviation of 1.
iii. The t distribution is more spread out and flatter at the center than is the standard normal distribution. However, as the sample size increases, the t distribution curve approaches the standard normal distribution.

Accepted Answer

- Statement (i) is correct because the degrees of freedom for a t-test are indeed n-1, where n is the sample size.- Statement (ii) is incorrect because while all t distributions have a mean of zero, they do not all have a standard deviation of 1. The standard deviation of a t distribution varies depending on the degrees of freedom and is greater than 1.- Statement (iii) is correct as it accurately describes the shape of the t distribution compared to the standard normal distribution and its convergence to the normal distribution as sample size increases.

Question 7

i. The t distribution is based on the assumption that the population of interest is normal or nearly normal. ii. The t distribution is a discrete distribution.
iii. There is not one t distribution, but rather a "family" of t distributions.

Accepted Answer

The t distribution does assume that the population is normal or nearly normal, and there is indeed a family of t distributions, each defined by its degrees of freedom. However, the t distribution is not discrete; it is a continuous distribution.

Question 8

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard error of $20. If 90% and 95% confidence intervals were developed to estimate the true cost of the DVD, what similarities would they have?

Accepted Answer

Both the 90% and 95% confidence intervals would use the same point estimate (the average price of $375) and the same standard error ($20) in their calculations. The Z-values (which determine the width of the confidence interval) would differ between the two confidence levels.

Question 9

A survey of 25 grocery stores revealed that the average price of a 4-litre bag of milk was $2.98 with a standard error of $0.10. If 90% and 95% confidence intervals were developed to estimate the true cost of a 4-litre bag of milk, what similarities would they have?

Accepted Answer

Both 90% and 95% confidence intervals would use the same point estimate ($2.98) and the same standard error ($0.10). The difference between the two confidence intervals would be the t-statistic used, which varies depending on the confidence level and degrees of freedom.

Question 10

i. The test statistic for a problem involving an unknown population standard deviation is the Student's t distribution. ii. The t distribution approaches the Z distribution as the sample size increases.
iii. As the sample size increases, the computed value of t decreases.

Accepted Answer

(i) When the population standard deviation is unknown, the Student's t distribution is used as the test statistic. (ii) The t distribution does indeed approach the Z distribution as the sample size increases, becoming practically the same at large sample sizes. (iii) The statement about the computed value of t decreasing as the sample size increases is a misunderstanding; the value of t depends on the sample data and the hypothesis being tested, not directly on the sample size. However, the critical values of t (for a given significance level) approach those of the Z distribution as sample size increases, which might be the source of confusion.

Question 11

i. The Student t distribution has a greater spread than does the z distribution. As a result, the critical values of t for a given level of significance are larger in magnitude than the corresponding z critical values. ii. The test statistic t has n-1 degrees of freedom.
iii. William S. Gosset, a brewmaster, developed the t test for the Guinness Brewery in Ireland, who published it in 1908 using the pen name "Student."

Accepted Answer

(i) is correct because the Student t distribution has a greater spread than the z distribution, so its critical values are larger in magnitude for a given level of significance.
(ii) is correct because the test statistic t has n-1 degrees of freedom, which is a fundamental property of the t-distribution.
(iii) is correct because William S. Gosset, who was a brewmaster, developed the t test for the Guinness Brewery in Ireland and published it in 1908 using the pen name "Student."

Question 12

Dr. Patton is a professor of English. Recently she counted the number of misspelled works in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a standard deviation of 2.44 words per essay. For her Tuesday class of 50 students, the mean number of misspelled words per essay was 6.05. Construct a 95% confidence interval for the mean number of misspelled words in the population of student essays.

Accepted Answer

The 95% confidence interval for the mean number of misspelled words can be calculated using the formula for the confidence interval of the mean with known standard deviation and sample size: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$, 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 ($\bar{x} = 6.05$, $\sigma = 2.44$, $n = 50$), we get the confidence interval as 5.374 to 6.726.

Question 13

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard error of $20. What is the 99% confidence interval to estimate the true cost of the DVD?

Accepted Answer

The 99% confidence interval is given by:x̄ ± z* (SE)where x̄ is the sample mean, z is the z-score corresponding to the desired confidence level, and SE is the standard error.In this case, x̄ = $375, SE = $20, and z = 2.576 (for a 99% confidence level).Therefore, the 99% confidence interval is:$375 ± 2.576 * $20 = $323.40 to $426.60

Question 14

Recently, a university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates returned the survey. The average salary was $55,000. The population standard deviation is known to be $2,500. What is the 95% confidence interval for the mean salary of all graduates from the English Department?

Accepted Answer

The 95% confidence interval is calculated using the formula: mean ± (Z * (σ/√n)). Here, the mean is $55,000, Z is 1.96 for 95% confidence, σ is $2,500, and n is 400. This results in the interval $54,755 to $55,245.

Question 15

A survey of 25 grocery stores revealed that the average price of a 4-litre bag of milk was $2.98 with a standard error of $0.10. What is the 95% confidence interval to estimate the true cost of a 4-litre bag of milk?

Accepted Answer

The 95% confidence interval is calculated as the sample mean ± (Z-score * standard error). For a 95% confidence level, the Z-score is approximately 1.96. Thus, the interval is $2.98 ± (1.96 * $0.10), which is $2.78 to $3.18.

Question 16

Dr. Patton is a professor of English. Recently she counted the number of misspelled works in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a standard deviation of 2.44 words per essay. For her Tuesday class of 44 students, the mean number of misspelled words per essay was 6.05. Construct a 90% confidence interval for the mean number of misspelled words in the population of student essays.

Accepted Answer

The answer of Dr. Patton is a professor of English....

Question 17

A survey of 144 retail stores revealed that the average price of a DVD was $375 with a standard error of $20. If 95% and 98% confidence intervals were developed to estimate the true cost of the DVD, what differences would they have?

Accepted Answer

The differences between the 95% and 98% confidence intervals would be in the interval widths and the Z-values. The Z-value changes because it is dependent on the level of confidence chosen; a 95% confidence interval uses a Z-value of approximately 1.96, while a 98% confidence interval uses a higher Z-value, typically around 2.33. This change in Z-value affects the width of the confidence interval, making the 98% confidence interval wider than the 95% confidence interval. The standard error remains the same in both cases as it is a measure of the variability of the sample mean and is not affected by the confidence level.

Question 18

Recently, a university surveyed recent graduates of the English Department for their starting salaries. Four hundred graduates returned the survey. The average salary was $55,000. The population standard deviation is known to be $2,500. What is the 90% confidence interval for the mean salary of all graduates from the English Department?

Accepted Answer

We will use the formula: 
Confidence interval = sample mean ± (z-score x (population standard deviation / square root of sample size))
Since the sample size is large (n = 400), we can assume normal distribution and use the z-score for a 90% confidence level, which is 1.645.

Confidence interval = 55,000 ± (1.645 x (2,500 / √400))
Confidence interval = 55,000 ± (1.645 x 125)
Confidence interval = 55,000 ± 205.625
Confidence interval = (55,000 - 205.625, 55,000 + 205.625)
Confidence interval = (54,794.375, 55,205.625)

Therefore, the answer is B, 54,794 to $55,206.

Question 19

Dr. Patton is a professor of English. Recently she counted the number of misspelled works in a group of student essays. She noted the distribution of misspelled words per essay followed the normal distribution with a standard deviation of 2.44 words per essay. For her Tuesday class of 50 students, the mean number of misspelled words per essay was 6.05. Construct a 99% confidence interval for the mean number of misspelled words in the population of student essays.

Accepted Answer

To construct a 99% confidence interval for the mean, we will use the formula:

CI = X̄ ± z*(σ/√n)

Where X̄ is the sample mean (6.05), σ is the population standard deviation (2.44), n is the sample size (50), and z is the critical value for a 99% confidence interval (2.576).

Plugging in the values, we get:

CI = 6.05 ± 2.576*(2.44/√50)
CI = 6.05 ± 0.693
CI = (5.357, 6.743)

So the best choice is B, which is the interval 5.160 to 6.940.

Question 20

i. The t distribution is positively skewed. ii. All t distributions have the same mean of zero and a standard deviation of 1.
iii. The t distribution is more spread out and flatter at the center than is the standard normal distribution. However, as the sample size increases, the t distribution curve approaches the standard normal distribution.

Accepted Answer

- The t distribution is symmetric and not positively skewed, so (i) is incorrect.- The t distribution does have a mean of zero, but its standard deviation is not 1; it varies depending on the degrees of freedom, so (ii) is incorrect.- Statement (iii) is correct; the t distribution is indeed more spread out and flatter at the center compared to the standard normal distribution, and it does approach the standard normal distribution as the sample size (and thus degrees of freedom) increases.

Quiz 8: Estimation and Confidence Intervals