Question 1

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, not positively skewed, and does not always have a mean of zero and a standard deviation of 1. Statement (iii) correctly describes the t distribution's characteristics compared to the standard normal distribution.

Question 2

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 answer of A survey of 25 grocery stores revealed...

Question 3

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 confidence intervals would have the same point estimate ($2.98) and the same standard error ($0.10).

Question 4

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

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 99% confidence interval for the mean number of misspelled words in the population of student essays.

Accepted Answer

The 99% confidence interval is calculated using the formula: mean ± (z * (standard deviation/√n)). With a mean of 6.05, standard deviation of 2.44, n=50, and z-value for 99% confidence approximately 2.576, the interval is 5.161 to 6.939.

Question 6

How does the t distribution differ from the standard z distribution?

Accepted Answer

The t distribution is a family of distributions that vary based on degrees of freedom, unlike the single standard normal z distribution.

Question 7

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

The Student t distribution does have a greater spread than the z distribution, leading to larger critical values for a given significance level. The test statistic t indeed has n-1 degrees of freedom. William S. Gosset developed the t-test and published it under the pen name "Student."

Question 8

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 assumes a normal or nearly normal population and is a continuous distribution with a family of distributions based on degrees of freedom.

Question 9

i. The test statistic for a problem involving a small sample of fewer than 30 and 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

The t distribution is used for small samples with unknown population standard deviation, it approaches the Z distribution as sample size increases, and the computed value of t decreases as sample size increases.

Question 10

What kind of distribution is the t distribution?

Accepted Answer

The t distribution is a continuous probability distribution used in statistics, especially for small sample sizes.

Question 11

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= (2.7474, 3.2126)Therefore, the 98% confidence interval is $2.75 to $3.21, which is closest to choice A.

Question 12

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 95% confidence interval would be narrower than the 98% confidence interval because the 95% confidence interval has a lower level of confidence, which means that it is less likely to include the true population mean. The standard errors would be the same for both confidence intervals because the standard error is a measure of the variability of the sample mean, which is not affected by the level of confidence. The z-values would be different for the two confidence intervals because the z-value is used to determine the critical value for the confidence interval, and the critical value is different for different levels of confidence.

Question 13

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 t distribution assumes a normal or nearly normal population, is continuous, and consists of a family of distributions based on degrees of freedom.

Question 14

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

Accepted Answer

The t distribution is symmetric, has a mean of 0, and is based on degrees of freedom.

Question 15

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 16

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

The point estimates and standard errors would be the same for both confidence intervals.

Question 17

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 18

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 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 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 20

A sample of 20 is selected from the population. What is the number of degrees of freedom used to determine the appropriate critical t-value?

Accepted Answer

The degrees of freedom for a sample is calculated as the sample size minus one, so for a sample of 20, it is 20 - 1 = 19.

Quiz 8: Estimation and Confidence Intervals