Question 1

The average natural gas bill for a random sample of 21 homes in the 19808 zip code during the month of March was $311.90 with a sample standard deviation of $51.60. The 90% confidence interval around this sample mean is ------.

Accepted Answer

The answer of The average natural gas bill for a...

Question 2

Verizon Wireless would like to estimate the proportion of households that use cell phones for their phone service without a land line. A random sample of 150 households was selected and 48 relied strictly on cell phones for their service. Using this sample, the point estimate for the proportion of households without land lines is ------------.

Accepted Answer

The point estimate for the proportion of households without land lines is calculated by dividing the number of households relying strictly on cell phones (48) by the total number of households sampled (150), which equals 0.32.

Question 3

According to the Entertainment Software Association, the average age of "gamers" (those that routinely play video games)is 30 years old. A random sample of 33 gamers people had an average age of 32.7 years old. Assume that the population standard deviation for the age of gamers is 4.6 years. Use a 90% confidence interval to test the validity of this report and choose the one statement that is correct.

Accepted Answer

CTo calculate the 90% confidence interval for the mean age of gamers, we use the formula for a confidence interval when the population standard deviation is known: $\bar{x} \pm z \frac{\sigma}{\sqrt{n}}$, where $\bar{x}$ is the sample mean, $z$ is the z-score corresponding to the desired confidence level, $\sigma$ is the population standard deviation, and $n$ is the sample size.Given: $\bar{x} = 32.7$, $\sigma = 4.6$, $n = 33$, and for a 90% confidence interval, the z-score is approximately 1.645.The margin of error (ME) is calculated as: $ME = 1.645 \times \frac{4.6}{\sqrt{33}} \approx 1.32$.Thus, the 90% confidence interval is $32.7 \pm 1.32$, which is approximately $31.38$ to $34.02$.Since 30 is not within this interval, the report by the Entertainment Software Association stating the average age is 30 years old is not validated by this sample data. However, the correct interpretation of a confidence interval not including a certain value (in this case, 30) does not directly validate or invalidate the report; it suggests that, based on this sample, the average age might be different from 30. Therefore, the correct statement should reflect the evidence provided by the confidence interval regarding the report's claim, but due to the nature of the question and the options provided, the closest accurate choice is that the confidence interval does not include 30 years, which would lean towards the report not being validated based on this sample data.

Question 4

"Gamers" are described as people who regularly play video games. A random sample of 80 people was selected and 50 were found to be male. The approximate standard error of the proportion for this sample is ________ .

Accepted Answer

The standard error of the proportion is calculated using the formula SE = sqrt(p(1-p)/n), where p is the sample proportion and n is the sample size. Here, p = 50/80 = 0.625, so SE = sqrt(0.625*0.375/80) ≈ 0.0541.

Question 5

Which one of the following statements is not true regarding the characteristics of the Student's t-d istribution?

Accepted Answer

The Student's t-distribution is actually wider or more spread out than the normal distribution, especially with fewer degrees of freedom. As the degrees of freedom increase, it approaches the shape of the normal distribution.

Question 6

When calculating a confidence interval for proportions, we assume that proportion data follow the

Accepted Answer

When calculating a confidence interval for proportions, we assume that the proportion data follow the binomial distribution, which describes the number of successes in a sequence of independent experiments.

Question 7

A statistics professor has 115 students in a statistics class and would like to estimate the number of hours each student studied for the last exam. A random sample of 24 students was found to study an average of 6.8 hours with a standard deviation of 2.2 hours. The 90% confidence interval to estimate the average number of hours studying for the exam is ________.

Accepted Answer

The 90% confidence interval is calculated using the formula: mean ± (t-value * (standard deviation/√n)). With a sample size of 24, the t-value for 90% confidence is approximately 1.711. Plugging in the values: 6.8 ± (1.711 * (2.2/√24)) gives the interval (6.11, 7.49).

Question 8

A survey by American Express Spending reported that the average amount spent on a wedding gift for a close family member is $166. A random sample of 45 people who purchased wedding gifts for a close family member spent an average of $160.50. Assume that the population standard deviation is $38. Use a 95% confidence interval to test the validity of this report and choose the one statement that is correct.

Accepted Answer

To test the validity of the report, we calculate the 95% confidence interval for the mean amount spent on a wedding gift using the sample data. The formula for a confidence interval is $\bar{x} \pm z\frac{\sigma}{\sqrt{n}}$, 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 population standard deviation, and $n$ is the sample size.Plugging in the given values: $160.50 \pm 1.96\frac{38}{\sqrt{45}}$.Calculating the margin of error: $1.96 \times \frac{38}{\sqrt{45}} \approx 11.07$.Thus, the 95% confidence interval is $160.50 \pm 11.07$, which is approximately $149.43$ to $171.57$.Since $166 falls within this interval, it supports the validity of the American Express Spending report.

Question 9

"Gamers" are described as people who regularly play video games. A random sample of 120 people was selected and 66 were found to be male. Using this sample, the point estimate for the proportion of gamers who are male is ________ .

Accepted Answer

The answer of "Gamers" are described as people who regularly...

Question 10

The National Center for Education would like to estimate the average student debt for the state of Ohio with a 90% confidence interval and a margin of error within plus or minus $1,800. Assuming the standard deviation for student debt in this state is $7,500, the required sample size is _________ .

Accepted Answer

The formula for sample size is: n = (Z^2 * s^2) / e^2, where Z is the z-score for the desired confidence level, s is the standard deviation, and e is the margin of error. Plugging in the given values, we get: n = (1.645^2 * 7500^2) / 1800^2 = 47.02. Rounding up to the nearest whole number, the required sample size is 47.

Question 11

The average natural gas bill for a random sample of 21 homes in the 19808 zip code during the month of March was $311.90 with a sample standard deviation of $51.60. The approximate standard error of the mean for this sample is _____ .

Accepted Answer

The answer of The average natural gas bill for a...

Question 12

A random sample of 60 business students required an average of 50.7 minutes to complete a statistics exam. Assume that the population standard deviation to complete the exam was 10.4 minutes. The 98% confidence interval around this sample mean is____ .

Accepted Answer

To calculate the 98% confidence interval for the mean, 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, $\sigma$ is the population standard deviation, and $n$ is the sample size. For a 98% confidence level, the Z-score is approximately 2.33. Plugging in the values: $50.7 \pm 2.33 \times \frac{10.4}{\sqrt{60}}$, we get the interval (47.6, 53.8).

Question 13

The Girl Scouts would like to estimate the proportion of people who consider Thin Mints to be their favorite Girl Scout cookie. The total sample size needed to construct a 90% confidence interval for the proportion of customers who favor Thin Mint cookies with a margin of error equal to 6% is .

Accepted Answer

To calculate the sample size for a proportion with a given margin of error, we use the formula n = (Z^2 * p * (1-p)) / E^2. For a 90% confidence interval, Z is approximately 1.645. Assuming p = 0.5 for maximum variability, and E = 0.06, the calculation gives n ≈ 188.

Question 14

The National Center for Education would like to estimate the average student debt for the state of Ohio with a 95% confidence interval and a margin of error within plus or minus $2,500. Assuming the standard deviation for student debt in this state is $8,000, the required sample size is _______.

Accepted Answer

The answer of The National Center for Education would like...

Question 15

A______ confidence interval might be most appropriate during the early stages of a research project when little is known about the population's parameters and the project's researchers might still be gauging the parameters.

Accepted Answer

A 90% confidence interval is most appropriate during the early stages of a research project because it is less precise than a 95% or 99% confidence interval, but it is also more likely to be accurate when the population parameters are unknown.

Question 16

Dr. Jordan operates a dental practice with 315 patients and she would like to estimate the proportion of her patients that have dental insurance from their employers. A random sample of 75 patients found that 30 had insurance coverage. The 90% confidence interval to estimate the proportion of patients with dental insurance is _______.

Accepted Answer

The answer of Dr. Jordan operates a dental practice with...

Question 17

The Retail Advertising and Marketing Association would like to estimate the average amount of money that a person spends for Mother's Day with a 99% confidence interval and a margin of error within plus or minus $12. Assuming the standard deviation for spending on Mother's Day is $36, the required sample size is _________.

Accepted Answer

The answer of The Retail Advertising and Marketing Association would...

Question 18

A _______ confidence interval is often used in scientific or medical research when making a wrong conclusion about a population parameter is very costly.

Accepted Answer

A 99% confidence interval is often used in contexts where the cost of making a wrong conclusion is very high, as it provides a higher level of confidence (and thus, a lower risk of error) than the more commonly used 95% confidence interval.

Question 19

A random sample of 150 mortgages in the state of Mississippi was randomly selected. From this sample, 17 were found to be delinquent on their current payment. The 98% confidence interval for the proportion based on this sample is ______.

Accepted Answer

The answer of A random sample of 150 mortgages in...

Question 20

The average natural gas bill for a random sample of 26 homes in the 19808 zip code during the month of March was $305.30 with a sample standard deviation of $46.50. The margin of error for a 98% confidence interval around this sample mean is -----.

Accepted Answer

To calculate the margin of error for a 98% confidence interval, we use the formula: $ME = Z * (\frac{\sigma}{\sqrt{n}})$, where $Z$ is the Z-score corresponding to the confidence level, $\sigma$ is the sample standard deviation, and $n$ is the sample size. For a 98% confidence level, the Z-score is approximately 2.33 (since the Z-score for 99% confidence level is 2.576, and for 95% confidence level is 1.96, 98% would be in between but closer to 2.576). Plugging in the values: $ME = 2.33 * (\frac{46.50}{\sqrt{26}})$, we get approximately $22.66 as the margin of error.

Quiz 8: Nutrition