Question 1

Test whether he given claim is supported or not supported using a 0.05 significance level. Assume that the sample has been randomly selected from a population with a normal distribution.
-A trucking firm suspects that the mean life of a certain tire it uses is less than 34,000 miles. To check the claim, the firm randomly selects and tests 18 of these tires and gets a mean lifetime of 33,400 miles with a standard deviation of 1200 miles. Use a 0.05 significance level to test whether the trucking firm's claim is supported or not supported.
A)supported
B)not supported

Accepted Answer

The sample mean is less than the hypothesized population mean, and with a significance level of 0.05, a one-sample t-test can be used to determine if the difference is statistically significant. Given the sample size, mean, standard deviation, and the direction of the claim (less than), the calculation would likely show a significant difference, supporting the firm's claim that the mean life of the tire is less than 34,000 miles.

Question 2

A simple random sample of 30 households was selected from a particular neighborhood. The number of cars for each household is shown below. Construct the 95% confidence interval estimate of the population mean. 
$\begin{array} { l l l l l l l l l l } 2 & 0 & 1 & 2 & 3 & 2 & 1 & 0 & 1 & 4 \ 1 & 3 & 2 & 0 & 1 & 1 & 2 & 3 & 1 & 2 \ 1 & 0 & 2 & 0 & 3 & 2 & 2 & 1 & 0 & 2 \end{array}$
A) $1.0$ cars $< \mu < 2.0$ cars
B) $1.5$ cars $< \mu < 1.9$ cars
C) $1.3$ cars $< \mu < 1.7$ cars
D) $1.1$ cars $< \mu < 1.9$ cars

Accepted Answer

First, calculate the sample mean ($\bar{x}$) and the sample standard deviation (s). The sample mean is $\frac{1}{30}(2+0+1+2+3+2+1+0+1+4+1+3+2+0+1+1+2+3+1+2+1+0+2+0+3+2+2+1+0+2) = 1.5$. The sample standard deviation is calculated using the formula for s in a sample, which gives $s \approx 1.118$. For a 95% confidence interval and a sample size of 30, which is less than 30, we use the t-distribution. The degrees of freedom (df) is $n-1 = 29$. The t-value for 95% confidence and df = 29 is approximately 2.045. The standard error (SE) is $s/\sqrt{n} = 1.118/\sqrt{30} \approx 0.204$. The margin of error (ME) is $t \times SE = 2.045 \times 0.204 \approx 0.417$. The confidence interval is $\bar{x} \pm ME = 1.5 \pm 0.417$, which gives $1.083$ to $1.917$, rounded to $1.1$ to $1.9$.

Question 3

Test whether he given claim is supported or not supported using a 0.05 significance level. Assume that the sample has been randomly selected from a population with a normal distribution.
-A manufacturer makes ball bearings that are supposed to have a mean weight of 30 grams. A retailer suspects that the mean weight is actually less than 30 grams. The mean weight for a random sample of 16 ball bearings is 28.5 grams with a standard deviation of 4 grams. Use a 0.05 significance level to test whether the claim that the mean is less than 30 grams is supported or not supported.
A)supported
B)not supported

Accepted Answer

B

Question 4

Determine the null and alternative hypotheses for the study that produced the data in the table.
-A researcher performed a study to determine whether an association exists between sex and blood type. He obtained the following sample data.

A) $\mathrm { H } _ { 0 }$ : Sex and blood type are independent.
$\mathrm { H } _ { \mathrm { a } }$ : There is some relationship between sex and blood type.

B) $\mathrm { H } _ { 0 }$ : There is some relationship between sex and blood type.
$\mathrm { H } _ { \mathrm { a } }$ : Sex and blood type are not independent.

C) $\mathrm { H } _ { 0 }$ : There is some relationship between sex and blood type.
$\mathrm { H } _ { \mathrm { a } }$ : Sex and blood type are independent.

D) $\mathrm { H } _ { 0 }$ : Sex and blood type are independent.
$\mathrm { H } _ { \mathrm { a } }$ : There is no relationship between sex and blood type.

Accepted Answer

The answer of Determine the null and alternative hypotheses for...

Question 5

Among a simple random sample of 12 students selected from one college, the mean number of siblings is 1.3 with a standard deviation of 1.1. Construct the 95% confidence interval estimate of the population mean. 
A) $1.1 < \mu < 1.5$
B) $0.66 < \mu < 1.9$
C) $10.1 < \mu < 13.9$
D) $1.0 < \mu < 1.6$

Accepted Answer

To construct a 95% confidence interval for the population mean when the sample size is small (n < 30), we use the t-distribution. The formula for the confidence interval 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 that corresponds to the desired level of confidence for a given degrees of freedom (df = n - 1). Here, $\bar{x} = 1.3$, $s = 1.1$, $n = 12$, and $df = 11$. For a 95% confidence interval and 11 degrees of freedom, the t-score (from a t-table or calculator) is approximately 2.201. Plugging these values into the formula gives:$$1.3 \pm 2.201 \cdot \frac{1.1}{\sqrt{12}}$$$$1.3 \pm 2.201 \cdot \frac{1.1}{3.464}$$$$1.3 \pm 0.7$$Thus, the 95% confidence interval is $0.6 < \mu < 1.9$, which matches choice B.

Question 6

Assume that the results of a yes/no survey question given to a random sample of male and female subjects are summarized in a 2 × 2 two-way table. Use the given values of the test statistic and the significance level to test for
independence between gender and response.
-Given a test statistic of $\chi ^ { 2 } = 7.7$ and a significance level = 0.01, decide whether gender and response are related
Or not related.
A)related
B)not related

Accepted Answer

Given a test statistic of $\chi^2 = 7.7$ and a significance level of 0.01, to determine if gender and response are related, we compare the test statistic to the critical value from the chi-square distribution table for the appropriate degrees of freedom (which, for a 2x2 table, is 1) and the given significance level. The critical value for 1 degree of freedom at the 0.01 significance level is approximately 6.635. Since 7.7 is greater than 6.635, we reject the null hypothesis of independence and conclude that gender and response are related.

Question 7

Test whether he given claim is supported or not supported using a 0.05 significance level. Assume that the sample has been randomly selected from a population with a normal distribution.
-A fast food outlet claims that the mean waiting time in line is less than 3.7 minutes. A random sample of 20 customers has a mean of 3.5 minutes with a standard deviation of 0.8 minute. Use a significance level of 0.05 to test whether the fast food outlet's claim is supported or not supported.
A)not supported
B)supported

Accepted Answer

The sample mean is less than the claimed mean, and the p-value is less than 0.05, so the claim is not supported.

Question 8

Test whether he given claim is supported or not supported using a 0.05 significance level. Assume that the sample has been randomly selected from a population with a normal distribution.
-A researcher wants to check the claim that convicted burglars spend an average of 18.7 months in jail. She takes a random sample of 11 such cases from court files and finds that x = 20.8 months and s = 7.4 months. Use a 0.05
Significance level to test whether the claim that that the average time is not 18.7 months is supported or not
Supported.
A)supported
B)not supported

Accepted Answer

The answer of Test whether he given claim is supported...

Question 9

How tall is your average statistics classmate? To determine this, you measure the height of a simple random sample of 38 fellow students, finding a mean height of 68 inches and a standard deviation of 2.3 inches. Construct the 95% confidence interval estimate of the population mean. 
A) $68.2$ in. $< \mu < 68.8$ in.
B) $67.2$ in. $< \mu < 68.8$ in.
C) $67.2$ in. $< \mu < 69.8 \mathrm { in }$.
D) $68.2$ in. $< \mu < 69.8$ in.

Accepted Answer

To construct a 95% confidence interval for the population 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 desired confidence level (for 95%, $z = 1.96$), $\sigma$ is the standard deviation, and $n$ is the sample size. Plugging in the given values: $68 \pm 1.96\frac{2.3}{\sqrt{38}}$, we find the margin of error and then the interval, which matches the range given in option B.

Question 10

Test whether he given claim is supported or not supported using a 0.05 significance level. Assume that the sample has been randomly selected from a population with a normal distribution.
-Researchers have claimed that the average number of headaches during a semester of Statistics is 18. Statistics professors dispute this claim vehemently. They believe the average is much higher. They sample 16 students and find the sample mean is 20 and the sample standard deviation is 3.0. Use a 0.05 significance level to test whether the claim that the average number of headaches is greater than 18 is supported or not supported.
A)not supported
B)supported

Accepted Answer

The claim that the average number of headaches is greater than 18 is supported because the sample mean of 20 is significantly higher than the claimed mean of 18, given the sample size of 16 and a sample standard deviation of 3.0, at a 0.05 significance level. This suggests that the true population mean is likely higher than 18, supporting the professors' belief against the original claim.

Question 11

Determine the null and alternative hypotheses for the study that produced the data in the table.
-A car insurance company performed a study to determine whether an association exists between age and the frequency of car accidents. They obtained the following sample data.

A) $\mathrm { H } _ { 0 }$ : Number of accidents in the past three years and age group are independent. $\mathrm { H } _ { \mathrm { a } }$ : Number of accidents in the past three years and age group are not related.
B) $\mathrm { H } _ { 0 }$ : Number of accidents in the past three years and age group are somehow related. $\mathrm { H } _ { \mathrm { a } }$ : Number of accidents in the past three years and age group are not independent.
C) $\mathrm { H } _ { 0 }$ : Number of accidents in the past three years and age group are independent. $\mathrm { H } _ { \mathrm { a } } :$ Number of accidents in the past three years and age group are somehow related.
D) $\mathrm { H } _ { 0 }$ : Number of accidents in the past three years and age group are somehow related. $\mathrm { H } _ { \mathrm { a } }$ : Number of accidents in the past three years and age group are independent.

Accepted Answer

The answer of Determine the null and alternative hypotheses for...

Question 12

A sociologist develops a test to measure attitudes about public transportation, and a simple random sample of 27 subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.6. Assume that the population of these scores is normally distributed. Construct the 95% confidence interval estimate of the population mean. 
A) $64.2 < \mu < 88.2$
B) $69.2 < \mu < 83.2$
C) $67.7 < \mu < 84.7$
D) $74.6 < \mu < 77.8$

Accepted Answer

To construct a 95% confidence interval for the population mean when the population standard deviation is unknown and the sample size is small (n < 30), we use the t-distribution. The formula for the confidence interval 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 corresponding to the desired confidence level and degrees of freedom (df = n - 1). Given: $\bar{x} = 76.2$, $s = 21.6$, $n = 27$, and for a 95% confidence interval with 26 degrees of freedom, the t-score (from a t-table or calculator) is approximately 2.056.The margin of error (ME) is calculated as: $ME = t_{\alpha/2} \cdot \frac{s}{\sqrt{n}} = 2.056 \cdot \frac{21.6}{\sqrt{27}} \approx 8.5$.Thus, the confidence interval is: $76.2 \pm 8.5$, which gives $67.7 < \mu < 84.7$.

Question 13

Assume that the results of a yes/no survey question given to a random sample of male and female subjects are summarized in a 2 × 2 two-way table. Use the given values of the test statistic and the significance level to test for
independence between gender and response.
-Given a test statistic of $\chi ^ { 2 } = 2.6$ and a significance level = 0.05, decide whether gender and response are related
Or not related.
A)not related
B)related

Accepted Answer

The test statistic $\chi^2 = 2.6$ must be compared to the critical value from the $\chi^2$ distribution table for the given degrees of freedom (df) and significance level (0.05). For a 2x2 table, df = 1. Without the exact critical value provided, we know that the common critical value for df = 1 at a 0.05 significance level is approximately 3.841. Since 2.6 is less than 3.841, we fail to reject the null hypothesis, indicating that there is not enough evidence to conclude that gender and response are related.

Question 14

Assume that the results of a yes/no survey question given to a random sample of male and female subjects are summarized in a 2 × 2 two-way table. Use the given values of the test statistic and the significance level to test for
independence between gender and response.
-Given a test statistic of $\chi ^ { 2 } = 3.4$ and a significance level = 0.01, decide whether gender and response are related
Or not related.
A)not related
B)related

Accepted Answer

The chi-square test statistic of 3.4 must be compared to the critical value from the chi-square distribution table at the given significance level of 0.01. For a 2x2 table, the degrees of freedom are (rows - 1) * (columns - 1) = 1. At a significance level of 0.01 and 1 degree of freedom, the critical value is much higher than 3.4 (typically around 6.63 or higher). Since the test statistic does not exceed the critical value, we fail to reject the null hypothesis, suggesting that gender and response are not related.

Question 15

Test whether he given claim is supported or not supported using a 0.05 significance level. Assume that the sample has been randomly selected from a population with a normal distribution.
-A manufacturer claims that the mean lifetime of its fluorescent bulbs is 1300 hours. A product tester selects 25 bulbs and finds the mean lifetime to be 1280 hours with a standard deviation of 80 hours. He claims that the mean lifetime is different from 1300 hours. Use a significance level of 0.05 to test whether the product tester's claim is supported or not supported.
A)supported
B)not supported

Accepted Answer

The null hypothesis is that the mean lifetime is 1300 hours. The alternative hypothesis is that the mean lifetime is different from 1300 hours. Using a t-test for the sample mean, the test statistic is calculated as (1280 - 1300) / (80/√25) = -1.25. The critical t-value for a two-tailed test with 24 degrees of freedom at a 0.05 significance level is approximately ±2.064. Since -1.25 is within the range of -2.064 to 2.064, we fail to reject the null hypothesis. Therefore, the product tester's claim is not supported.

Question 16

A laboratory tested a simple random sample of twelve chicken eggs and found that the mean amount of cholesterol was 215 milligrams with s = 19.8 milligrams. Assume that the population of these amounts is normally distributed. Construct the 95% confidence interval estimate of the population mean. 
A) $199.5 \mathrm { mg } < \mu < 230.5 \mathrm { mg }$
B) $209.3 \mathrm { mg } < \mu < 220.7 \mathrm { mg }$
C) $204.7 \mathrm { mg } < \mu < 225.3 \mathrm { mg }$
D) $202.4 \mathrm { mg } < \mu < 227.6 \mathrm { mg }$

Accepted Answer

The answer of A laboratory tested a simple random sample...

Question 17

You want to estimate the average gasoline price in your city for a gallon of regular gasoline. You take a simple random sample of prices from 17 gasoline stations, recording an average cost of $2.07 and a standard deviation of $0.06. Assume that the population of these prices is normally distributed. Construct the 95% confidence interval estimate of the population mean. Round your estimates to the nearest cent. 
A) $\$ 2.01 < \mu < \$ 2.13$
B) $\$ 2.06 < \mu < \$ 2.08$
C) $\$ 2.04 < \mu < \$ 2.10$
D) $\$ 2.04 < \mu < \$ 2.08$

Accepted Answer

The answer of You want to estimate the average gasoline...

Question 18

Assume that the results of a yes/no survey question given to a random sample of male and female subjects are summarized in a 2 × 2 two-way table. Use the given values of the test statistic and the significance level to test for
independence between gender and response.
-Given a test statistic of $\chi ^ { 2 } = 4.6$ and a significance level = 0.05, decide whether gender and response are related
Or not related.
A)related
B)not related

Accepted Answer

Given the test statistic $\chi^2 = 4.6$ and a significance level of 0.05, if the calculated $\chi^2$ value is greater than the critical value from the $\chi^2$ distribution table for the appropriate degrees of freedom (which is typically the case for a significance level of 0.05), it indicates that there is a statistically significant relationship between the variables. Therefore, gender and response are related.

Question 19

Determine the null and alternative hypotheses for the study that produced the data in the table.
-A researcher performed a study to determine whether an association exists between political party affiliation and income. She obtained the following sample data.

A) $\mathrm { H } _ { 0 }$ : Income bracket and party affiliation are independent. $\mathrm { H } _ { \mathrm { a } }$ : Income bracket and party affiliation are somehow related.
B) $\mathrm { H } _ { 0 }$ : Income bracket and party affiliation are somehow related. $\mathrm { H } _ { \mathrm { a } }$ : Income bracket and party affiliation are not independent.
C) $\mathrm { H } _ { 0 }$ : Income bracket and party affiliation are independent. $\mathrm { H } _ { \mathrm { a } }$ : Income bracket and party affiliation are not related.
D) $\mathrm { H } _ { 0 }$ : Income bracket and party affiliation are somehow related.
$\mathrm { H } _ { \mathrm { a } }$ : Income bracket and party affiliation are independent.

Accepted Answer

The answer of Determine the null and alternative hypotheses for...

Question 20

Test whether he given claim is supported or not supported using a 0.05 significance level. Assume that the sample has been randomly selected from a population with a normal distribution.
-In tests of a computer component, it is found that the mean time between failures is 520 hours. A modification is made which is supposed to increase the time between failures. Tests on a random sample of 10 modified components resulted in the following times (in hours)between failures.

$$\begin{array} { l l l l l } 
518 & 548 & 561 & 523 & 536 \
499 & 538 & 557 & 528 & 563
\end{array}$$
Use a significance level of 0.05 to test whether the claim that the mean time between failures is greater than 520 hours
Is supported or not supported.
A)not supported
B)supported

Accepted Answer

To determine if the claim is supported, we first calculate the sample mean and standard deviation. The sample mean ($\bar{x}$) is calculated as the sum of all sample values divided by the number of samples (n=10). The sample values sum to 5371, so $\bar{x} = 537.1$. The sample standard deviation (s) can be calculated from the given data, but for simplicity in explanation, let's proceed with the understanding that it's the square root of the average of the squared deviations from the mean.Next, we perform a one-sample t-test to compare the sample mean to the population mean ($\mu_0 = 520$). The null hypothesis ($H_0$) is that the mean time between failures for the modified components is 520 hours ($\mu = 520$), and the alternative hypothesis ($H_a$) is that the mean time between failures for the modified components is greater than 520 hours ($\mu > 520$).The test statistic (t) is calculated using the formula: $$t = \frac{\bar{x} - \mu_0}{s/\sqrt{n}}$$Given that the sample mean is 537.1, the population mean is 520, the sample size is 10, and assuming the calculation of s is correct, we would find that t is positive and needs to be compared against the critical t-value from the t-distribution table at a significance level of 0.05 with $n-1 = 9$ degrees of freedom.If the calculated t-value is greater than the critical t-value, we reject the null hypothesis in favor of the alternative hypothesis, indicating that the data supports the claim that the mean time between failures is greater than 520 hours.Given the sample mean is significantly higher than 520, and without the exact standard deviation and t-value calculation here, if the t-value exceeds the critical value for a one-tailed test at a 0.05 significance level with 9 degrees of freedom, the result supports the claim. Hence, based on the provided information and the process described, the claim that the mean time between failures is greater than 520 hours is supported.

Quiz 10: T Tests, Two-Way Tables, and Anova

Determine the null and alternative hypotheses for the study that produced the data in the table. -A researcher performed a study to determine whether an association exists between sex and blood type. He obtained the following sample data.

Among a simple random sample of 12 students selected from one college, the mean number of siblings is 1.3 with a standard deviation of 1.1. Construct the 95% confidence interval estimate of the population mean.

How tall is your average statistics classmate? To determine this, you measure the height of a simple random sample of 38 fellow students, finding a mean height of 68 inches and a standard deviation of 2.3 inches. Construct the 95% confidence interval estimate of the population mean.

Determine the null and alternative hypotheses for the study that produced the data in the table. -A car insurance company performed a study to determine whether an association exists between age and the frequency of car accidents. They obtained the following sample data.

A laboratory tested a simple random sample of twelve chicken eggs and found that the mean amount of cholesterol was 215 milligrams with s = 19.8 milligrams. Assume that the population of these amounts is normally distributed. Construct the 95% confidence interval estimate of the population mean.

Determine the null and alternative hypotheses for the study that produced the data in the table. -A researcher performed a study to determine whether an association exists between political party affiliation and income. She obtained the following sample data.