1. The mean income per person in the United States is $38,500, and the distribution of incomes follows a normal distribution. A random sample of 14 residents of Wilmington, Delaware, had a mean of $45,500 with a standard deviation of $9,400. At the 0.050 level of significance, is that enough evidence to conclude that residents of Wilmington, Delaware, have more income than the national average?
  1. State the null hypothesis and the alternate hypothesis.
  1. State the decision rule for 0.050 significance level.(Round your answer to 3 decimal places.)
  1. Compute the value of the test statistic.(Round your answer to 2 decimal places.)
  1. Is there enough evidence to substantiate that residents of Wilmington, Delaware, have more income than the national average at the 0.050 significance level?

2.According to the Census Bureau, 3.36 people reside in the typical American household. A sample of 25 households in Arizona retirement communities showed the mean number of residents per household was 2.71 residents. The standard deviation of this sample was 1.10 residents. At the .10 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.36 persons?

  1. State the null hypothesis and the alternate hypothesis.(Round your answer to 2 decimal places.)
  1. State the decision rule for .10 significance level.(Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
  1. Compute the value of the test statistic.(Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
  1. Is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.36 persons?

3..In order to conduct a hypothesis test of the population mean, a random sample of 12 observations is drawn from a normally distributed population. The resulting mean and the standard deviation are calculated as 14.7 and 2.0, respectively. UseTable 2.
Use the critical value approach to conduct the following tests atα= 0.05.
H0:μ≤ 14.0 againstHA:μ> 14.0
a-1. / Calculate the value of the test statistic.(Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
Test statistic
a-2. / Calculate the critical value.(Round your answer to 3 decimal places.)
Critical value
a-3. / What is the conclusion?
RejectH0since the value of the test statistics is smaller than the critical value.
RejectH0since the value of the test statistics is greater than the critical value.
Do not rejectH0since the value of the test statistics is smaller than the critical value.
Do not rejectH0since the value of the test statistics is greater than the critical value.
H0:μ= 14.0 againstHA:μ≠ 14.0
b-1. / Calculate the value of the test statistic.(Round your intermediate calculations to 4 decimal places and final answer to 2 decimal places.)
Test statistic
b-2. / Calculate the critical value(s).(Round your answers to 3 decimal places.)
Critical value(s) / ±
b-3. / What is the conclusion?
RejectH0since the value of the test statistics is greater than the critical value.
RejectH0since the value of the test statistics is smaller than the critical value.
Do not rejectH0since the value of the test statistics is smaller than the critical value.
Do not rejectH0since the value of the test statistics is greater than the critical value.

4. What is the probability of making a Type II error if the null hypothesis is actually true?

Multiple choice:00.05 1α

5. In hypothesis testing, what is the level of significance?

  • Multiple choice:

It is selected before a decision rule can be formulated.

A value between 0 and 1.

The risk of rejecting the null hypothesis when it is true.

A value symbolized by the Greek letterα.

All of the answers apply.

.

6. Given the following hypotheses:

H0: μ ≤ 10

H1: μ > 10

A random sample of 10 observations is selected from a normal population. The sample mean was 19 and the sample standard deviation 3.4. Using the 0.05 significance level:

  1. State the decision rule.(Round your answer to 3 decimal places.)
  1. Compute the value of the test statistic.(Negative answers should be indicated by a minus sign. Round your answer to 3 decimal places.)
  1. What is your decision regarding the null hypothesis?

7. Given the following hypotheses:

H0: μ = 470

H1: μ ≠ 470

A random sample of 13 observations is selected from a normal population. The sample mean was 475 and the sample standard deviation 9. Using the 0.05 significance level:

  1. State the decision rule.(Negative amount should be indicated by a minus sign. Round your answers to 3 decimal places.)
  1. Compute the value of the test statistic.(Round your answer to 3 decimal places.)
  1. What is your decision regarding the null hypothesis?

Do not reject

Reject

8. The American Water Works Association reports that the per capita water use in a single-family home is 74 gallons per day. Legacy Ranch is a relatively new housing development. The builders installed more efficient water fixtures, such as low-flush toilets, and subsequently conducted a survey of the residences. Twenty-five owners responded, and the sample mean water use per day was 69 gallons with a standard deviation of 8.3 gallons per day.

At the 0.05 level of significance, is that enough evidence to conclude that residents of Legacy Ranch use less water on average?

  1. What is the decision rule?(Negative amount should be indicated by a minus sign. Round your answer to 3 decimal places.)
  1. Compute the value of the test statistic.(Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
  1. What is your decision regardingH0?

Reject

Fail to reject

9. A hypothesis regarding the weight of newborn infants at a community hospital is that the mean is 6.6 pounds. A sample of seven infants is randomly selected and their weights at birth are recorded as 9.0, 7.3, 6.0, 8.8, 6.8, 8.4, and 6.6 pounds. What is the alternate hypothesis?

Multiple Choice

Top of Form

H1:µ≥ 6.6

H1:µ >7.6

H1:µ= 6.6

H1:µ≠ 6.6

10.Bottom of Form

The mean length of a candy bar is 43 millimeters. There is concern that the settings of the machine cutting the bars have changed. Test the claim at the 0.02 level that there has been no change in the mean length. The alternate hypothesis is that there has been a change. Twelve bars (n= 12) were selected at random and their lengths in millimeters recorded. The lengths (in millimeters) are 42, 39, 42, 45, 43, 40, 39, 41, 40, 42, 43, and 42. The mean of the sample is 41.5 and the standard deviation is 1.784. If the computedt= −2.913, has there been a statistically significant change in the mean length of the bars?

Multiple Choice

Top of Form

No, because the computedtlies in the area to the right of −2.718.

Yes, because 43 is greater than 41.5.

No, because the information given is not complete.

Yes, because the computedtlies in the rejection region.

Bottom of Form

11. It is claimed that in a bushel of peaches, fewer than 10% are defective. A sample of 400 peaches is examined and 50 are found to be defective. What is the alternate hypothesis for a one-sided test?

Multiple Choice

Top of Form

H1:π> 0.10

H1:π≥ 0.10

H1:π≤ 0.10

H1:π< 0.10

12. The following hypotheses are given.

H0:π≤ 0.81
H1:π> 0.81

13. A sample of 80 observations revealed thatp= 0.95. At the 0.01 significance level, can the null hypothesis be rejected?

  1. State the decision rule.(Round your answer to 2 decimal places.)
  1. Compute the value of the test statistic.(Round your answer to 2 decimal places.)
  1. What is your decision regarding the null hypothesis?

RejectH0.

Do not rejectH0.

14. Chicken Delight claims that 92% of its orders are delivered within 10 minutes of the time the order is placed. A sample of 80 orders revealed that 70 were delivered within the promised time. At the 0.01 significance level, can we conclude that less than 92% of the orders are delivered in less than 10 minutes?

  1. What is the decision rule?(Negative amount should be indicated by a minus sign. Round your answer to 2 decimal places.)
  1. Compute the value of the test statistic.(Negative amount should be indicated by a minus sign. Round the intermediate values and final answer to 2 decimal places.)
  1. What is your decision regarding the null hypothesis?

15. After a losing season, there is a great uproar to fire the head football coach. In a random sample of 350 college alumni, 147 favor keeping the coach. Test at the .02 level of significance whether the proportion of alumni who support the coach is less than 50 percent.
(a) / State the null hypothesis and the alternate hypothesis.(Round your answers to 2 decimal places.)
H0: π≥
H1: π
(b) / State the decision rule for .02 significance level.(Negative amount should be indicated by a minus sign.Round your answer to 2 decimal places.)
RejectH0ifzis <
(c) / Compute the value of the test statistic.(Negative amount should be indicated by a minus sign.Round your answer to 3 decimal places.)
Value of the test statistic
(d) / Test at the .02 level of significance whether the proportion of alumni who support the coach is less than 50 percent.
(Click to select) Do not reject Reject H0. There is (Click to select) sufficient insufficient evidence to conclude the population proportion
of alumni supporting the coach is less than .50.
16. According to an ABC News survey, 40 percent of Americans do not eat breakfast. A sample of 30 college students found 16 had skipped breakfast that day. Use the .01 significance level to check whether college students are more likely to skip breakfast.
(a) / State the null hypothesis and the alternate hypothesis.(Round your answers to 2 decimal places.)
H0:π≤
H1:π
(b) / State the decision rule for .01 significance level.(Round your answers to 3 decimal places.)
RejectH0ifzis >
(c) / Compute the value of the test statistic.(Round your answer to 2 decimal places.)
zvalue is
(d) / Is it reasonable to conclude that the data shows college students are more likely to skip breakfast? Use the .01 significance level.
(Click to select) Reject Do not reject H0. The data (Click to select) shows do not shows college students are more likely to skip
breakfast.

Bottom of Form

17. The U.S. Department of Transportation estimates that 10% of Americans carpool. Does that imply that 10% of cars will have two or more occupants? A sample of 300 cars traveling southbound on the New Jersey Turnpike yesterday revealed that 63 had two or more occupants. At the 0.01 significance level, can we conclude that 10% of cars traveling on the New Jersey Turnpike have two or more occupants?

  1. State the null hypothesis and the alternate hypothesis.(Round your answers to 2 decimal places.)
  1. State the decision rule.

H0is rejected if z < 2.326

H0is rejected if z > 2.326

  1. Compute the value of the test statistic.(Round your answer to 2 decimal places.)
  1. What is your decision regardingH0?

RejectH0

Do not rejectH0

  1. Can we conclude that 10% of cars traveling on the New Jersey Turnpike have two or more occupants? Yes or No