MATH 230 Chapter 9 STATS and PROB Assignment 2016

1) Consider the following.
At most 60% of Americans vote in presidential elections.
State the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the appropriate parameter (? or p).


H0: ? < 60, Ha: ? > 60

H0: p = 0.60, Ha: p > 0.60

H0: p < 0.60, Ha: p > 0.60

H0: ? ? 60, Ha: ? > 60

H0: p ? 0.60, Ha: p > 0.60

H0: ? > 60, Ha: ? ? 60

H0: ? = 60, Ha: ? > 60

H0: p > 0.60, Ha: p ? 0.60


2) Consider the following.
Fewer than 5% of adults in Los Angeles ride the bus to work.
State the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the appropriate parameter (? or p).


H0: p < 0.05, Ha: p ? 0.05

H0: ? = 5, Ha: ? ? 5

H0: p ? 0.05, Ha: p > 0.05

H0: ? ? 5, Ha: ? > 5

H0: ? < 5, Ha: ? ? 5

H0: ? ? 5, Ha: ? < 5

H0: p = 0.05, Ha: p ? 0.05

H0: p ? 0.05, Ha: p < 0.05


3Consider the following.
The average number of cars a person owns in his or her lifetime is not more than 10.
State the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the appropriate parameter (? or p).


H0: p = 10, Ha: p ? 10

H0: p ? 10, Ha: p < 10

H0: p ? 10, Ha: p > 10

H0: ? < 10, Ha: ? ? 10

H0: ? ? 10, Ha: ? < 10

H0: ? ? 10, Ha: ? > 10

H0: p < 10, Ha: p ? 10

H0: ? = 10, Ha: ? ? 10

4) Consider the following.
Europeans have an average paid vacation each year of six weeks.
State the null hypothesis, H0, and the alternative hypothesis, Ha, in terms of the appropriate parameter (? or p).


H0: ? = 6, Ha: ? ? 6

H0: ? ? 6, Ha: ? > 6

H0: p ? 6, Ha: p < 6

H0: p ? 6, Ha: p = 6

H0: p = 6, Ha: p ? 6

H0: p ? 6, Ha: p > 6

H0: ? ? 6, Ha: ? < 6

H0: ? ? 6, Ha: ? = 6


5) Consider the following.
At most 60% of Americans vote in presidential elections. State the Type I and Type II errors in complete sentences.
Type I error: We believe that more than 60% of Americans vote in presidential elections when it is really at most 60%. Type II error: We believe that at most 60% of Americans vote in presidential elections when it is really more than 60%.
Type I error: We believe that less than 60% of Americans vote in presidential elections when it really equals 60%. Type II error: We believe that 60% of Americans vote in presidential elections when it is really less than 60%.
Type I error: We believe that 60% of Americans vote in presidential elections when it is really less than 60%. Type II error: We believe that less than 60% of Americans vote in presidential elections when it really equals 60%.
Type I error: We believe that at most 60% of Americans vote in presidential elections when it is really more than 60%. Type II error: We believe that more than 60% of Americans vote in presidential elections when it is really at most 60%.


6) Consider the following.
Fewer than 5% of adults in Los Angeles ride the bus to work.
State the Type I and Type II errors in complete sentences.
Type I error: We believe that at least 5% of adults in Los Angeles ride the bus to work when the percentage really is fewer than 5%. Type II error: We believe that fewer than 5% of adults in Los Angeles ride the bus to work when the percentage really is greater than or equal to 5%.
Type I error: We believe that the percentage of adults in Los Angeles who ride the bus to work is not 5% when it really is 5%. Type II error: We believe that the percentage of adults in Los Angeles who ride the bus to work is 5% when it really is not 5%.
Type I error: We believe that fewer than 5% of adults in Los Angeles ride the bus to work when the percentage really is greater than or equal to 5%. Type II error: We believe that at least 5% of adults in Los Angeles ride the bus to work when the percentage really is fewer than 5%. Type I error: We believe that the percentage of adults in Los Angeles who ride the bus to work is 5% when it really is not 5%. Type II error: We believe that the percentage of adults in Los Angeles who ride the bus to work is not 5% when it really is 5%.


7) Consider the following.
The average number of cars a person owns in his or her lifetime is not more than 10.
State the Type I and Type II errors in complete sentences.
Type I error: We believe that the average number of cars a person owns in his or her lifetime is not more than 10 when it really is more than 10. Type II error: We believe that the average number of cars a person owns in his or her lifetime is more than 10 when it really is not more than 10.
Type I error: We believe that the average number of cars a person owns in his or her lifetime is more than 10 when it really is not more than 10. Type II error: We believe that the average number of cars a person owns in his or her lifetime is not more than 10 when it really is more than 10.
Type I error: We believe that the average number of cars a person owns in his or her lifetime is not 10 when it really is 10. Type II error: We believe that the average number of cars a person owns in his or her lifetime is 10 when it really is not 10.
Type I error: We believe that the average number of cars a person owns in his or her lifetime is 10 when it really is not 10. Type II error: We believe that the average number of cars a person owns in his or her lifetime is not 10 when it really is 10.


8) Consider the following.
About half of Americans prefer to live away from cities, given the choice.
State the Type I and Type II errors in complete sentences.
Type I error: We believe that the percentage of Americans who, given the choice, prefer to live away from cities is around 50% when it really is not around 50%. Type II error: We believe that the percentage of Americans who, given the choice, prefer to live away from cities is not around 50% when it really is around 50%.
Type I error: We believe that the percentage of Americans who, given the choice, prefer to live away from cities is more than 50% when it really is less than 50%. Type II error: We believe that the percentage of Americans who, given the choice, prefer to live away from cities is less than 50% when it really is more than 50%.
Type I error: We believe that the percentage of Americans who, given the choice, prefer to live away from cities is not around 50% when it really is around 50%. Type II error: We believe that the percentage of Americans who, given the choice, prefer to live away from cities is around 50% when it really is not around 50%.
Type I error: We believe that the percentage of Americans who, given the choice, prefer to live away from cities is less than 50% when it really is more than 50%. Type II error: We believe that the percentage of Americans who, given the choice, prefer to live away from cities is more than 50% when it really is less than 50%.


9) A particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000. A survey of owners of that tire design is conducted. Of the 26 tires in the survey, the average lifespan was 46,700 miles with a standard deviation of 9800 miles. Do the data support the claim at the 5% level?
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

• Part (a)
State the null hypothesis.
H0: ? < 50,000
H0: ? ? 50,000
H0: ? = 50,000
H0: ? ? 50,000

• Part (b)
State the alternative hypothesis.
Ha: ? = 50,000
Ha: ? ? 50,000
Ha: ? ? 50,000
Ha: ? < 50,000

• Part (c)
In words, state what your random variable X represents.
X represents the average life span, in years, of the tires surveyed.
X represents the life span, in miles, for a tire.
X represents the number of tires surveyed.
X represents the average life span, in miles, of the tires surveyed.

• Part (d)
State the distribution to use for the test. (Round your standard deviation to two decimal places.)
X ~ __ ( __, __ )
Part (e)
What is the test statistic? (Round your answer to two decimal places.)
=
• Part (f)
What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem.
If H0 is true, then there is a chance equal to the p-value that the average life span of a tire is not 46,700 miles or less.
If H0 is false, then there is a chance equal to the p-value that the average life span of a tire is not 46,700 miles or less.
If H0 is false, then there is a chance equal to the p-value that the average life span of a tire is 46,700 miles or less.
If H0 is true, then there is a chance equal to the p-value that the average life span of a tire is 46,700 miles or less.
• Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

• Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
? =
(ii) Decision:
reject the null hypothesis do not reject the null hypothesis

(iii) Reason for decision:
Since ? > p-value, we do not reject the null hypothesis.
Since ? < p-value, we do not reject the null hypothesis.
Since ? > p-value, we reject the null hypothesis.
Since ? < p-value, we reject the null hypothesis.

(iv) Conclusion:
There is sufficient evidence to conclude that the average life span of the tires is less than 50,000.
There is not sufficient evidence to conclude that the average life span of the tires is less than 50,000.

• Part (i)
Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to the nearest whole number.)


10) From generation to generation, the average age when smokers first start to smoke varies. However, the standard deviation of that age remains constant at around 2.1 years. A survey of 45 smokers of this generation was done to see if the average starting age is at least 19. The sample average was 18.2 with a sample standard deviation of 1.3. Do the data support the claim at the 5% level?
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)


• Part (a)
State the null hypothesis.
H0: ? ? 19
H0: ? = 19
H0: ? ? 19
H0: ? < 19

• Part (b)
State the alternative hypothesis.
Ha: ? < 19
Ha: ? = 19
Ha: ? ? 19
Ha: ? ? 19

• Part (c)
In words, state what your random variable X represents.
X represents the average age when smokers first start to smoke.
X represents the average age of the sample of smokers.
X represents the average number of cigarettes for each smoker.
X represents the age of a person when he or she first began smoking.

• Part (d)
State the distribution to use for the test. (Round your standard deviation to four decimal places.)
X~ __ ( __ , __ )
Part (e)
What is the test statistic? (Round your answer to two decimal places.)
=
• Part (f)
What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem.
If H0 is false, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is not 18.2 years or less.
If H0 is false, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is 18.2 years or less.
If H0 is true, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is 18.2 years or less.
If H0 is true, then there is a chance equal to the p-value that the average age of people when they first begin to smoke is not 18.2 years or less.
• Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

• Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
? =
(ii) Decision:
reject the null hypothesis
do not reject the null hypothesis

(iii) Reason for decision:
Since ? > p-value, we reject the null hypothesis.
Since ? < p-value, we reject the null hypothesis.
Since ? > p-value, we do not reject the null hypothesis.
Since ? < p-value, we do not reject the null hypothesis.

(iv) Conclusion:
There is sufficient evidence to conclude that the starting age for smoking in this generation is lower than 19.
There is not sufficient evidence to conclude that the starting age for smoking in this generation is lower than 19.

• Part (i)
Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to two decimal places.)


11) The cost of a daily newspaper varies from city to city. However, the variation among prices remains steady with a standard deviation of 6¢. A study was done to test the claim that the average cost of a daily newspaper is 35¢. Ten costs yield an average cost of 30¢ with a standard deviation of 4¢. Do the data support the claim at the 1% level?
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

• Part (a)
State the null hypothesis.
H0: ? < 35
H0: ? ? 35
H0: ? = 35
H0: ? ? 35
• Part (b)
State the alternative hypothesis.
Ha: ? = 35
Ha: ? ? 35
Ha: ? ? 35
Ha: ? < 35
• Part (c)
In words, state what your random variable X represents.
X represents how much the cost of a daily newspaper varies from the average cost of all daily newspapers.
X represents the number of cities that publish daily newspapers.
X represents the average cost of a daily newspaper.
X represents the cost of a daily newspaper.
• Part (d)
State the distribution to use for the test. (Round your standard deviation to four decimal places.)
X~ __ ( __ , __ )
Part (e)
What is the test statistic? (Round your answer to two decimal places.)
=
• Part (f)
What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem.
If H0 is false, then there is a chance equal to the p-value that the average cost of a daily newspaper is 30¢ or less OR 40¢ or more.
If H0 is false, then there is a chance equal to the p-value that the average cost of a daily newspaper is not 30¢ or less OR 40¢ or more.
If H0 is true, then there is a chance equal to the p-value that the average cost of a daily newspaper is 30¢ or less OR 40¢ or more.
If H0 is true, then there is a chance equal to the p-value that the average cost of a daily newspaper is not 30¢ or less OR 40¢ or more.
• Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

• Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
? =
(ii) Decision:
reject the null hypothesis
do not reject the null hypothesis
(iii) Reason for decision:
Since ? > p-value, we reject the null hypothesis.
Since ? < p-value, we do not reject the null hypothesis.
Since ? < p-value, we reject the null hypothesis.
Since ? > p-value, we do not reject the null hypothesis.

(iv) Conclusion:
There is sufficient evidence to conclude that the average cost of a daily newspaper is not equal to 35¢.
There is not sufficient evidence to conclude that the the average cost of a daily newspaper is not equal to 35¢.

• Part (i)
Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to two decimal places.)


12) The average work week for engineers in a start-up company is believed to be about 60 hours. A newly hired engineer hopes that it's shorter. She asks 10 engineering friends in start-ups for the lengths of their average work weeks. Based on the results that follow, should she count on the average work week to be shorter than 60 hours? Conduct a hypothesis test at the 5% level.
Data (length of average work week): 70; 45; 55; 60; 65; 60; 55; 60; 50; 55.
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

• Part (a)
State the null hypothesis.
H0: ? ? 60
H0: ? < 60
H0: ? ? 60
H0: ? ? 60
• Part (b)
State the alternative hypothesis.
Ha: ? < 60
Ha: ? ? 60
Ha: ? = 60
Ha: ? > 60
• Part (c)
In words, state what your random variable X represents.
X represents the average number of engineers needed in a start-up company.
X represents the number of hours an engineer works in one week.
X represents the number of engineers in a start-up company.
X represents the average length of a work week for engineers.

• Part (d)
State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom.): ____
• Part (e)
What is the test statistic? (Round your answer to two decimal places.)
=
• Part (f)
What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem.
If H0 is true, then there is a chance equal to the p-value that the average work week for engineers in a start-up business is not the mean of the sample or less.
If H0 is false, then there is a chance equal to the p-value that the average work week for engineers in a start-up business is the mean of the sample or less.
If H0 is true, then there is a chance equal to the p-value that the average work week for engineers in a start-up business is the mean of the sample or less.
If H0 is false, then there is a chance equal to the p-value that the average work week for engineers in a start-up business is not the mean of the sample or less.
• Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

• Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
? =
(ii) Decision:
reject the null hypothesis
do not reject the null hypothesis

(iii) Reason for decision:
Since ? > p-value, we do not reject the null hypothesis.
Since ? < p-value, we reject the null hypothesis.
Since ? > p-value, we reject the null hypothesis.
Since ? < p-value, we do not reject the null hypothesis.

(iv) Conclusion:
There is sufficient evidence to conclude that the average number of hours engineers work each week is less than 60 hours.
There is not sufficient evidence to conclude that the average number of hours engineers work each week is less than 60 hours.

• Part (i)
Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to two decimal places.)


13) According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100:114 (46.7% girls). Suppose you don't believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 60 girls and 90 boys born of the 150. Based on your study, do you believe that the percent of girls born in China is 46.7? Conduct a hypothesis test at the 5% level.
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

• Part (a)
State the null hypothesis.
H0: p = 0.467
H0: p ? 0.467
H0: p ? 0.467
H0: p ? 0.467
• Part (b)
State the alternative hypothesis.
Ha: p < 0.467
Ha: p > 0.467
Ha: p = 0.467
Ha: p ? 0.467
• Part (c)
In words, state what your random variable P' represents.
P' represents the ratio of girls to boys in China.
P' represents the number of girls born in China.
P' represents the percent of girls born in China.
P' represents the percent of boys born in China.
• Part (d)
State the distribution to use for the test. (Round your standard deviation to four decimal places.)
P' ~ ___ ( __ , __)
Part (e)
What is the test statistic? (Round your answer to two decimal places.)
=
• Part (f)
What is the p-value? (Round your answer to four decimal places.)



Explain what the p-value means for this problem.
If H0 is true, then there is a chance equal to the p-value that the sample ratio is not 60 out of 150 or less OR 80 out of 150 or more.
If H0 is false, then there is a chance equal to the p-value that the sample ratio is not 60 out of 150 or less OR 80 out of 150 or more.
If H0 is false, then there is a chance equal to the p-value that the sample ratio is 60 out of 150 or less OR 80 out of 150 or more.
If H0 is true, then there is a chance equal to the p-value that the sample ratio is 60 out of 150 or less OR 80 out of 150 or more.
• Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

• Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
? =
(ii) Decision:
reject the null hypothesis
do not reject the null hypothesis

(iii) Reason for decision:
Since ? > p-value, we do not reject the null hypothesis.
Since ? < p-value, we do not reject the null hypothesis.
Since ? > p-value, we reject the null hypothesis.
Since ? < p-value, we reject the null hypothesis.

(iv) Conclusion:
There is sufficient evidence to conclude that the percent of girls born in China is not equal to 46.7%.
There is not sufficient evidence to conclude that the percent of girls born in China is not equal to 46.7%.

• Part (i)
Construct a 95% confidence interval for the true proportion. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to four decimal places.)


14) Toastmasters International cites a February 2001 report by Gallup Poll that 40% of Americans fear public speaking.† A student believes that less than 40% of students at her school fear public speaking. She randomly surveys 361 schoolmates and finds that 135 report they fear public speaking. Conduct a hypothesis test at the 5% level to determine if the percent at her school is less than 40%.
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)
• Part (a)
State the null hypothesis.
H0: p < 0.40
H0: p ? 0.40
H0: p ? 0.40
H0: p ? 0.40

• Part (b)
State the alternative hypothesis.
Ha: p ? 0.40
Ha: p = 0.40
Ha: p ? 0.40
Ha: p < 0.40
• Part (c)
In words, state what your random variable P' represents.
P' represents the average number of students who fear public speaking.
P' represents the proportion of students who will never speak in public.
P' represents the number of students who fear public speaking.
P' represents the proportion of students who fear public speaking.
• Part (d)
State the distribution to use for the test. (Round your standard deviation to four decimal places.)
P' ~ __ ( __ , __ )
• Part (e)
What is the test statistic? (Round your answer to two decimal places.)
=
• Part (f)
What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem.
If H0 is true, then there is a chance equal to the p-value that 135 or less out of 361 students fear public speaking.
If H0 is false, then there is a chance equal to the p-value that 135 or less out of 361 students fear public speaking.
If H0 is false, then there is a chance equal to the p-value that 135 or less out of 361 students do not fear public speaking.
If H0 is true, then there is a chance equal to the p-value that 135 or less out of 361 students do not fear public speaking.
• Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

• Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
? =
(ii) Decision:
reject the null hypothesis do not reject the null hypothesis

(iii) Reason for decision:
Since ? < p-value, we do not reject the null hypothesis.
Since ? < p-value, we reject the null hypothesis.
Since ? > p-value, we do not reject the null hypothesis.
Since ? > p-value, we reject the null hypothesis.

(iv) Conclusion:
There is sufficient evidence to conclude that the percent of Americans that are afraid of public speaking is less than 40%.
There is not sufficient evidence to conclude that the percent of Americans that are afraid of public speaking is less than 40%.

• Part (i)
Construct a 95% confidence interval for the true proportion. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to four decimal places.)


15) An article in the San Jose Mercury News stated that students in the California state university system take an average of 4.5 years to finish their undergraduate degrees. Suppose you believe that the average time is longer. You conduct a survey of 40 students and obtain a sample mean of 5.1 with a sample standard deviation of 1.2. Do the data support your claim at the 1% level?
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

• Part (a)
State the null hypothesis.
H0: ? ? 4.5
H0: ? ? 4.5
H0: ? < 4.5
H0: ? ? 4.5
• Part (b)
State the alternative hypothesis.
Ha: ? ? 4.5
Ha: ? = 4.5
Ha: ? > 4.5
Ha: ? ? 4.5
• Part (c)
In words, state what your random variable X represents.
X represents the average number of students in the California university system.
X represents the average length of time it takes students to finish their undergraduate degrees.
X represents the length of time it takes a student to finish his or her undergraduate degree.
X represents the average number of students who receive an undergraduate degree in California.
• Part (d)
State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom.) : ¬____
• Part (e)
What is the test statistic? (Round your answer to two decimal places.)
=
• Part (f)
What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem.
If H0 is true, then there is a chance equal to the p-value the average time needed to complete an undergraduate degree is not 5.1 years or more.
If H0 is false, then there is a chance equal to the p-value that the average time needed to complete an undergraduate degree is 5.1 years or more.
If H0 is true, then there is a chance equal to the p-value that the average time needed to complete an undergraduate degree is 5.1 years or more.
If H0 is false, then there is a chance equal to the p-value the average time needed to complete an undergraduate degree is not 5.1 years or more.
• Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

• Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
? =
(ii) Decision:
reject the null hypothesis do not reject the null hypothesis

(iii) Reason for decision:
Since ? < p-value, we do not reject the null hypothesis.
Since ? > p-value, we do not reject the null hypothesis.
Since ? < p-value, we reject the null hypothesis.
Since ? > p-value, we reject the null hypothesis.

(iv) Conclusion:
There is sufficient evidence to conclude that the average time it takes to finish the undergraduate degrees is longer than 4.5 years.
There is not sufficient evidence to conclude that the average time it takes to finish the undergraduate degrees is longer than 4.5 years.
• Part (i)
Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to two decimal places.)


16) In 1955, Life Magazine reported that a 25-year-old mother of three worked (on average) an 80 hour week. Recently, many groups have been studying whether or not the women's movement has, in fact, resulted in an increase in the average work week for women (combining employment and at-home work). Suppose a study was done to determine if the average work week has increased. 74 women were surveyed with the following results. The sample average was 83; the sample standard deviation was 10. Does it appear that the average work week has increased for women at the 5% level?
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

• Part (a)
State the null hypothesis.
H0: ? ? 80
H0: ? < 80
H0: ? ? 80
H0: ? ? 80
• Part (b)
State the alternative hypothesis.
Ha: ? = 80
Ha: ? ? 80
Ha: ? > 80
Ha: ? ? 80
• Part (c)
In words, state what your random variable X represents.
X represents the number of hours a woman works in a week.
X represents the number of women who work over 80 hours a week.
X represents the average number of women who work over 80 hours a week.
X represents the average number of hours women work in one week.

• Part (d)
State the distribution to use for the test. (Enter your answer in the form z or tdf where df is the degrees of freedom.) : ¬¬¬____
• Part (e)
What is the test statistic? (Round your answer to two decimal places.)
=
• Part (f)
What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem.
If H0 is false, then there is a chance equal to the p-value that the average number of hours women work each week is 83 hours. or more
If H0 is false, then there is a chance equal to the p-value that the average number of hours women work each week is not 83 hours or more.
If H0 is true, then there is a chance equal to the p-value that the average number of hours women work each week is not 83 hours or more.
If H0 is true, then there is a chance equal to the p-value that the average number of hours women work each week is 83 hours or more.
• Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value.

• Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
? =
(ii) Decision:
reject the null hypothesis do not reject the null hypothesis

(iii) Reason for decision:
Since ? < p-value, we do not reject the null hypothesis. Since ? < p-value, we reject the null hypothesis. Since ? > p-value, we do not reject the null hypothesis. Since ? > p-value, we reject the null hypothesis.

(iv) Conclusion:
There is sufficient evidence to conclude that the average number of hours women work each week is more than 80 hours.
There is not sufficient evidence to conclude that the average number of hours women work each week is more than 80 hours.
• Part (i)
Construct a 95% confidence interval for the true mean. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your lower and upper bounds to two decimal places.)

17) In 2004, 68% of online courses taught at community colleges nationwide were taught by full-time faculty. To test if 68% also represents California's percent for full-time faculty teaching the online classes, Long Beach City College (LBCC), CA, was randomly selected for comparison. In 2004, 34 of the 44 online courses LBCC offered were taught by full-time faculty. Conduct a hypothesis test at the 5% level to determine if 68% represents CA.†
Note: If you are using a Student's t-distribution for the problem, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

• Part (a)
State the null hypothesis.
H0: p = 0.68
H0: p ? 0.68
H0: p ? 0.68
H0: p ? 0.68
• Part (b)
State the alternative hypothesis.
Ha: p = 0.68
Ha: p > 0.68
Ha: p ? 0.68
Ha: p < 0.68
• Part (c)
In words, state what your random variable P' represents.
P' represents the average number of California's online classes taught by full-time faculty.
P' represents the proportion of all faculty members that teach California's online classes.
P' represents the number of California's online classes taught by full-time faculty. P' represents the proportion of California's online classes taught by full-time faculty.
• Part (d)
State the distribution to use for the test. (Round your standard deviation to four decimal places.)
P' ~ __ ( __, __ )

• Part (e)
What is the test statistic? (Round your answer to two decimal places.)
=
• Part (f)
What is the p-value? (Round your answer to four decimal places.)

Explain what the p-value means for this problem.
If H0 is true, then there is a chance equal to the p-value that the proportion of online courses taught by full-time faculty is at least as different as the sample proportion is from 68%.
If H0 is true, then there is a chance equal to the p-value that the proportion of online courses taught by full-time faculty is not at least as different as the sample proportion is from 68%.
If H0 is false, then there is a chance equal to the p-value that the proportion of online courses taught by full-time faculty is at least as different as the sample proportion is from 68%.
If
H0
is false, then there is a chance equal to the p-value that the proportion of online courses taught by full-time faculty is not at least as different as the sample proportion is from 68%.

• Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis and shade the region(s) corresponding to the p-value. (Upload your file below.)

This answer has not been graded yet.

• Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis), the reason for it, and write an appropriate conclusion.
(i) Alpha:
? =
(ii) Decision:
reject the null hypothesis do not reject the null hypothesis

(iii) Reason for decision:
Since ? < p-value, we reject the null hypothesis. Since ? < p-value, we do not reject the null hypothesis. Since ? > p-value, we do not reject the null hypothesis. Since ? > p-value, we reject the null hypothesis.

(iv) Conclusion:
There is sufficient evidence to conclude that the percent of online courses taught at community colleges in the state is not equal to 68%.
There is not sufficient evidence to conclude that the percent of online courses taught at community colleges in the state is not equal to 68%.
• Part (i)
Construct a 95% confidence interval for the true proportion. Sketch the graph of the situation. Label the point estimate and the lower and upper bounds of the confidence interval. (Round your answers to four decimal places.)