A simple random sample with n=56 provided a sample mean of 22.5 and a sample standard deviation of 4.4

a.Develop a 90% confidence interval for the populationmean.

b.Develop a 95% confidence interval for the populationmean.

c.Develop a 99% confidence interval for the populationmean.

d.What happens to the margin of error and the confidenceinterval as the confidence level is increased?

Below are sets of three variables. On the lines after each set, do the following: Write a hypotheses relating the first two variables. Identify independent and dependent variables. State how you expect the third variable to affect the hypothesized relationship. Draw an arrow diagram including all three variables. Determine whether the third variable is antecedent, intervening, or alternative. Primary caregiver for children (primary caregiver, not primary caregiver); support for Family Medical Leave Law (thermometer scale for support); gender (female, male) Intention to vote in upcoming election; respondent’s general interest in politics; predicted outcome of election (“too close to call,” “somewhat competitive,” “lopsided victory”) Type of lightbulb purchased (regular, energy efficient); difference in cost of regular and high-efficiency lightbulbs (small difference, large difference); concern about global climate change. The only answer given is that there aren't three variables. I really need help with this question.

To better understand how husbands and wives feel about their finances, a magazine conducted a national poll of 1,006 married adults age 25 and older with household incomes of $50,000 or more. Consider the following example set of responses to the question "Who is better at getting deals?"

(a)

Develop a joint probability table and use it to answer the following questions. (Round your answers to four decimal places.)

(b)

According to the marginal probabilities, what is the most likely response?

I ammy spouse    we are equal

(c)

Given that the respondent is a husband, what is the probability that he feels he is better at getting deals than his wife? (Round your answer to four decimal places.)

(d)

Given that the respondent is a wife, what is the probability that she feels she is better at getting deals than her husband? (Round your answer to four decimal places.)

(e)

Given a response "My spouse" is better at getting deals, what is the probability that the response came from a husband? (Round your answer to four decimal places.)

(f)

Given a response "We are equal," what is the probability that the response came from a husband?

What is the probability that the response came from a wife?

True or False The critical value is the number of standard errors on either side of the sample proportion.

True or False The margin of error is smaller for a 95% confidence interval than for a 90% confidence interval.

True or False Type II Error is caused by rejecting the null hypothesis when the null hypothesis was actually true.

True or False Type I and Type II errors are related. The only way to decrease both is to decrease the sample size.

True or False A small P-value indicates that the observation obtained is improbable given the null hypothesis and thus provides evidence against the null hypothesis.

True or False The purpose of hypothesis testing is to prove our belief for a given scenario.

True or False A Type I error is always worse than a Type II error.

True or False The alternative hypothesis, HA, contains the values of the parameter we consider plausible when we reject the null hypothesis, H0.

According to a survey in a​ country, 15 ​% of adults do not own a credit card. Suppose a simple random sample of 200 adults is obtained. Complete parts​ (a) through​ (d) below.

​(a) Describe the sampling distribution of ModifyingAbove p with caret ​, the sample proportion of adults who do not own a credit card. Choose the phrase that best describes the shape of the sampling distribution of ModifyingAbove p with caret below.

A. Approximately normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis greater than or equals 10

B. Approximately normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis less than 10

C. Not normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis less than 10

D. Not normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis greater than or equals 10 Determine the mean of the sampling distribution of ModifyingAbove p with caret . mu Subscript ModifyingAbove p with caret Baseline equals nothing ​(Round to two decimal places as​ needed.)

Determine the standard deviation of the sampling distribution of ModifyingAbove p with caret . sigma Subscript ModifyingAbove p with caret equalsnothing ​(Round to three decimal places as​ needed.)

​(b) What is the probability that in a random sample of 200 ​adults, more than 17 ​% do not own a credit​ card?

The probability is    ? (Round to four decimal places as​ needed.)

Interpret this probability.

If 100 different random samples of __ 200 adults were​ obtained, one would expect ____ to result in more than 17% not owning a credit card.

​(Round to the nearest integer as​ needed.)

​(c) What is the probability that in a random sample of 200 adults, between 12% and 17​% do not own a credit​ card?

The probability is ____? ​(Round to four decimal places as​ needed.)

Interpret this probability.

If 100 different random samples of 200 adults were​ obtained, one would expect ____ to result in between 12% and 17% not owning a credit card.

​(Round to the nearest integer as​ needed.)

​(d) Would it be unusual for a random sample of 200 adults to result in 24 or fewer who do not own a credit​ card? Why? Select the correct choice below and fill in the answer box to complete your choice.

​(Round to four decimal places as​ needed.)

A.The result is not unusual because the probability that ModifyingAbove p with caret is less than or equal to the sample proportion is ___ which is greater than​ 5%.

B.The result isunusual because the probability that ModifyingAbove p with caret is less than or equal to the sample proportion is _____ which is less than​ 5%.

C.The result is notunusual because the probability that ModifyingAbove p with caretis less than or equal to the sample proportion is _____ which is less than​ 5%.

D.The result is unusual because the probability that ModifyingAbove p with caret is less than or equal to the sample proportion is _____ which is greaterthan​ 5%.

Suppose a mutual fund qualifies as having moderate risk if the standard deviation of its monthly rate of return is less than 3​%. A​ mutual-fund rating agency randomly selects 24 months and determines the rate of return for a certain fund. The standard deviation of the rate of return is computed to be 2.68​%. Is there sufficient evidence to conclude that the fund has moderate risk at the alpha equals 0.01 level of​ significance? A normal probability plot indicates that the monthly rates of return are normally distributed.

X^2 = ? (Round to three decimal places as needed.)

Use technology to determine the P-value for the test statistic

What is the correct conclusion at the a = 0.01 level of significance?

10. In a study conducted to determine whether the role that sleep disorders play in academic performance, researcher conducted a survey of 1800 college students to determine if they had a sleep disorder. Of the 500 students with a sleep disorder, the mean GPA was 2.51 with a standard deviation of 0.85. Of the 1300 students without a sleep disorder, the mean GPA is 2.85 with a standard deviation of 0.78. Test the claim that sleep disorder adversely affects one’s GPA at the 0.05 level of significance?

11. In one experiment, the participant must press a key on seeming a blue screen and reaction time (in seconds) to press the key is measured. The same person is then asked to press a key on seeing a red screen, again with reaction time measured. The results for six randomly sampled study participants are as follows:

Participant

1

2

3

4

5

6

Blue

0.582

0.481

0.841

0.267

0.685

0.450

Red

0.408

0.407

0.542

0.402

0.456

0.522

Construct a 99% confidence interval about the population mean difference. Assume the differences are approximately normally distributed.

HOMEWORK 3 (part3):

3) You now wish to concern yourself with a comparison of the proportions of the supporters of the candidate based upon gender concerning their assertions of party loyalty. Specifically, you wish to know whether the proportion of men that supports the candidate that describes itself as party loyalists is less than the proportion of women that feels the same. The sample data concerning whether the supporter describes himself or herself as a party loyalist is also shown in appendix two below. At both the 2% and 5% levels of significance, is the proportion of male supporters that describes itself as party loyalists less than the proportion of female supporters of the candidate that describes itself as party loyalists? If the procedure you have chosen for this problem allows it (using PHStat) to construct confidence intervals for the difference in the proportions of male and female supporters that describes itself as party loyalists, construct 98% and 95% confidence intervals for the difference in the proportions, and explain their meanings in the context of the problem.

Appendix Two:

Male Supporter Loyalty? (Y = party loyalist, N = not a party loyalist)

Y   N         Y         Y         Y         N         Y         N         Y         N         Y         N

Y   Y         Y         Y         Y         Y         Y         Y         N         Y         Y         Y

Y   Y         N         Y         Y         Y

Female Supporter Loyalty? (Y = party loyalist, N = not a party loyalist)

Y   Y         Y         Y         N         Y         N         Y         Y         Y         N         Y

Y   Y         N         Y         Y         Y         Y         Y         Y         Y         Y         Y

Y   Y         Y         Y         Y         Y

Suppose that you are testing the hypotheses H0​: p=0.18 vs. HA​: p=/ 0.18. A sample of size 150 results in a sample proportion of 0.25.

​a) Construct a 99​% confidence interval for p. ​

b) Based on the confidence​ interval, can you reject H0 at a =0.01​? Explain.

​c) What is the difference between the standard error and standard deviation of the sample​ proportion?

​d) Which is used in computing the confidence​ interval?

The average time to run the 5K fun run is 20 minutes and the standard deviation is 2.4 minutes. 8 runners are randomly selected to run the 5K fun run. Round all answers to 4 decimal places where possible and assume a normal distribution.

What is the distribution of X X ? X X ~ N(,)

What is the distribution of ¯ x x¯ ? ¯ x x¯ ~ N(,)

What is the distribution of ∑ x ∑x ? ∑ x ∑x ~ N(,)

If one randomly selected runner is timed, find the probability that this runner's time will be between 20.3272 and 21.1272 minutes. For the 8 runners, find the probability that their average time is between 20.3272 and 21.1272 minutes.

Find the probability that the randomly selected 8 person team will have a total time more than 152.8.

For part e) and f), is the assumption of normal necessary? NoYes

The top 15% of all 8 person team relay races will compete in the championship round. These are the 15% lowest times. What is the longest total time that a relay team can have and still make it to the championship round? minutes

A component is purchased from 3 suppliers, A, B, and C, where the suppliers have

respective defective rates of 2%, 6%, and 4%. Of all the components purchased, 20%

comes from supplier A, 50% from supplier B, and 30% from supplier C, that is, each

shipment comes from each of these suppliers with these probabilities. The company uses

the following quality control policy. A sample of 15 units is randomly selected from each

shipment of components. If at most 2 defective units are found in the sample, then the

entire shipment is accepted; otherwise, the entire shipment is rejected.

Determine the following:

(a-1) probability that a shipment will be accepted given that it came from Supplier A.

(a-2) probability that a shipment will be accepted given that it came from Supplier B.

(a-3) probability that a shipment will be accepted given that it came from Supplier C.

(a-4) Using the law of total probability and your responses to parts (a-1)-(a-3) above,

determine the probability that a shipment will be accepted.

(b-1) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier A.

(b-2) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier B.

(b-3) expected number of defectives in a random sample of 15 units, if the shipment came

from Supplier C.

(b-4) Using the law of total expectation and your responses to parts (b-1)-(b-3) above,

determine the expected number of defectives in a random sample of 15 units.

A restaurant manager asks all of the customers this month to take an online survey (and get a free appetizer their next visit). It turns out that 241 customers actually take the survey, and of these 209 were ''extremely satisfied'' with their visit. Set up an approximate 95% confidence interval for the proportion of all the restaurant's customers that say they were ''extremely satisfied'' with their visit..

Select only one of the boxes below.

A. The Normal curve cannot be used to make the requested confidence interval.

B. Making the requested confidence interval does not make any sense.

C. It is appropriate to compute a confidence interval for this problem using the Normal curve.

If it is appropriate to compute a confidence interval for this problem using the Normal curve, then enter the confidence interval below. Otherwise, enter [0,0] for the confidence interval.

15.3-11. Management of the Telemore Company is considering developing
and marketing a new product. It is estimated to be twice
as likely that the product would prove to be successful as unsuccessful.
It it were successful, the expected profit would be
$1,500,000. If unsuccessful, the expected loss would be
$1,800,000. A marketing survey can be conducted at a cost of
$300,000 to predict whether the product would be successful. Past
experience with such surveys indicates that successful products
have been predicted to be successful 80 percent of the time, whereas
unsuccessful products have been predicted to be unsuccessful 70
percent of the time.
(a) Develop a decision analysis formulation of this problem by
identifying the alternative actions, the states of nature, and the
payoff table when the market survey is not conducted.
T (b) Assuming the market survey is not conducted, use Bayes’
decision rule to determine which decision alternative should
be chosen.
T (c) Find EVPI. Does this answer indicate that consideration
should be given to conducting the market survey?
T (d) Assume now that the market survey is conducted. Find the
posterior probabilities of the respective states of nature for
each of the two possible predictions from the market survey.
(e) Find the optimal policy regarding whether to conduct the market
survey and whether to develop and market the new product.

Variable Mean Median Q1 Q3 Two-Star 65.08 69.50 46.50 79.25 Three-Star 89.74 87.00 70.75 107.00 Four-Star 127.74 125.50 90.75 150.00 B. Variable Range Interquartile Range Variance Standard deviation Coefficient of variation Two - star 94.00 32.75 473.80 21.77 33.45 Three-star 126.00 36.25 761.66 27.60 30.75 Four-star 134.00 59.25 1583.60 39.79 31.15 Based on the results, what conclusions can you reach concerning these variables?

6. List below are the numbers of words spoken in a day by each member of six different couples.

Use a 0.05 significance level to test the claim that among couples, males speak more words in a

day than females. Assume that the population differences is normally distributed.

Male

5638

21,319

17,572

26,429

46,978

25,835

Female

5198

11,661

19,624

13,397

31,553

18,667

7. A sociologist wanted to estimate the difference in the amount of daily leisure time (in hours) of adults who do not have children under the age of 18 years and adults who have children under the age of 18 years. A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.62 hours, with a standard deviation of 2.43 hours. A random sample of 40 adults with children under the age of 18 years results in a mean daily leisure time of 4.10 hours, with a standard deviation of 1.82 hours. Construct a 95% confidence interval for the mean difference in leisure time between adults with no children and adults with children.