One of the questions in a study of marital satisfaction of dual-career couples was to rate the statement, "I'm pleased with the way we divide the responsibilities for childcare." The ratings went from 1 (strongly agree) to 5 (strongly disagree). The table below contains ten of the paired responses for husbands and wives. Conduct a hypothesis test at the 5% level to see if the mean difference in the husband's versus the wife's satisfaction level is negative (meaning that, within the partnership, the husband is happier than the wife).

Wife's score

3

2

2

3

4

2

1

1

2

4

Husband's score

2

2

1

3

2

1

1

1

2

4

NOTE: If you are using a Student'st-distribution for the problem, including for paired data, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)

Part 1) State the distribution to use for the test. (Enter your answer in the formzort_dfwheredfis the degrees of freedom.)

Part 2) What is the test statistic? (If using thezdistribution round your answer to two decimal places, and if using thetdistribution round your answer to three decimal places.)

z or t=_____

Part 3) What is thep-value? (Round your answer to four decimal places.)

2

Perhaps unsurprisingly, the events A, B, and C are dependent.That is, a congressperson having or not having one of these characteristics in fact, does affect the likelihood of having the other characteristics.Here, we will determine exactly how that affect happens.You do not need toa anycomputations for this section, the info has been either given or computed in the previous part.

Compare P( A | B ) with P( A ).Which is bigger?Summarize what this tells us.Fill in the specific information into the sentence structure:"If a member of congress is [characteristic B] this [increases/decreases] the probability that they are [characteristic A]."

Repeat this to compare P( B | A ) with P( B ).

Repeat this to compare P( B | C ) with P( B ).

Repeat this to compare P( C | B ) with P( C ).

Repeat this to compare P( C | A ) with P( C ).

Repeat this to compare P( A | C ) with P( A ).

3. Seattle Temperatures

Below are some links to histograms about Seattle Weather Data from the years 1949 - 2018.Throughout this assignment, assume that all of this data is approximately normally distributed.

Seattle Daily High Temperatures

Seattle Daily Low Temperatures

1. From the links above, locate the graph showing high temperatures in Seattle for your birthday month and locate the graph showing low temperatures in Seattle for your birthday month.

1. On each graph, sketch an approximate normal distribution onto the histogram, and use this to guess the mean and standard deviation of the temperature distribution.Do not do ny computation for this problem, I do really want you to guess.Use your pictures to show your reader how you guessed these values.

1. Use this website to find the high and low temperatures on your most recent birthday:https://www.timeanddate.com/weather/usa/seattle/historic

1. Use your estimates for the mean and standard deviation for the high temperature distribution to compute the z-score for the high temperature for your most recent birthday.

• Is the z-score a large number or a small number?
• Is it positive or negative?
• What does this tell you about the high temperature on your most recent birthday compared to historical data?

1. Repeat part d using your estimates for the low-temperature to compute and analyze the z-score for the low temperature on your most recent birthday.

3

A nutritionist wants to determine how much time nationally people spend eating and drinking. Suppose for a random sample of

1083

people age 15 or?older, the mean amount of time spent eating or drinking per day is

1.08

hours with a standard deviation of

0.75

hour. Complete parts?(a) through?(d) below.

?(a) A histogram of time spent eating and drinking each day is skewed right. Use this result to explain why a large sample size is needed to construct a confidence interval for the mean time spent eating and drinking each day.

A.

The distribution of the sample mean will always be approximately normal.

B.

The distribution of the sample mean will never be approximately normal.

C.

Since the distribution of time spent eating and drinking each day is not normally distributed?(skewed right), the sample must be large so that the distribution of the sample mean will be approximately normal.

Your answer is correct.

D.

Since the distribution of time spent eating and drinking each day is normally?distributed, the sample must be large so that the distribution of the sample mean will be approximately normal.

?(b) There are more than 200 million people nationally age 15 or older. Explain why?this, along with the fact that the data were obtained using a random?sample, satisfies the requirements for constructing a confidence interval.

A.

The sample size is greater than?5% of the population.

B.

The sample size is less than?5% of the population.

Your answer is correct.

C.

The sample size is less than?10% of the population.

D.

The sample size is greater than?10% of the population.

?(c) Determine and interpret a

90?%

confidence interval for the mean amount of time Americans age 15 or older spend eating and drinking each day.

Select the correct choice below and fill in the answer?boxes, if?applicable, in your choice.

?(Type integers or decimals rounded to three decimal places as needed. Use ascending?order.)

A.

The nutritionist is

90?%

confident that the amount of time spent eating or drinking per day for any individual is between

nothing

and

nothing

hours.

B.

There is a

90?%

probability that the mean amount of time spent eating or drinking per day is between

nothing

and

nothing

hours.

C.

The nutritionist is

90?%

confident that the mean amount of time spent eating or drinking per day is between

nothing

and

nothing

hours.

D.

The requirements for constructing a confidence interval are not satisfied.

4

Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below 6 mg/dl. Recently, the patient's total calcium tests gave the following readings (in mg/dl). Assume that the population ofxvalues has an approximately normal distribution.

10.1

9.2

10.9

9.5

9.4

9.8

10.0

9.9

11.2

12.1

(a)

Use a calculator with mean and sample standard deviation keys to find the sample mean readingand the sample standard deviations. (in mg/dl; round your answers to two decimal places.)

=mg/dl

s=mg/dl

(b)

Find a 99.9% confidence interval for the population mean of total calcium in this patient's blood. (in mg/dl; round your answer to two decimal places.)

lower limitmg/dl

upper limitmg/dl

(c)

Based on your results in part (b), do you think this patient still has a calcium deficiency? Explain.

Yes. This confidence interval suggests that the patient may still have a calcium deficiency.Yes. This confidence interval suggests that the patient no longer has a calcium deficiency.No. This confidence interval suggests that the patient may still have a calcium deficiency.No. This confidence interval suggests that the patient no longer has a calcium deficiency.

A random sample of nj = 16 communities in western Kansas gave the following rates of hay fever per 1000 population for people under 25 years of age. 121 115 124 99 134 121 110 116 113 96 116 116 135 96 96 116 A random sample of n2 = 14 communities in western Kansas gave the following rates of hay fever per 1000 population for people over 50 years old. 113 86 106 102 113 94 94 108 103 99 78 105 88 100 Assume that the hay fever rate in each age group has an approximately normal distribution. Using the method outlined in Brase and Brase, do the data indicate that the age group over 50 has a lower rate of hay fever? Use a -0,05. Do you reject or fail to reject the null hypothesis? Are the data statistically significant at the a - 0.05 level of significance? Since the p-value is greater than the level of significance, the data are not statistically significant. Thus, we fail to reject the null hypothesis. Since the p-value is less than the level of significance, the data are not statistically significant. Thus, we fail to reject the null hypothesis. Since the p-value is greater than the level of significance, the data are statistically significant. Thus, we fail to reject the null hypothesis. Since the p-value is less than the level of significance, the data are statistically significant. Thus, we reject the null hypothesis. Since the p-value is greater than the level of significance, the data are not statistically significant. Thus, we reject the null hypothesis.2. Suppose you ask a person at many random times to state how happy they are [on some scale}. These scores fora singie person are then averaged to get an average happiness snore [AHIL If you nd the AH score for different people. you'll curiously get different results, but let's suppose that the AH scores for random people are normaily distributed with mean 65 [on some particular happiness instrument}. a. Suppose you are told that the happiest 1% of people score 83 or above. What is the standard deviation of the AH distribution using this happiness instrument? b. A new happiness instrument is designed where AH is normallyI distributed with mean 50 and standard deviation 5. How marryr random peopie would you have to talk to before you met a total ofthree people who scored below 43 on this new instrument