1. While working for a polling company at C-Span, you are asked  to determine the percentage of the population that feel that the president has done a favorable job in handling the Pandemic.  You will take a SRS of adults in the USA.  We would like to be 95 % confident with a margin of error of 3 points.

1. If the last estimate done by FOX News was known to be 58%, what size of sample would be needed?   
2. Since you believe that the responses will be between 30% and 70% and you are not sure that previous estimate was accurate what estimate should you use? And what sample size would be needed?
3. If you would like to be 99% confident with a margin of error of 3 points and unsure of the last estimate, what sample size would be needed?
4. Comment about the difference between answers of b and c.  Also why would a polling company use 95% rather than 99%?

Write a report to your boss at C-Span, explaining to her what you are doing.

The health of college students is monitored by periodically weighing them in. A sample of 32 students has a mean weight of 145.9 lb. Assuming that the population standard deviation is known to be 81.2 lb., use a 0.10 significance level to test the claim that the population mean of all such students weights is different than 150 lb. WHAT ARE THE STEPS ON STATCRUNCH?

The average height of 49 randomly selected men is 175 cm with σ = 7.
When calculating a 95% confidence interval. The point estimate would be? The critical value? The standard error? The margin of error?

A certain type of cable has a mean breaking point of 159 pounds with a standard deviation of 6 pounds. What weight should be specified so that 95​% of the cables can be expected not to break supporting that​ weight?

8. High density lipoprotein (HDL) in healthy males follows a normal distribution with a mean of 50 and a standard deviation of 8. What proportion of healthy males has HDL exceeding 60?

A)0.50

B)1.00

C)0.894

D)0.106

In the recent Census, three percent of the U.S. population reported being of two or more races. However, the percent varies tremendously from state to state. Suppose that two random surveys are conducted. In the first random survey, out of 1,000 North Dakotans, only nine people reported being of two or more races. In the second random survey, out of 500 Nevadans, 17 people reported being of two or more races. Conduct a hypothesis test to determine if the population percents are the same for the two states or if the percent for Nevada is statistically higher than for North Dakota.

1. Find the p-value. (Round your answer to four decimal places.)

Not everyone pays the same price for the same model of a new car. Prices paid for a particular model of a new car take on a normal distribution. The mean is $17,000 and the standard deviation is $500. Start by drawing a picture of the normal distribution and then labeling this information.

a. What percentage of buyer paid between $16,000 and $18,000 for a new car?

b. What percentage of buyer paid between $17,000 and $18,000 for a new car?

c. What percentage of buyer paid between $16,500and $17,000 for a new car?

d. What percentage of buyer paid between $17,000 and $18,500 for a new car?

1. : The president of a company that manufactures car seats has been concerned about the number and cost of machine breakdowns. The problem is that the machines are old and becoming quite unreliable. However, the cost of replacing them is quite high, and the president is not certain that the cost can be made up in today’s slow economy. To help make a decision about replacement, he gathered data about last month’s costs for repairs in dollars (y) and the ages in months (x) of the plant’s 20 welding machines as recorded in repairs.mtw.

a) Find the estimated regression equation to predict the repair cost of a machine from its age.

b) Interpret the slope in the estimated regression equation.

c) Find and interpret the correlation between the repair cost of a machine and its age.

d) Find the coefficient of determination, and discuss what this statistic tells you.

e) At 5% significance level, test to determine whether the age of a machine and its monthly cost of repair are linearly related.

f) Predict the average monthly repair cost of welding machines that are 120 months old.

g) Predict with 95% confidence the monthly repair cost of a welding machine that is 120 months old.

Ages X
110
113
114
134
93
141
115
115
115
142
96
139
89
93
91
109
138
83
100
137

Cost Y
655.34
753.36
785.04
886.28
685.24
952.32
649.48
677.96
866.90
1052.74
724.84
897.52
670.54
701.88
583.62
935.60
948.96
708.30
840.22
832.08

Let X and Y be independent positive random variables. Let Z=X/Y. In what follows, all occurrences of x, y, z are assumed to be positive numbers.

1. Suppose that X and Y are discrete, with known PMFs, pX and pY. Then,

   pZ|Y(z|y)=pX(?).

   What is the argument in the place of the question mark?

     

2. Suppose that X and Y are continuous, with known PDFs, fX and fY. Provide a formula, analogous to the one in part (a), for fZ|Y(z|y) in terms of fX. That is, find A and B in the formula below.

   fZ|Y(z|y)=AfX(B).

   A=

   B=

3. Which of the following is a formula for fZ(z)?

   fZ(z)=
   (Choose all that apply.)

   fZ(z)=∫∞0fY,Z(y,z)dy

   fZ(z)=∫∞0fY,Z(y,z)dz

   fZ(z)=∫∞0fY(y)fZ,Y(z,y)dy

   fZ(z)=∫∞0fY(y)fZ|Y(z|y)dy

   fZ(z)=∫∞0fY(y)fX(yz)dy

   fZ(z)=∫∞0yfY(y)fX(yz)dy

1. With the virus outbreak, the average number of times Justin washes his hands during the day is 14 with a standard deviation of 3. Assuming that this number is normally distributed, what is the probability that tomorrow he will wash her hands between 16 and 22 times? Use the z-table to answer the question (Select the answer that is closest to the answer that you calculated.)

   		0.33
   		0.79
   		0.95
   		0.5
   		0.67

2. The mean age of presidents at inauguration is 55 years. The age of presidents at inauguration is normally distributed with a standard deviation of 6.6 years. Donald Trump was 70 years old when he was inaugurated. What proportion of presidents were younger than Donald Trump at their inauguration? Use the z-table to answer this question (Select the answer that is closest to the answer that you calculated.)

   		2.273
   		0.988
   		0.012
   		0.786
   		None of the above

3. Use this information to answer the following two questions: Varshini wants to read 4 books this month, but her busy work schedule may get in the way of her goal. Let X represent the number of books she will read this month. The table below shows the probabilities associated with the number of books she will read this month.

   X	P(X = x)
   0	0.17
   1	0.23
   2
   3	0.32
   4	0.16

   What is the probability that she reads less than 3 books?

   		0.48
   		0.52
   		0.84
   		0.74
   		This cannot be determined based on the data provided.

Part 2: More Review of Confidence Intervals

The following questions might be more challenging, but we want you to wrestle through them and ask for clarification along the way. Talking through these problems with a neighbor can help, and we hope that, ultimately, working through these problems will strengthen your understanding of the big ideas behind confidence intervals.

1. A 95% confidence interval is constructed in order to estimate the average number of minutes college students spend on Facebook per day. The interval ends up being from 30.1 minutes to 47.1 minutes. Based on this interval, we know the sample mean must be __________________ and the margin of error must be _____________ . (Note that your answers should be in the form of numbers, and it might help to review the general format of the confidence interval presented earlier in this lab activity).

2. A 95% confidence interval for the proportion of college students who have texted during a class was 0.75 to 0.95. Which of the following is the 90% confidence interval from the same sample?

1. 0.05 to 0.25
2. 0.731 to 0.969
3. 0.766 to 0.934
4. 0.777 to 0.9

3. A 90% confidence interval is constructed in order to estimate the average number of hours college students spend studying per week. The resulting interval has a margin of error of 3 hours. Which of the following could be the margin of error for a 95% confidence interval based on the same sample of data?

A. 2 hours

B. 3 hours

C. 4 hours

D. 8 hours

E. This cannot be answered without knowing the sample size.

4. Which of the following statements about confidence intervals is not correct?  

1. A confidence interval is an interval of values computed from sample data that is likely to include the population parameter.
2. The general format of a confidence interval is “sample statistic ± margin of error.”
3. Doubling the population size will result in a more narrow confidence interval.
4. If you construct a confidence interval for a population mean, the size of the sample mean has no effect on the size of the margin of error.

5. Suppose that a survey is planned to estimate the proportion of the population of OSU students who are left-handed. The sample data will be used to form a confidence interval. Which one of the following combinations of sample size and confidence level will give the widest confidence interval?

1. n = 400, confidence level = 90%
2. n = 400, confidence level = 95%
3. n = 1000, confidence level = 90%
4. n = 1000, confidence level = 95%

6. A 95% confidence interval is calculated for the percentage of OSU students who believe the parking options offered at OSU are satisfactory. The resulting confidence interval is 59.5% to 64.4%. Based on this information, which of the following is not true?

1. The confidence interval was produced by a process that will capture the true population percentage 95% of the time.
2. We are 95% confident that the interval 59.5% to 64.4% contains the true population percentage of OSU students who believe the parking options offered at OSU are satisfactory.
3. We are 95% confident that the interval 59.5% to 64.4% contains the sample percentage of OSU students who believe the parking options offered at OSU are satisfactory.
4. The sample percentage was about 62%.

7. Based on a random sample of data, an administrator at Sweet Valley High School estimates, at a 99% confidence level, that 22% ± 8% of Sweet Valley High School students plan to take summer classes. If the school has 1420 students, this means the possible number of students who plan to take summer classes is from

1. 199 to 426 students.
2. 195 to 430 students.
3. 114 to 312 students.
4. 47 to 178 students.

A sample of blood pressure measurements is taken for a group of​ adults, and those values​ (mm Hg) are listed below. The values are matched so that

1010

subjects each have a systolic and diastolic measurement. Find the coefficient of variation for each of the two​ samples; then compare the variation.

The coefficient of variation for the systolic measurements is

14.6 14.6​%.

​(Type an integer or decimal rounded to one decimal place as​ needed.)

The coefficient of variation for the diastolic measurements is

15.8 15.8​%.

​(Type an integer or decimal rounded to one decimal place as​ needed.)

Compare the variation.

The coefficients of variation for each data set are

within 5 percentage points of each other.

within 5 percentage points of each other.

more than 5 percentage points apart.

​Therefore, the systolic measurements vary

about the same as

significantly more than

about the same as

significantly less than

the diastolic measurements.

Question:

(Bayesian) Suppose that X is Poisson(λ + 1), and the prior distribution of λ is binomial(2,1/3).

(a) Find the Bayesian estimate of λ for mean square loss based on the single observation X, if X = 1.

(b) Find the Bayesian estimate of λ for mean square loss based on the single observation X, if X = 2

Hints:

Because of its prior distribution, λ can take only three values, 0,1,2.

Don’t expect its posterior distribution to be any distribution we have seen before;

You will have to compute (numerically) the posterior probabilities of the three possible values.

Random samples of students at 118 four-year colleges were interviewed in 1999 and 2008. Of the students who reported drinking alcohol, the percentage who reported bingeing at least three times in the last two weeks was 139 of 336 surveyed in 1999 and 181 of 842 surveyed in 2008.

1. Estimate the difference in proportions between 2008 and 1999.
2. What is the standard error for this difference?
3. What is the 95% confidence interval for this difference?
4. Does this indicate a difference between 2008 and 1999?
5. If you wanted the evaluate these results using a hypothesis test, what would be the null and alternative hypotheses?
6. Is this a one tailed or two tailed test? Draw the rejection regions.
7. What is the relevant test statistic?
8. What is the probability associated with that test statistic ?
9. Should you accept or reject the null hypothesis?