A random sample of size n=500 yielded p̂ =0.08
a) Construct a 95% confidence interval for p.
b) Interpret the 95% confidence interval.
c) Explain what is meant by the phrase "95% confidence interval."

Please try to type your solution for this question, so I can read it without a problem. I truly appreciate you for typing in advance.

The Question:

For some Mechanical Engineer students, data for 224 student’s GPAs in their first 4 semesters of college were compared using several predictors, namely HSM, HSS, HSE, SATM, and SATV. Here significant level α is 0.05.

Then, the researcher fit a regression model and got the following results (intercept is included in the model):

ANOVA Table

Parameter Estimates Table

1. What are the values for (A), (B), (C), (D), (E), (F) and (G)?

2. State the multiple regression model for this problem.

3. What % of the variation in GPA are we explaining by the multiple regression?

4. Specify the prediction equation (Include all the regressors)

5. Determine if this is a significant model, i.e. check ANOVA table. (If significant, then p-value < α.) Here significant level α is 0.05.

6. Determine if each predictor variable is significant to the model, give the reason, based on the JMP output (i.e. compare the test statistics or p-value)

Describe the sampling distribution of ModifyingAbove p with caret. Assume the size of the population is 25 comma 000. nequals900​, pequals0.235 Describe the shape of the sampling distribution of ModifyingAbove p with caret. Choose the correct answer below. A. The shape of the sampling distribution of ModifyingAbove p with caret is 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. B. The shape of the sampling distribution of ModifyingAbove p with caret is 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. C. The shape of the sampling distribution of ModifyingAbove p with caret is approximately normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis less than 10. D. The shape of the sampling distribution of ModifyingAbove p with caret is not normal because n less than or equals 0.05 Upper N and np left parenthesis 1 minus p right parenthesis less than 10. Determine the mean of the sampling distribution of ModifyingAbove p with caret. mu Subscript ModifyingAbove p with caret Baseline equals 0.235 0.235 ​(Round to three decimal places as​ needed.) Determine the standard deviation of the sampling distribution of ModifyingAbove p with caret. sigma Subscript ModifyingAbove p with caret Baseline equals nothing ​(Round to three decimal places as​ needed.)

this question, but calculate the 90% confidence interval for the coefficient for cable by hand (but use the SE from the software output) and do the test whether age and number of TVs should be dropped by hand (use ANVOA table to get p-value and confirm with software).

The data in the table below contains observations on age, sex (male = 0, female = 1), number of television sets in the household, cable (no = 0, yes = 1), and number of hours of television watched per week. Using hours of television watched per week as the response, you can use Minitab's Regress or R's lm() command [e.g., model <- lm(hours~age+sex+num.tv+cable)] to fit a least squares regression model to all the other given variables.

What is the estimated value of the regression coefficient for age?  [2 pt(s)]

What is the estimated value of the regression coefficient for cable?  [2 pt(s)]

According to the regression equation, which of the following statements is true?
Females watch television 3.577 hours more per week than males.
Males watch television 3.577 hours more per week than females.
From the data given, one can not tell which gender watches more television per week.
Females watch television approximately the same number of hours per week as males.
[2 pt(s)]

What is the value of the test-statistic for the overall regression significance test?  [3 pt(s)]

What are the degrees of freedom associated with the test-statistic? Numerator:  Denominator:  [1 pt(s)]

Select the interval below that contains the p-value for this test.
p-value ≤ 0.001
0.001 < p-value ≤ 0.01
0.01 < p-value ≤ 0.05
0.05 < p-value ≤ 0.1
0.1 < p-value ≤ 0.25
p-value > 0.25
[3 pt(s)]

Compute a 90% confidence interval for the coefficient for cable. Lower Bound:  Upper Bound:  [3 pt(s)]

Compute a 95% confidence interval for the mean number of hours watched by 18-year old females with cable and 2 TV sets. Lower Bound:  Upper Bound: [3 pt(s)]

Compute a 95% prediction interval for the mean number of hours watched by 18-year old females with cable and 2 TV sets. Lower Bound:  Upper Bound: [3 pt(s)]

Test whether age and number of TV sets are needed in the model or should be dropped. What is the value of the test-statistic?  [3 pt(s)]

What are the degrees of freedom associated with this test? Numerator:  Denominator:  [1 pt(s)]

Select the interval below that contains the p-value for this test.
p-value ≤ 0.001
0.001 < p-value ≤ 0.01
0.01 < p-value ≤ 0.05
0.05 < p-value ≤ 0.1
0.1 < p-value ≤ 0.25
p-value > 0.25
[3 pt(s)]

You are conducting a multinomial Goodness of Fit hypothesis test for the claim that the 4 categories occur with the following frequencies:

H o : p A = 0.15 ; p B = 0.4 ; p C = 0.3 ; p D = 0.15 Complete the table.

Report all answers accurate to three decimal places.

1)

2) What is the chi-square test-statistic for this data?

χ 2 =

Significance level alpha is 0.005.

3) What is the p-value?

p-value =

4) What would be the conclusion of this hypothesis test?

A)Reject the Null Hypothesis

B)Fail to reject the Null Hypothesis

Report all answers accurate to three decimal places.

Which of the following are suitable null hypotheses. If not, explain why.

a. Comparing two groups

Consider comparing the average blood pressure of a group of subjects, both before and after they are placed on a low salt diet. In this case, the null hypothesis is that a low salt diet does reduce blood pressure, i.e., that the average blood pressure of the subjects is the same before and after the change in diet.

b. Classification.

Assume there are two classes, labeled + and -, where we are most interested in the positive class, e.g., the presence of a disease. H₀ is the statement that the class of an object is negative, i.e., that the patient does not have the disease.

c. Association Analysis

For frequent patterns, the null hypothesis is that the items are independent and thus, any pattern that we detect is spurious.

d. Clustering

The null hypothesis is that there is cluster structure in the data beyond what might occur at random.

e. Anomaly Detection

Our assumption, H₀, is that an object is not anomalous.

Volunteers who had developed a cold within the previous 24 hours were randomized to take either zinc or placebo lozenges every 2 to 3 hours until their cold symptoms were gone. Twenty-five participants took zinc lozenges, and 23 participants took placebo lozenges. The mean overall duration of symptoms for the zinc lozenge group was 4.5 days, and the standard deviation of overall duration of symptoms was 1.1 days. For the placebo group, the mean overall duration of symptoms was 8.6 days, and the standard deviation was 1.8 days.

(a) Calculate a 95% confidence interval for the mean overall duration of symptoms if everyone in the population were to take the zinc lozenges. (Round your answers to two decimal places.) to days

(b) Calculate a 95% confidence interval for the mean overall duration of symptoms if everyone in the population were to take the placebo lozenges. (Round your answers to two decimal places.) to days

(c) On the basis of the intervals computed in parts (a) and (b), is it reasonable to conclude generally that taking zinc lozenges reduces the overall duration of cold symptoms more than if taking a placebo? Explain why you think this is or is not an appropriate conclusion. Correct: Your answer is correct. reasonable to conclude that taking zinc lozenges reduces the duration of symptoms. The confidence intervals computed in parts (a) and (b) Correct: Your answer is correct. , so a reasonable conclusion is that in a population of cold sufferers the mean duration of symptoms if zinc lozenges are taken is Correct: Your answer is correct. than the mean duration if placebo is taken.

(d) In their paper, the researchers say that they checked whether it was reasonable to assume that the data were sampled from a normal curve population and decided that it was. How is this relevant to the calculations done in parts (a) and (b)? For relatively Correct: Your answer is correct. samples such as these, a necessary condition (in theory) is that the population of measurements has a normal distribution.

A random sample of 28 statistics tutorials was selected from the past 5 years and the percentage of students absent from each one recorded. The results are given below. Assume the percentages of students' absences are approximately normally distributed. Use Excel to estimate the mean percentage of absences per tutorial over the past 5 years with 90% confidence. Round your answers to two decimal places and use increasing order.

Number of Absences

13.9

16.4

12.3

13.2

8.4

4.4

10.3

8.8

4.8

10.9

15.9

9.7

4.5

11.5

5.7

10.8

9.7

8.2

10.3

12.2

10.6

16.2

15.2

1.7

11.7

11.9

10.0

12.4

A random variable X has a distribution p(X=k) = A / (k(k+1)), k = 1,2,...,4, where A is an constant. Then compute the value of p(1<=X<=3)

The answer will be either: 2/3, 3/4, 5/6, or 15/16

A discrete random variable X is uniformly distributed among −1,0,...,12. Then, what is its PMF for k=−1,0,...,12

The answer will be either: p(X = k) = 1/12, 1/13, 1/14, or 1

The annualized cost of acquiring capacity for the new barracuda drives at Seagate is calculated as $25 per unit and the contibution margin for the product is $45 per unit.

a.) what service level(percentile of demand met) should seagate target for in building capacity? how much capacity will it build? the demand forcast for the barracuda drives is uniform between 100 and 200.

b.) how does this service level change, if Seagate outsources manufacturing to china, to arrive at an annualized cost of capacity of $10 per unit? How much capacity will it build?

c.) How does this service level change if the inbound shipment cost per unit from China to US reduces the margin per unit to $35? How much capacity will it build?

d.) Conduct the capacity calculations in all 3 scenarios above when the forcast for the barracuda drives is expected to follow the following distributions.

Determine the area under the standard normal curve that lies between the following values z=0.2 and z=1.4

Run a regression analysis on the following bivariate set of data with y as the response variable.

Verify that the correlation is significant at an α=0.05α=0.05. If the correlation is indeed significant, predict what value (on average) for the explanatory variable will give you a value of 67.7 on the response variable.

What is the predicted explanatory value?

A hospital wants to determine if the type of treatment for pneumonia is a factor in recovery time? The table below shows the number of days to recovery for several randomly selected pneumonia patients that had various types of treatment.

Assume that all distributions are normal, the three population standard deviations are all the same, and the data was collected independently and randomly. Use a level of significance of α=0.1α=0.1.

H0: μ1=μ2=μ3H0: μ1=μ2=μ3
H1:H1: At least two of the means differ from each other.

1. For this study, we should use Select an answer 1-PropZInt χ²GOF-Test 1-PropZTest 2-PropZTest χ²-Test 2-SampTTest T-Test ANOVA TInterval 2-SampTInt 2-PropZInt
2. The test-statistic for this data =  (Please show your answer to 3 decimal places.)
3. The p-value for this sample =  (Please show your answer to 4 decimal places.)
4. The p-value is Select an answer less than (or equal to) alpha greater than alpha  αα
5. Base on this, we should Select an answer fail to reject the null hypothesis accept the null hypothesis reject the null hypothesis  hypothesis
6. As such, the final conclusion is that...
   • There is sufficient evidence to support the claim that treatment type is a factor in recovery time for pneumonia.
   • There is insufficient evidence to support the claim that treatment type is a factor in recovery time for pneumonia.

Suppose you have four extra tickets to a football game and you are trying to determine which of your five closest friends to take with you: Mark, Samuel, Elizabeth, James, and Mary. You decided to randomly choose four of them to attend the game with you. What is the probability that you will choose at least one friend whose name starts with the letter M? 0 2/5 4/5 1 2