The Wall Street Journal reported that long term Treasury bonds had a mean return of 24.03% in 2008. Assume that the returns for the long term Treasury bonds were distributed as a normal random variable, with a mean of 24.03 and a standard deviation of 10. If you select an individual Treasury bond from this population, what is the probability that it would have a return of...

a. less than 0 (a loss)?
b. between 10 and 20 ?
c. greater than 10 ?

A report included the following information on the heights (in.) for non-Hispanic white females.

(a)

Calculate a confidence interval at confidence level approximately 95% for the difference between population mean height for the younger women and that for the older women. (Use μ_20–39 − μ_{60 and older}.) ,

We are 95% confident that the true average height of younger women is greater than that of older women by an amount within the confidence interval.We are 95% confident that the true average height of younger women is greater than that of older women by an amount outside the confidence interval.    We cannot draw a conclusion from the given information.We are 95% confident that the true average height of younger women is less than that of older women by an amount within the confidence interval.Interpret the interval.

(b)

Let

μ₁

denote the population mean height for those aged 20–39 and μ₂ denote the population mean height for those aged 60 and older. Interpret the hypotheses

H₀: μ₁ − μ₂ = 1

and

H_a: μ₁ − μ₂ > 1.

The null hypothesis states that the true mean height for older women is 1 inch higher than for younger women. The alternative hypothesis states that the true mean height for older women is more than 1 inch higher than for younger women.The null hypothesis states that the true mean height for younger women is more than 1 inch higher than for older women. The alternative hypothesis states that the true mean height for younger women is 1 inch higher than for older women.    The null hypothesis states that the true mean height for younger women is 1 inch higher than for older women. The alternative hypothesis states that the true mean height for younger women is more than 1 inch higher than for older women.The null hypothesis states that the true mean height for older women is more than 1 inch higher than for younger women. The alternative hypothesis states that the true mean height for older women is 1 inch higher than for younger women.

Carry out a test of these hypotheses at significance level 0.001. Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.)

z=

P-value=

(c)

Based on the P-value calculated in (b) would you reject the null hypothesis at any reasonable significance level? Explain your reasoning.

Reject H₀. The data suggests that the difference in the true average heights exceeds 1.Fail to reject H₀. The data suggests that the difference in the true average heights exceeds 1.    Reject H₀. The data does not suggest that the difference in the true average heights exceeds 1.Fail to reject H₀. The data does not suggest that the difference in the true average heights exceeds 1.

(d)

What hypotheses would be appropriate if μ₁ referred to the older age group, μ₂ to the younger age group, and you wanted to see if there was compelling evidence for concluding that the population mean height for younger women exceeded that for older women by more than 1 in.?

H₀: μ₁ − μ₂ = 1
H_a: μ₁ − μ₂ < 1H₀: μ₁ − μ₂ = −1
H_a: μ₁ − μ₂ < −1    H₀: μ₁ − μ₂ = 1
H_a: μ₁ − μ₂ > 1H₀: μ₁ − μ₂ = −1
H_a: μ₁ − μ₂ > −1

You may need to use the appropriate table in the Appendix of Tables to answer this question.

A deficiency of the trace element selenium in the diet can negatively impact growth, immunity, muscle and neuromuscular function, and fertility. The introduction of selenium supplements to dairy cows is justified when pastures have low selenium levels. Authors of a research paper supplied the following data on milk selenium concentration (mg/L) for a sample of cows given a selenium supplement (the treatment group) and a control sample given no supplement, both initially and after a 9-day period.

(a)

Use the given data for the treatment group to determine if there is sufficient evidence to conclude that the mean selenium concentration is greater after 9 days of the selenium supplement. (Use α = 0.05. Use μ_d = μ_initial − μ_9-day.)

Find the test statistic. (Round your answer to two decimal places.)

t =

Find the df.

df =

Find the P-value. (Use technology to calculate the P-value. Round your answer to three decimal places.)

P-value =

(b)

Are the data for the cows in the control group (no selenium supplement) consistent with the hypothesis of no significant change in mean selenium concentration over the 9-day period? (Use α = 0.05. Use μ_d = μ_initial − μ_9-day.)

Find the test statistic. (Round your answer to two decimal places.)

t =

Find the df.

df =

Find the P-value. (Use technology to calculate the P-value. Round your answer to three decimal places.)

P-value =

A process has an in-control fraction nonconforming of p = 0.02.

a) What sample size would be required for the fraction nonconforming control chart if it is desired to have a probability of a least one nonconforming unit in the sample to be at least 0.9?

b) Now assume n = 200. Establish a control chart for the fraction nonconforming.

a. Generate a model for y as a function of x

b. Is this model useful? Justify your conclusion based on

i) R2 adjusted,

ii) Hypothesis test for model coefficient,

iii) overall model adequacy test and

iv) regression assumptions

c. If needed, modify model as appropriate and generate the new model.

Highlighted the parts that I need most. Please be detailed with explanation and use Excel. Thank you.

All work needs to be shown

Consider all observations as one sample of X (1st column) and Y (2nd column) values. Answer the following questions: (20 points)

a) Calculate the correlation coefficient r

b) Fit the regression model (predicting Y from X) and report the estimated intercept and slope.

c) Test whether the slope equals 0. Report your hypothesis, test statistic, p-value.  

All work needs to be shown

A regional planner employed by a public university is studying the demographics of nine counties in the eastern region of an Atlantic seaboard state. She has gathered the following data:

Include the aspect that the county is "coastal" or not in a multiple linear regression analysis using a "dummy" variable. (Negative amounts should be indicated by a minus sign. Round your answers to 2 decimal places.)

Income= _____________ + ______________ Median Age + ____________ Coastal

Test each of the individual coefficients to see if they are significant. (Negative amounts should be indicated by a minus sign. Leave no cells blank - be certain to enter "0" wherever required. Round your answers to 2 decimal places.)

Predictor   t   p-value
Constant      
Median Age      
Coastal      

If a patient has a disease there is a 93.8% chance that the test gives a correct positive response and if a patient does not have this disease, there is a 95.6% chance that the test gives a correct negative response. About 20 out of every 1000 members have the disease.

i) A member has the disease given that they have a positive test response

ii) A member has the disease given that they have a negative test response

TOPIC: Hypothesis Testing (Statistic Question)

In 2015, I taught my first semester at Yale University, a standard deviation was calculated based on the scores on the SDF, the question was asking whether or not I encourage student questions in my classroom. The standard deviation for the 2015 class (economics class) was 1.03 with 31 responses. In 2018, the standard deviation for that same question came out to be 0.88 with 24 responses in my calculus class. [2015 = economics class, 2018 = calculus class] Questions are below.

(a). Please test the hypothesis that the Standard Deviation of the 2015 class on this question is larger than it is for the 2018 class at a significance level of 0.05. Please do problem by hand and find the correct critical value using the R Studio (software program). If you statisticians are familiar with that software.

(b). Please test the hypothesis that the Standard Deviation of the 2015 class on this question is different than it is for the 2018 class at a significance level of 0.05. Please do problem by hand and find the correct critical value using the R Studio (software program).

THANK YOU CHEGG EXPERTS

DirectTV is interested in conducting a study to determine the percentage of all their subscribers would be willing to pay $90 per month for a premium cable package. What is the minimum size sample needed to estimate the population proportion with a margin of error of 0.03 or less at 96% confidence?

During the investigation of an alleged unfair trade practice, the Federal Trade Commission takes a random sample of 50 “3-ounce” candy bars from a large shipment. If the mean and the standard deviation of their weights are, respectively, 2.92 ounces and 0.21 ounce, determine at the level of 0.01 significance whether the commission has grounds upon which to proceed against the manufacturer on the unfair practice of short-weight selling. State hypotheses, P-value, and conclusion.

2. The median of an income distribution is $30,000. The mean income is $35,000 and the standard deviation is $25,000. For each part, if possible compute the quantity, if not possible give your reason.

a) Is the distribution symmetric? Why or why not?

b) The probability that a randomly selected income being less than $35,000

c) The probability that a randomly selected income being less than $30,000

d) The probability that the average of 4 randomly selected incomes being between $30,000 and $40,000

e) The probability that the average of 100 randomly selected incomes being between $30,000 and $40,000.

An admissions director wants to estimate the mean age of all students enrolled at a college. The estimate must be within 1 years of the population mean. Assume the population of ages is normally distributed and the population standard deviation is 9.5 years. Determine the minimum sample size required to construct a 80% confidence interval for the population mean age. Determine the minimum sample size required to construct a 95% confidence interval for the population mean age. Which level of confidence requires a larger sample size? 80% 95%

A university offers finance courses numbered 1,2,3,4,5 and accounting courses numbered 1,2,3,4,5,6. Let F be the event of selecting a finance course, A the event of selecting an accounting course, E the event of selecting an even numbered course, and O the event of selecting an odd course.

Selecting the accounting course number 3 is an example of which of the following events? Select all correct answers.

• A AND O

• F OR E

• F AND O

• F OR O

• E′

• A′

In 2016, the World Health Organization estimated that the average life expectancy at birth worldwide was 72 years[1]. (This includes all countries of the world, not just the countries in the sample.)

Complete the steps below to carry out a one-mean hypothesis test to test the claim that the average life expectancy has increased beyond the global average using a 5% significance level.

Let mean = the average life expectancy of a person at birth (globally).

1. State your Null and Alternative Hypothesis symbolically. (You may need to copy and paste symbols.)
2. Verify that the conditions of the one-mean hypothesis test are satisfied.

3. calculate the test statistic and p-value. Round your answer to the nearest thousandth. Insert the results from the hypothesis test from StatCrunch in the space provided.

Test Statistic:

p-value:

4. Determine if you should Reject null hypothesis or fail to reject the null hypothesis. Use the significance level and your p-value to explain how you made your decision.

5. Using sentences, write your conclusion using the context of the problem.

[1] Source: World Health Organization.