X₁, X₂, and X3are independent random variables with a uniform (Beta - 1, Beta + 1) distribution.

The distribution is f_x(x) = 1/2 for Beta - 1 < x < Beta + 1

Beta(hat) = (X₁ + X₂ + X₃)/3

Find the expected value and variance of beta(hat).

A psychologist is interested in the conditions that affect the number of dreams per month that people report in which they are alone. We will assume that based on extensive previous research, it is known that in the general population the number of such dreams per month follows a normal curve, with mean (μ) = 5 and standard deviation (σ) = 4. The researcher wants to test the prediction that the number of such dreams will be greater among people who have recently experienced a traumatic event. Thus, the psychologist studies 36 individuals who have recently experienced a traumatic event, having them keep a record of their dreams for a month. Their mean number of alone dreams is 8. Should you conclude that people who have recently had a traumatic experience have a significantly different number of dreams in which they are alone? Carry out the full five steps of hypothesis testing for a two-tailed test. In step V, sketch the comparison distribution. Label the mean with both a Z and a raw score. Draw in the significance level 0.05 (cutoff) and label with Z-scores. Label the percent of the area cut-off. Indicate the position of the sample mean and label it with both a Z and a raw score.

Step I: Restate the Question. Population 1:

Population 2 :

Research hypothesis :

Null hypothesis (2 points):

Step II: Give the characteristics of the comparison distribution .

Step III: Give the cut-off sample score(s) .

Step IV: Determine the sample score on the comparison distribution .

Step V(a): Sketch the comparison distribution (cut-off & computed values) and give the decision .

Step V(b): Compute the 99% confidence intervals .

lower limit=

upper limit=

.The Quicky Zippy Lube Station has a mean time for an oil change as 12.6 minutes with a standard deviation of 4.15 minutes. (a) What is the probability that an oil change will take more than 16.0 minutes? (b) What is the probability that a random sample of 22 oil changes yields an average between 10.0 and 14.0 minutes?

A psychology major recently conducted a survey with a random sample of 38 undergraduate students. She found that 18% of those surveyed stated that their preferred superpower would be teleportation. An article in a popular magazine stated that 33% of college students would choose teleportation as their preferred superpower. At Alpha = 0.10, test the claim that the proportion of students preferring teleportation is less than 33%?

In South Korea, “Jeju” is a famous holiday destination with breathtaking scenic views. A travel agent in Jeju wants to know the satisfaction levels of tourists who visit the island. The travel agent surveyed 45 tourists at random and constructed a 95% confidence interval for the proportion of tourists who were extremely satisfied with their Jeju visit to be (0.5289, 0.8044).

Which of the following statements are true? Select all the that apply.

Question 5 options:

Automobile insurance companies take many factors into consideration when setting rates. These factors include age, marital status and miles driven per year. To determine the effect of gender, a random sample of young (under 25, with at least 2 years of driving experience) male and female drivers was surveyed. Each was asked how many miles he or she had driven in the past year. The distances (in thousands of miles) are stored in stacked format (column 1= driving distances and column 2 identifies gender where 1=male and 2=female). (Assume equal variances.) (a) Can we conclude that male & female drivers differ in the numbers of miles driven per year? (you need to compute the sample means and sample standard deviations for each gender in the Excel Workbook that comes with this exercise. Use ”=AVERAGE()” and ”=STDEV()” formulas. This also applies to exercises further below.) (b) Estimate with 95% confidence the difference in mean distance driven by male and female drivers.

I don't know if you have the stats for this but I tried to put them in here and it told me this is too long. I do not have Minitab so I would need it in Excel please and thank you so much. It will not let me put all the info in here. Says its too long.

Assume that a simple random sample has been selected from a normally distributed population. Find the test statistic, P-value, critical value(s), and state the final conclusion.

28) Test the claim that for the population of history exams, the mean score is 80. Sample data are summarized as n = 16, x = 84.5, and s = 11.2. Use a significance level of ΅ = 0.01.

Use the sample data and confidence level given below to complete parts​ (a) through​ (d).

In a study of cell phone use and brain hemispheric​ dominance, an Internet survey was​ e-mailed to 2330 subjects randomly selected from an online group involved with ears. 1176 surveys were returned. Construct a 99​% confidence interval for the proportion of returned surveys.

a) Find the best point estimate of the population proportion p.

___ (Round to three decimal places as​ needed.)

​b) Identify the value of the margin of error E.

___ (Round to three decimal places as​ needed.)

​c) Construct the confidence interval.

___ < p < ___ (Round to three decimal places as​ needed.)

​d) Write a statement that correctly interprets the confidence interval. Choose the correct answer below.

A. One has 99% confidence that the sample proportion is equal to the population proportion.

B. 99​% of sample proportions will fall between the lower bound and the upper bound.

C. There is a 99% chance that the true value of the population proportion will fall between the lower bound and the upper bound.

D. One has 99​% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion.

Solve problem #1 by hand using the 7-step method. To receive full credit, show all hand calculations.

1. Rats who grow up in an uncrowded environment (1 rat per cage) live a mean number of 38 months.
A psychologist wants to know the effect of crowding on life expectancy of rats. A group of 31 rats
are raised in crowded conditions (3 rats per cage) and their age at the time of their natural death is recorded.

The data are: 43 39 38 37 36 39 35 36 40 40 41 36 38 39 41 39
            41 40 37 35 38 38 39 37 43 39 37 38 40 39 32

a) Use the 7 step method at = .05 to test the psychologist's prediction.

Let X denote the amount of space occupied by an article placed in a 1-ft³ packing container. The pdf of X is below.

f(x) =

(a) Graph the pdf.

Obtain the cdf of X.

Graph the cdf of X.

(b) What is P(X ≤ 0.65) [i.e., F(0.65)]? (Round your answer to four decimal places.)


(c) Using the cdf from (a), what is P(0.25 < X ≤ 0.65)? (Round your answer to four decimal places.)


What is P(0.25 ≤ X ≤ 0.65)? (Round your answer to four decimal places.)


(d) What is the 75th percentile of the distribution? (Round your answer to four decimal places.)


(e) Compute E(X) and σ_X. (Round your answers to four decimal places.)

(f) What is the probability that X is more than 1 standard deviation from its mean value? (Round your answer to four decimal places.)

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Use either the traditional method or P-value method as indicated. Identify the null and alternative hypotheses, test statistic, critical value(s) or P-value (or range of P-values) as appropriate, and state the final conclusion that addresses the original claim.

26) A test of sobriety involves measuring the subject's motor skills. Twenty randomly selected sober subjects take the test and produce a mean score of 41.0 with a standard deviation of 3.7. At the 0.01 level of significance, test the claim that the true mean score for all sober subjects is equal to 35.0. Use the traditional method of testing hypotheses.

Problem 2

The Manager at Brackley Fun Park advertises that the typical family visiting the park spends at least one hour in the park during weekends. A sample of 35 visitors during the weekends in the month of July revealed that the mean time spent in the Park was 62 minutes with a standard deviation of 8 minutes.

1. Using the 0.01 significance level and a one-tailed test, is it reasonable to conclude that the mean time in the Park is greater than 60 minutes? Show all steps in your test of hypothesis.
2. Repeat the analysis at the 0.05 significance level. (You may show only the calculations that change).
3. Repeat with a survey mean of 65 minutes at the .01 significance level. (You may show only the calculations that change).
4. What do you conclude from your analysis?

The height of women ages​ 20-29 is normally​ distributed, with a mean of 64.7 inches. Assume sigma equals 2.7 inches. Are you more likely to randomly select 1 woman with a height less than 66 inches or are you more likely to select a sample of 30 women with a mean height less than 66 ​inches? Explain.

What is the probability of randomly selecting 1 woman with a height less than 66 ​inches?

What is the probability of selecting a sample of 30 women with a mean height less than 66 inches?

Are you more likely to randomly select 1 woman with a height less than 66 inches or are you more likely to select a sample of 30 women with a mean height less than 66 inches? Choose the correct answer below.

A- It is more likely to select 1 woman with a height less than 66 inches because the probability is higher.

B- It is more likely to select 1 woman with a height less than 66 inches because the probability is lower

C- It is more likely to select a sample of 30 women with a mean height less than 66 inches because the sample of 30 has a higher probability

D- It is more likely to select a sample of 30 women with a mean height less than 66 inches because the sample of 30 has a lower probability.

According to a poll of adults about 41% work during their summer vacation. Suppose that this claim about the population proportion is true. Now if we take a sample of 75 adults, and find the sample proportion [^(p)] of adults who work during summer vacation.

What is the expected value of sample proportion [^(p)]?

What is the standard deviation of sample proportion [^(p)]?

The shape of the sampling distribution of sample proportion [^(p)] will be roughly like a
normal distribution
uniform distribution
Poisson distribution
student's t distribution
binomial distribution

P(0.30 ≤ [^(p)] ≤ 0.47) =

Is the sample large enough to compute the above probability?
Yes, because np ≥ 10 and n(1−p) ≥ 10.
No, the sample size is not sufficiently large. We actually had to assume normally distributed population.
Yes, because n ≥ 30
Yes, because np ≥ 10.
Yes, because the number of successes and failures are both larger than 10.

Find c such that P([^(p)] ≥ c) = 0.7