Assume adult IQ scores are normally distributed with a mean of 100 and a standard deviation of 15

a) What is the probability that a randomly selected adult has an IQ that is less than 115

b) Find the probability that an adult has an IQ greater than 131.5 (requirement to join MENSA)

c) Find the probability that a randomly selected adult has an IQ between 110 and 120

d) Find the IQ separating the top 15% from the others e) Find the IQ score separating the bottom 10% from the others

In a statistic class, 11 scores were randomly selected with the following results were obtained: 68, 74, 66, 37, 52, 71, 90, 65, 76, 73, 22. What are the outer fences?

A)-6.0, 140.0

B)-10.0, 92.0

C)2.0, 128.0

D)2.0, 162.0

E)37.0, 107.0

This question will be perfomed entirely in R. Consider the following sample from an unkown distribution:

sample_data <- c(1.15, 0.5, 28.03 , 0.085, 1.82, 25.30, 0.7, 0.02, 0.01 ,13.23)

1.a) Calculating the p-value using the bootstrap hypothesis testing method to determine whether the mean is greater than 1 (one-sided test). Use 10,000 bootstrap samples. Set the seed to 248 before beginning the bootstrap process, i.e. include this line of code at the beginning of your script.

set.seed(248)
Include a histogram of the generated bootstrap samples of X̄⋆, does it appear to be symmetric? Do you

reject the null at 0.05 significance level? (10 points)
1.b) Find the rejection region (for X̄) based on your bootstrap samples at a 0.05 significance level. (5 points)

1.c) Perform a t-test for the same hypothesis as in 1.a) using the t.test function in R. Make sure that you apply the correct arguments.

Do you reject the null at a 0.05 significance level based on the t-test?

Compare the p-value from 1.a) to the p-value from the t-test, do they imply different conlcusion? Which p-value would you trust more? Support your anwser. (10 points)

According to the college board, scores by women on the SAT-I test were normally distributed with a mean of 998 and standard deviation of 202. Score by women on the ACT test are normally distributed with a mean of 20.9 and a standard deviation of 4.6. Assume that the two tests use different scales to measure the same aptitude

a) If a woman gets an SAT-I score in the 67thh percentile, find her actual SAT-I score and her equivalent ACT score

b) If a woman gets an SAT score of 1220, find her equivalent ACT score

Answer True or False for the following:

1. Chi-square requires assumptions about the shape of the population distribution from which a sample is drawn.

2. Goodness of fit refers to how close the observed data are to those predicted from a hypothesis.

3. The null hypothesis (H0) states that no association exists between the two cross-tabulated variables in the population, and therefore the variables are statistically independent.

4. High chi square values indicate a high probability that the observed deviations are due to random chance alone

5. A scatter plot shows the direction of a relationship between the variables.

A firm works 5 days a week. Every employee must work exactly 2 full days and 3 half-days each week. A half-day can be either morning or afternoon, and two half-days cannot be held on the same day. How many possible different weekly schedules are there? if the firm has 374 employees, how many people must have the same work schedule for a particular week? What is the smallest number of employees needed to guarantee at least 7 workers have exactly the same schedule?

the midpoint of a class is the sum of its lower and upper limits divided by two

Using the data found in Table 4 and Bayes’ Formula, determine the probability that a randomly selected patient will have Strep Throat given the SARTD test result was positive. Use the CDC stated prevalence of 25%. Round answer to nearest hundredth of a percent (i.e. 45.67%).

Then using the same Table 4, and Bayes’ Formula, determine the probability that a randomly selected patient will not have Strep Throat given the SARTD test result was negative. Use the CDC stated prevalence of 25%. Round answer to nearest hundredth of a percent.

Many people grab a granola bar for breakfast or for a snack to make it through the afternoon slump at work. A Kashi GoLean Crisp Chocolate Caramel bar weights 45 grams. The mean amount of protein in each bar is 8 grams. Suppose the distribution of protein in a bar is normally distributed with a standard deviation of 0.27 grams and a random Kashi bar is selected.

c) Find a symmetric interval about the mean such that 98.64% of all amounts of protein lie in this interval.
d) Suppose the amount of protein is at least 8.1 grams. What is the probability that it is more than 8.3 grams?

Professor Fair believes that extra time does not improve grades on exams. He randomly divided a group of 300 students into two groups and gave them all the same test. One group had exactly 1 hour in which to finish the test, and the other group could stay as long as desired. The results are shown in the following table. Test at the 0.01 level of significance that time to complete a test and test results are independent.

(i) Give the value of the level of significance.


State the null and alternate hypotheses.

H₀: The distributions for a timed test and an unlimited test are the same.
H₁: The distributions for a timed test and an unlimited test are different.H₀: Time to take a test and test score are not independent.
H₁: Time to take a test and test score are independent.    H₀: Time to take a test and test score are independent.
H₁: Time to take a test and test score are not independent.H₀: The distributions for a timed test and an unlimited test are different.
H₁: The distributions for a timed test and an unlimited test are the same.

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


(iii) Find or estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(iv) Conclude the test.

Since the P-value < α, we reject the null hypothesis.Since the P-value is ≥ α, we do not reject the null hypothesis.    Since the P-value < α, we do not reject the null hypothesis.Since the P-value ≥ α, we reject the null hypothesis.

(v) Interpret the conclusion in the context of the application.

At the 1% level of significance, there is insufficient evidence to claim that time to do a test and test results are not independent.At the 1% level of significance, there is sufficient evidence to claim that time to do a test and test results are not independent.    

Condé Nast Traveler conducts an annual survey in which readers rate their favorite cruise ship. All ships are rated on a 100-point scale, with higher values indicating better service. A sample of 36 ships that carry fewer than 500 passengers resulted in an average rating of 85.33 , and a sample of 43 ships that carry 500 or more passengers provided an average rating of 81.3. Assume that the population standard deviation is 4.55 for ships that carry fewer than 500 passengers and 3.95 for ships that carry 500 or more passengers.

A.) What is the point estimate of the difference between the population mean rating for ships that carry fewer than 500 passengers and the population mean rating for ships that carry 500 or more passengers?

B.) At 95% confidence, what is the margin of error?

C.) What is a 95% confidence interval estimate of the difference between the population mean ratings for the two sizes of ships?

[ ] to [ ]

Describe the connection between a correlation and a bivariate regression analysis. In your discussion, specifically note: 1) statistical significance, 2) sign, and 3) use or application.

A researcher thinks that listening to classical music reduces anxiety. She measures the anxiety of 10 persons then plays Mozart's "Eine Kleine Nachtmusik". Following that the researcher measures their anxiety again. (Note that anxiety is measured on a scale from 1 to 7, with higher numbers indicating increased anxiety.)

Does the study support her hypothesis? Compute the upper bound of the confidence interval using the following data:

mean of the difference scores (subtract pretest from posttest): -1.6

standard error of the difference scores: 0.4

The formula for the CI upper bound is [standard error of the difference scores]*[t critical value]+[mean of the difference scores]

How do you find the t critical value? and what is the value for the upper bound?

Consider the following results for two independent random samples taken from two populations.

Sample 1:

n1 = 40

x̅1 = 13.9

σ1 = 2.3

Sample 2:

n2 = 30

x̅2 = 11.1

σ2 = 3.4

What is the point estimate of the difference between the two population means? (to 1 decimal)

Provide a 90% confidence interval for the difference between the two population means (to 2 decimals).

Provide a 95% confidence interval for the difference between the two population means (to 2 decimals).