You and your spouse each take two gummy vitamins every day. You share a single bottle of 60 vitamins, 30 of one flavor and 30 of another. You each prefer a different flavor, but it seems childish to fish out two of each type (but not to take gummy vitamins). So you just take the first four that fall out and then divide them up according to your preferences. For example, if there are two of each flavor, you and your spouse get the vitamins you prefer, but if three of your preferred flavor come out, you get two of the ones you like and your spouse gets one of each. Of course, you start a new bottle every 15 days. On average, over a 15 day period, how many of the vitamins you take are the flavor you prefer?

I need the code in SAS and R and outputs please

2. The data below come from a study investigating a method of measuring body composition, and give the body fat percentage (% fat), age and sex for 18 adults aged between 23 and 61 years. Source: Mazess, R.B., Peppler, W.W., and Gibbons, M. (1984) Total body composition by dual-photon (153GD) absorptiometry. American Journal of Clinical Nutrition, 40, 834-839.

a Enter the data into SAS using a DATALINES statement in the DATA step. Use PROC PRINT to print the resulting data set. Report your output.

b Create a data frame in R for the body composition data (from part a). Print the data frame and report the output.

Let Z be a normal random variable with mean µ = 0 and standard deviation σ = 1, that is, Z ∼ N(0, 1). Find each of the following:

(a) P(Z ≤ −1.13).

(b) P(Z ≥ −2.18).

(c) P(2.13 ≤ Z ≤ 2.57).

(d) P(−2.3 ≤ Z ≤ −1.1).

(e) P(0 ≤ Z ≤ 1.54).

(f) P(−1.54 ≤ Z ≤ 1.54).

(g) N(1.1243).

(h) N(−1.1243).

1. A telephone company claims that less than 15% of all college students have their own cell phone plan. A random sample of 70 students revealed that 8 of them had their own plan. Test the company's claim at the 0.05 level of significance.

2. A college statistics instructor claims that the mean age of college statistics students at a local Dallas-based institution is 23. A random sample of 35 college statistics students revealed a mean age of 25.1. The population standard deviation is known to be 4.1 years. Test his claim at the 0.1 level of significance.

3. A random sample of 85 adults ages 18-24 showed that 11 had donated blood within the past year, while a random sample of 254 adults who were at least 25 years old had 18 people who had donated blood within the past year. At the 0.05 level of significance, test the claim that the proportion of blood donors is not equal for these two age groups.

Dr. Mack Lemore, an expert in consumer behavior, wants to estimate the average amount of money that people spend in thrift shops. He takes a small sample of 8 individuals and asks them to report how much money they had in their pockets the last time they went shopping at a thrift store. Here is the data:

29.34, 24.88, 28.5, 12.02,   13.45,   13.41, 10.03,   25.09.

He wishes to test the null hypothesis that the average amount of money people have in their pockets is equal to $20. Calculate the test statistic to two decimal places. Take all calculations toward the answer to three decimal places.

Below,

n

is the sample size,

p

is the population proportion and

p

is the sample proportion. Use the Central Limit Theorem and the TI-84 calculator to find the probability. Round the answer to at least four decimal places.

=n111

=p0.54

Match an appropriate correlation for given set of two variables.

1. Number of siblings & Gender Which one: (Chi-squared, point biserial correlation, spearman's rho, pearson's r, kendalls tau)

2. Math scores & Reading score Which one: (Chi-squared, point biserial correlation, spearman's rho, pearson's r, kendalls tau)

3. Birth order & Education level  Which one: (Chi-squared, point biserial correlation, spearman's rho, pearson's r, kendalls tau)

4. Job satisfaction & Annual income Which one: (Chi-squared, point biserial correlation, spearman's rho, pearson's r, kendalls tau)

A coffee shop claims that its fresh-brewed drinks have a mean caffeine content of 80 milligrams per 5 ounces. A city health agency believes that the coffee shop’s fresh- brewed drinks have higher caffeine content. To test this claim the health agency takes a random sample of 100 five-ounce servings and found the average mean caffeine content of the sample was 87 milligrams with standard deviation of 25 milligrams. Does this provide enough evidence at the 1% significance level to claim that the coffee shop’s fresh- brewed drinks have higher caffeine content?

For the test of significance questions, clearly indicate each of the formal steps in the test of significance.

Step 1: State the null and alternative hypothesis.

Step 2: Calculate the test statistic.

Step 3: Find the p-value.

Step 4: State your conclusion. (Do not just say “Reject H0” or “Do not reject H0”, state the conclusion in the context of the problem.)

A major insurance company looked at teen cell phone use, specifically while driving. Eighty-two percent were using phones in the sample of fifty drivers.

a. Construct the 95% confidence interval for the proportion of all teenagers that have used cell phones while driving. (Round answers to 2 decimal places.)

b. What is the margin of error with 95% confidence? (Round answer to 2 decimal places.)

In a recent year, the Better Business Bureau settled 75% of complaints they received. (Source: USA Today, March 2, 2009) You have been hired by the Bureau to investigate complaints this year involving computer stores. You plan to select a random sample of complaints to estimate the proportion of complaints the Bureau is able to settle. Assume the population proportion of complaints settled for the computer stores is the 0.75, as mentioned above. Suppose your sample size is 276. What is the probability that the sample proportion will be at least 2 percent more than the population proportion?

Based on historical data, your manager believes that 27% of the company's orders come from first-time customers. A random sample of 191 orders will be used to estimate the proportion of first-time-customers. What is the probability that the sample proportion is between 0.31 and 0.48?

According to a 2009 Reader's Digest article, people throw away about 9% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 67 grocery shoppers to investigate their behavior. What is the probability that the sample proportion does not exceed 0.14?

Briefly explain to a non-statistician why ANOVA (which is a test of means) is called analysis of variance rather than analysis of means? what variances do we compare in anova?

You are considering the risk-return profile of two mutual funds for investment. The relatively risky fund promises an expected return of 7.2% with a standard deviation of 15.7%. The relatively less risky fund promises an expected return and standard deviation of 3.9% and 6%, respectively. Assume that the returns are approximately normally distributed.

a-1. Calculate the probability of earning a negative return for each fund. (Round final answer to 4 decimal places.)Probability: Riskier fund and Less risky fund

a-2. Which mutual fund will you pick if your objective is to minimize the probability of earning a negative return?

Less risky or fund Riskier fund

b-1. Calculate the probability of earning a return above 8.3% for each fund. (Round final answer to 4 decimal places.) Probability Riskier fund and Less risky fund

b-2. Which mutual fund will you pick if your objective is to maximize the probability of earning a return above 8.3%?

Riskier fund or Less risky fund

The mean cost of a meal for two in a midrange restaurant in City A is $48. How do prices for comparable meals in Hong Kong compare? The file HongKongMeals contains the costs for a sample of 42 recent meals for two in Hong Kong midrange restaurants.

(a)

With 95% confidence, what is the margin of error for the estimated mean cost in dollars for a mid-range meal for two in Hong Kong? (Round your answer to the nearest cent.)

$

(b)

What is the 95% confidence interval estimate of the population mean cost in dollars for a mid-range meal for two in Hong Kong? (Round your answers to the nearest cent.)

$  to $

A task requires the completion of four activities. A teacher would like to know if differences in the sequence of the four activities results in different task completion times. The teacher selects three students and demonstrates the activities in random order to the students. Then each student completes the task with each of the activity sequences. The completion times are recorded. The following table shows the minutes for each student to complete each task. Time (minutes) Sequence Allen Carla Henry A 21.7 20.5 22.0 B 18.0 18.6 20.4 C 19.9 20.4 22.5 D 20.1 18.8 21.2 Identify the right option for Treatments (Time). Null hypothesis: H0: μ1 = μ2 = μ3 H0: μ1 ≠ μ2 ≠ μ3 a b Alternate hypothesis: All means are the same. Not all means are the same. Identify the right option for Blocks (Sequence). Null hypothesis: H0: μ1 ≠ μ2 ≠ μ3 ≠ μ4 H0: μ1 = μ2 = μ3 = μ4 a b Alternate hypothesis: Not all means are the same. All means are the same. At 0.05 significance level, state the decision rule for for both treatment (Time) and blocks (Sequence). (Round your answers to 2 decimal places.) Complete the ANOVA table. (Round your SS, MS, and F values to 2 decimal places.) At the 0.05 significance level, is there a difference in the treatments and blocks?

Consider a short multiple choice quiz with three items, and each item has four choices (only one of the choices is correct). Suppose that you are taking this quiz but you are completely and utterly unprepared for it. That means that you only option for the quiz is to guess the answers. Suppose you are thinking about the first item: what's the probability that you'll answer it incorrectly (hint: think about how many options there are, and how many of them are wrong)?

Now, list the "sample space" of all possible outcomes on this exam. Hint: here is one possible outcome: Correct, Correct, Wrong (or, CCW for short).

Using the format in #2 above, we actually haven't listed the entire sample space, because there are multiple ways to answer each item incorrectly! In order to calculate the probability of a particular outcome, we either need to list every single way to answer a problem wrong, or we can use the multiplication rule. These outcomes are all independent because guessing on problem #1 has no effect on your guess for problems #2 or #3. For example, P(Correct on the 1st, and Correct on the 2nd, and Wrong on the 3rd) = P(Correct on the 1st) * P(Correct on the 2nd) * P(Wrong on the Third). Find P(CCW).

Now, use your list from #2 and write the probability for each outcome next to it, using the same type of calcuation you did in #3.

Next, use your work from #4 to fill in the following table with the relevant probabilities, where x represents the number of items you answer correctly. Notice that the number of correct items will sometimes include more than one of the listed outcomes (what do you do with those numbers?)

In #5 above, you should have a probability distribution (which means that all probabilities are between 0 and 1, and the sum of the probabilities is 1). Check to make sure this is correct, fix any errors, and then explain why this situation fits the criteria for a binomial distribution.

Using the table you made in #5, determine the probability of getting at most 2 items correct on the quiz.

We don't need to calculate each probability by hand. It can be done in a spreadsheet: Using excel or sheets, you can type the command "BINOM.DIST" (in sheets, there is no dot) to help compute these probabilities. For example, the command "=BINOM.DIST(0,3,.25,0)" tells you the probability of getting 0 correct (# of "successes") out of three (# of trials) where the probability of getting one item correct is .25 (probability of a "success"), and we want exactly that number of correct items ("0" at the end of the command refers to the fact that we aren't adding up any probabilities, that the "cumulative" feature of the command is turned off). Verify, using this command, that each of the values in your table above is correct.

Now, you can also check the calculation you did in #7, by typing the following command: "BINOM.DIST(2,3,.25,1)", which means you want to get two correct (# of successes) out of three items (# of trials), where the probability of getting one item correct is .25 (probability of a success), and we are adding up the probabilities for all values below 2 as well (the cumulative spot is turned on when you type "1").

Finally, create a histogram of the distribution and copy it into your response.