2. I stand outside the student union and conduct a poll of passing students about their favorite fast food place. There are 6 response options = McDonald's, Wendy's, Burger King, Taco Bell, Popeye's, and Arby's. I also collect the student's year in school for demographic purposes - 5 levels - First Year, Sophomore, Junior, Senior, and Grad Student. I build a two-way table of this data to prepare to conduct a Chi-square analysis. How many degrees of freedom would my analysis have?

Group of answer choices:

6

35

20

5

1

3. The expected values for a Chi-Square Test of Independence come from:

Group of answer choices:

the population values

the marginals

a chi-square table

4. I conduct the Chi-Square test of independence for my Fast Food poll and obtain an observed Chi-square value of 22.55. The chisq-test() in R also reports a p-value of 0.3114. How do I interpret this result if my alpha is 0.05?

Group of answer choices

I fail to reject the null hypothesis and therefore determine that there is a significant association between fast food preference and class year.

I fail to reject the null hypothesis and therefore conclude that there is no association between class year and favorite fast food restaurant.

I reject the null hypothesis and conclude that there is a significant association between fast food preference and class year.

I reject the null hypothesis and conclude that there is no association between the variables fast food preference and class year.

5. My student union poll included another question regarding the preference for different dog breeds. I find a significant association between preferred dog breeds and gender of the students. I calculate a Cramer's V test and get a result of 0.05. What conclusion would I make about this result?

Group of answer choices:

The Cramer's V score disproves our statistical significant finding.

The Cramer's V value further proves that the result is significant.

The result was statistically significant, but not substantively significant.

6. I decide to conduct another poll outside the student union, and I want to ensure that my poll will have a low probability of Type II error and will be able to detect a difference with a large effect size. I run the following code:

pwr.chisq.test(w = 0.3, N=NULL, df = 20, sig.level = 0.05, power = 0.8)

I get the following output in R:

Chi squared power calculation

w = 0.3
N = 232.8977
df = 20
sig.level = 0.05
power = 0.8

What does this output tell me about how I need to design my next poll.

Group of answer choices

My new poll needs a power of 0.8 to have an effect size of 0.3.

A sample size of 230 should be sufficient for my poll.

Since I set my sample size at 233 I will achieve a power of 0.8.

I need a sample size of 233 students to obtain a result with the power I desire to have in my analysis.

7. The area under the curve of a normal distribution is equal to:

Group of answer choices

the mean of the distribution

a probability of 1.0

the standard deviation of the distribution.

the z-score

8. In my student union poll I asked students what they scored on the SAT. I know that the mean score of the UMD population is 1340 with a standard deviation of 222. My friend wants to know how her score of 1280 stacks up to the distribution of all scores at UMD.

What is her z-score?

Group of answer choices

-0.53

-1

0.27

-0.27

9. Another friend asked me to calculate his z-score so he could see how he compared to the distribution of SAT scores among UMD students. I found that his z-score was 0.33. What is the interpretation of his z-score?

Group of answer choices

He scored 3 SDs higher than the mean.

He scored better than 33% of students at UMD.

His SAT score shows he was 1/3 of an SD above the mean score.

He did worse than 33% of students at UMD.

10. We have more fitness test data from Vitor (who is male) and Manuela (who is female), who are applying to a military academy. Vitor did 50 push-ups in a minute, while Manuela only did 45.

We know that among previous applicants to the academy, the distribution of number of sit-ups is as follows:

Males have a mean of 60 and a standard deviation of 6.5.

Females have a mean of 40 and a standard deviation of 4.3.

What is the z-score for Manuela's result on the test?

Group of answer choices

1.16

1.69

-0.92

3.49

11. Vitor did 50 push-ups in a minute, while Manuela only did 45.

We know that among previous applicants to the academy, the distribution of number of sit-ups is as follows:

Males have a mean of 60 and a standard deviation of 6.5.

Females have a mean of 40 and a standard deviation of 4.3.

What is the z-score for Vitor's result on the test?

Group of answer choices:

0

2.33

-1.16

-1.54

12. Vitor did 50 push-ups in a minute, while Manuela only did 45.

We know that among previous applicants to the academy, the distribution of number of sit-ups is as follows:

Males have a mean of 60 and a standard deviation of 6.5.

Females have a mean of 40 and a standard deviation of 4.3.

Relative to their gender, who did better on the push-up test, Vitor or Manuela?

Group of answer choices:

Manuela

Vitor

State which sampling design should be used in the following situations, and justify your reasoning:

i. You need to collect statistics on types of software applications used by employees of a company. Separate lists of employees are available classified by job categories.

ii. Patients arriving at an emergency room facility are to be sampled to study their demographics, type of injury, etc. It is decided to sample 10% of the patients.

iii. An engineering school plans to conduct a mail survey of its alumni who graduated after 1980 to collect data on their current employment status. A complete alphabetical list of the alumni is available.

iv. A national sample of college students is to be taken, but a comprehensive list of all college students doesn't exist. A list of all colleges (private and public) exists. For any particular college, a list of all students can be obtained.

The following data have to do with the relationship between maternal smoking (# of cigarettes smoked per day,

which is variable X) and infant birth weight (which is variable Y). (∑X, ∑X², ∑Y, ∑Y², and ∑XY have already been

calculated for you and are shown below in red font.)

Cigarettes Per Day (X) X²    Infant Birth Weight (Y) Y² XY

2                                                        4                                            7.5                                 56.25                             15.0

6                                                      36                                            7.2                                 51.84                             43.2

10                                                100                                            6.9                                 47.61                             69.0

12                                                  144                                            6.2                                 38.44                             74.4

14                                                196    5.8 33.64 81.2

∑X = 44                                        ∑X² = 480                              ∑Y = 33.6                      ∑Y² = 227.78             ∑XY = 282.8

What is the y-intercept (a) for this data?

Suppose two independent random samples of sizes n1 = 9 and n2 = 7 that have been taken from two normally distributed populations having variances σ12 and σ22 give sample variances of s12 = 117 and s22 = 19. (a) Test H0: σ12 = σ22 versus Ha: σ12 ≠ σ22 with σ = .05. What do you conclude? (Round your answers to 2 decimal places.) F = F.025 = H0:σ12 = σ22 (b) Test H0: σ12 < σ22versus Ha: σ12 > σ22 with σ = .05. What do you conclude? (Round your answers to 2 decimal places.) F = F.05 = H0: σ12 < σ22

You are opening a doughnut franchise, but first, you must pass quality control on sugar glaze thickness before you can open. You randomly select 10 doughnuts from your production and send the following glaze thickness data to corporate headquarters for analysis: (units in mm)

1.21, 1.25, 1.23, 1.21, 1.28, 1.22, 1.27, 1.29, 1.20, 1.26

1. At α = 0.05, your sample data must statistically demonstrate that the mean glaze thickness is at least 1.20 mm. Determine whether or not your doughnuts meet this criterion.

1. Corporate also demands that the mean glaze thickness not by more than 0.025 inches. What sample size would you use in order to hold the probability of selling “bad” donuts to no more than 5%? Solve this problem analytically and graphically. (Hint: this is a β question)

Please solve by hand and not in Excel.

A college looked at a random sample of students who worked the previous summer.  

Use the following information to answer questions 1 - 7.

The following data show the average earnings and sample standard deviations for a random sample of males and a random sample of females.

The following degrees of freedom may be helpful: 25.49

1) Find the point estimate of the difference between the mean summer earnings for males and the mean summer earnings for females.

2) Find the 95% confidence interval for the difference between the means of the two populations.

3) The college is interested in showing that the mean summer earnings for the males is greater than the mean summer earnings for females. What null and alternate hypothesis should the college use?

4) The college is interested in showing that the mean summer earnings for males is greater than the mean summer earnings for females. What is the value of the test statistic?
a) 3.17
b) -3.17
c) 2.10
d) -2.10
e) 1.16
f) -1.16
g) none of the above

5) The null hypothesis in problem 3 is to be tested at the 5% level of significance. The rejection region (regions) from the table is (are):
a)           z≤-1.96 or z≥1.96
b)           z≤1.96
c)            z≤-1.96
d)            z≤-1.645 or z≥1.645
e)            t≥1.708
f)            t≤-1.708
g)            t≤-1.708or t≥1.708
h)            t≤-2.060
i)             t≥2.060
j)             t≤-2.060 or t≥2.060

6) If the null hypothesis is tested in problem 4 at the 5% level, the null hypothesis should be:
a) rejected
b) not rejected
c) impossible to determine

7) Find the p-value for the hypothesis test in problem 4.
a) p-value>.10
b) .05 c) .025 d) .01 e) .005 f) p-value<.005

Steps on how to get the answers would be appreciated. Thanks.

1. (a) A statistician randomly sampled 100 observations and found

= 106 and s = 35. Calculate the t-statistic and p-value for testing

H₀: μ = 100 vs H_A: μ > 100.

Carry out the test at the 1% level of significance.

(b) Repeat part (a), with s = 25.

(c) Repeat part (a), with s = 15.

(d) Discuss what happens to the t-statistic and the p-value when the standard deviation decreases.

2. Repeat Question 1 using H_A: μ ≠ 100.

What are the advantages and disadvantages of strip mines vs. underground mines?

The number of chocolate chips in a bag of chocolate chip cookies is approximately normally distributed with a mean of 1262 chips and a standard deviation of 118 chips. ​(a) Determine the 26th percentile for the number of chocolate chips in a bag. ​(b) Determine the number of chocolate chips in a bag that make up the middle 98​% of bags. ​(c) What is the interquartile range of the number of chocolate chips in a bag of chocolate chip​ cookies?

Show work please. So i can understand. Also how do you use and find the standard normal table or x table?

1. Have you ever played rock-paper-scissors (or Rochambeau)? It’s considered a “fair game” in that the two players are equally likely to win (like a coin toss). Both players simultaneously display one of three hand gestures (rock, paper, or scissors), and the objective is to display a gesture that defeats that of your opponent. The main gist is that rocks break scissors, scissors cut paper, and paper covers rock. We investigated some results of the game rock-paper-scissors, where the researchers had 119 people play rock-paper-scissors against a computer. They found 66 players (55.5%) started with rock, 39 (32.8%) started with paper, and 14 (11.8%) started with scissors.We want to see if players start with scissors with a probability that is different from 1/3.

a) Identify the parameter in the question. (hint: the long-run proportion of …) (1 pts)

b) State the null hypothesis in words. (1 pts)

The null would be a player that starts with scissors is 33%

c) State the null hypothesis using symbols. (0.5 pts)

H_O π=33%

d) State the alternative hypothesis in words. (1 pts)

The alternative would be a player that has scissors wouldn’t be 33%

e) State the alternative hypothesis in symbols. (0.5 pts)

H_a  π≠33%

f) What is the value for the statistics? (1 pts) Assign a symbol to this value. (0.5 pts)

z=-4.9

g) Using an “one proportion” applet, find the p-value. Report your p-value here: ___0____

(1 pts)

h) Based on this p-value, do we have strong evidence against the null hypothesis?  (1 pts)

Yes.

i) Summarize the conclusion in the context of the problem. (1 pts)

Lemington’sis trying to determine how many Jean Hudson dresses to order for the spring season. Demand for the dresses is assumed to follow a normal distribution with mean of 400 and standard deviation of 100. The contract between Jean Hudson and Lemington’s works as follows. At the beginning of the season, Lemington’s reserves x units of capacity. Lemington’s must take delivery for at least 0.8x dresses and can, if desired, take delivery on up to x dresses. Each dress sells for £160 and Hudson charges £50 per dress. If Lemington’s does not take delivery on all x dresses, it owes Hudson a £5 penalty for each unit of reserved capacity that is unused. For example, if Lemington’s orders 450 dresses and demand is for 400 dresses, Lemington’s will receive 400 dresses and owe Jean 400(£50) + 50(£5). How many units of capacity should Lemington’s reserve to maximise its expected profit? a) Set up a simulation model to help the company make the decision of how many units of capacity to reserve. b) Discuss on the obtained results and make suggestions. For example, what if the demand for the dress has different distributions?

An urn contains n white balls and m black balls. ( m and n are both positive numbers.)

(a) If two balls are drawn without replacement , what is the probability that both balls are the same color?

(b) If two balls are drawn with replacement (i.e., One ball is drawn and it’s color recorded and then put back. Then the second ball is drawn.) What is the probability that both balls are the same color.

(c) Show that the probability in part (b) is always larger than the one in part (a)

Please show in EXCEL how to express with function formula.

As we discussed during Week 3, researchers decide to ask 4 tie-purchasing customers whether they bought a bow tie or normal tie. According to national data, 3% of all ties purchased are bow ties.

Not surprisingly, this problem can be easily framed as a Binomial Distribution problem, as it meets all of the conditions outlined in our Week 4 (Monday) class. Further, we can identify the purchase of a bow tie with fixed probability of .03 as a success; the purchase of a standard tie with fixed probability of .97 as a failure. In other words:

p = .03, q=(1-p) = .97, n=4.

1.         On your Excel sheet, type in “PDF by hand” in cell A39. Then in cells A40 through A 44, use the COMBIN and POWER (or “^”) functions to compute the probability of x = 0, 1, 2, 3, 4 in each of the cells. In other words, in cell A40, compute P(X=0); cell A41, P(X=1); and so on. (That is, you’re using Excel to apply the binomial formula shown in class.)

              Hint: For P(X=0), enter the following:

“=COMBIN(4,0)*POWER(0.03,0)*POWER(1-0.03,4-0)” or

=COMBIN(4,0)*.03^0* (1-.03)^(4-0)”

(Note: for this problem, and all others in this Exercise, please round the numbers to 4 digits using Excel. You can do this after the fact from the toolbar, as we’ve shown before in class.)

Finally, in cell A46, use the SUM function to sum the individual probabilities you’ve computed in (a).

2.         In cell A48, compute the probability that at least one of the four persons purchased a bow tie by adding up each of the relevant probabilities, rounding to 4 digits.   (Note: do not type in any values. To receive credit, you must reference the relevant cells in your equations.)

              Then, in cell A49, compute the probability that at least one of the four persons purchased a bow tie by taking the complement of the event, X = 0. (Note: again, do not type in any specific values. Reference the relevant cells from the values you’ve already computed.) Round to 4 digits. In cell B49, type in “1-P(X=0)”.

3.         In cells B40:B44, compute the cumulative probabilities for x = 0, 1, 2, 3, 4, respectively, by entering appropriate commands that reference your computed pdf probabilities in cells A40 through A44. Again, round to 4 digits. Label the column “CDF by hand” in cell B39.

              Then, in cells, C40, C41, …, C44 type in “x=0, x=1, x=2, x=3, x=4”, respectively.

4.         Now, in cell D39 type PDF BINOM. Now use the BINOM.DIST function to populate cells D40:D44 with the PDFs of x = 0, 1, 2, 3, 4. (Remember, the PDF requires the “FALSE” entry.) Make sure you round your answers to four digits.

5.         Finally, using BINOM.DIST, populate cells E40:E44 with the CDFs of x = 0, 1, 2, 3, 4, rounding your answers to four digits. (Remember, the CDF requires the BINOM.DIST “TRUE” entry. Further, you must use BINOM.DIST, i.e., do not reference the output found in any other column.) As in (d), label the column by typing “CDF BINOM” in cell E39.

              In cell E46, reference via an “=” sign the appropriate cell that identifies the probability that fewer than 2 persons purchased a bow tie. In cell F46 type in “P(X≤1)”.

As we discussed during Week 3, researchers decide to ask 4 tie-purchasing customers whether they bought a bow tie or normal tie. According to national data, 3% of all ties purchased are bow ties.

Not surprisingly, this problem can be easily framed as a Binomial Distribution problem, as it meets all of the conditions outlined in our Week 4 (Monday) class. Further, we can identify the purchase of a bow tie with fixed probability of .03 as a success; the purchase of a standard tie with fixed probability of .97 as a failure. In other words:

p = .03, q=(1-p) = .97, n=4.

1.         On your Excel sheet, type in “PDF by hand” in cell A39. Then in cells A40 through A 44, use the COMBIN and POWER (or “^”) functions to compute the probability of x = 0, 1, 2, 3, 4 in each of the cells. In other words, in cell A40, compute P(X=0); cell A41, P(X=1); and so on. (That is, you’re using Excel to apply the binomial formula shown in class.)

              Hint: For P(X=0), enter the following:

“=COMBIN(4,0)*POWER(0.03,0)*POWER(1-0.03,4-0)” or

=COMBIN(4,0)*.03^0* (1-.03)^(4-0)”

(Note: for this problem, and all others in this Exercise, please round the numbers to 4 digits using Excel. You can do this after the fact from the toolbar, as we’ve shown before in class.)

Finally, in cell A46, use the SUM function to sum the individual probabilities you’ve computed in (a).

2.         In cell A48, compute the probability that at least one of the four persons purchased a bow tie by adding up each of the relevant probabilities, rounding to 4 digits.   (Note: do not type in any values. To receive credit, you must reference the relevant cells in your equations.)

              Then, in cell A49, compute the probability that at least one of the four persons purchased a bow tie by taking the complement of the event, X = 0. (Note: again, do not type in any specific values. Reference the relevant cells from the values you’ve already computed.) Round to 4 digits. In cell B49, type in “1-P(X=0)”.

3.         In cells B40:B44, compute the cumulative probabilities for x = 0, 1, 2, 3, 4, respectively, by entering appropriate commands that reference your computed pdf probabilities in cells A40 through A44. Again, round to 4 digits. Label the column “CDF by hand” in cell B39.

              Then, in cells, C40, C41, …, C44 type in “x=0, x=1, x=2, x=3, x=4”, respectively.

4.         Now, in cell D39 type PDF BINOM. Now use the BINOM.DIST function to populate cells D40:D44 with the PDFs of x = 0, 1, 2, 3, 4. (Remember, the PDF requires the “FALSE” entry.) Make sure you round your answers to four digits.

5.         Finally, using BINOM.DIST, populate cells E40:E44 with the CDFs of x = 0, 1, 2, 3, 4, rounding your answers to four digits. (Remember, the CDF requires the BINOM.DIST “TRUE” entry. Further, you must use BINOM.DIST, i.e., do not reference the output found in any other column.) As in (d), label the column by typing “CDF BINOM” in cell E39.

              In cell E46, reference via an “=” sign the appropriate cell that identifies the probability that fewer than 2 persons purchased a bow tie. In cell F46 type in “P(X≤1)”.

Your job is to fully staff your facility at the lowest cost. The facility must have at least 2 people w working from 6am-8pm Monday thru Friday and at least 1 person working from 10am to 6 pm on Saturdays. Nobody works on Sundays. Full time staff must work 8 hours a day, five days a week. Part time staff work 4 hours a day, 5 days a week. Nobody is allowed to work overtime. All employees receive the same hourly rate. What is the number of part time and full time employees are needed in order to fully staff the operation with the smallest total labor cost?

FT employees needed?:

PT employees needed?: