1) Assume that for a recent 41-year period there were 5469 earthquakes which were considered as “strong” earthquakes. Using a Poisson distribution, find the probability that in a given year, there are exactly 150 earthquakes that are considered “strong”.

2) Assume that the mean number of aircraft accidents in the United States is 8.5 per month and that a Poisson distribution applies. Find P(5), the probability of having 5 accidents in a month. Is it unlikely to have a month with 5 accidents?

Using SPSS:

You are conducting an experiment to see if growing up bilingual has any effect on a person’s verbal ability later in life. You recruit two groups of subjects: Group 1 consists of native English speakers that speak no other languages. Group 2 consists of people who speak English and at least one other language fluently, and they acquired their second language before the age of 12. Group 1 has 41 subjects, and Group 2 has 37. Verbal ability is measured through a standardized test, that is scored on a scale of 0-100. Each subject is administered the test, and their scores are recorded.

[40 pts] Next, you want to compare the scores of the two groups. You believe that the bilingual subjects (Group 2) will score higher on average than the monolingual subjects (Group 1). Use group1.sav and group2.1.sav for this problem.

What test should you perform here? [5 pts]

What are the distributions of each group? [5 pts]

What are the null/alternative hypotheses? [5 pts]

Run the test in SPSS. [5 pts]

Report your results, as in part 1. [10 pts]

Make a plot comparing the distributions of the two groups. [10 pts]

group1.sav:

84.0781458648940
78.1659813381156
83.2097990726180
74.6179953832211
80.2209401911208
84.9246643735632
79.5972632452217
80.3692891088453
75.5014243441437
76.8739343384229
83.9250985668693
75.3718964391512
78.9411130336794
80.1669903661018
78.2992377573003
78.7896332931477
71.0962392023740
84.1626256134368
82.4740174461893
78.3526586448991
73.3404985790769
83.2337839161124
84.6766160746263
70.7541667161686
83.1011338187307
77.5797973931312
81.3983154504291
78.7066558698611
86.2556403970559
74.8616971481931
81.7496553846257
76.1337325292457
76.4781081018565
79.1532261410977
80.6304842615301
84.1827007993516
80.7528310629980
71.3705246775774
78.5325983274464
78.0476329999499
87.2320045671994

group2.1.sav:

80.8028582039345
92.1782185508421
81.7624767266380
87.6090549775763
81.7988630107953
79.6059009502792
91.0122984144023
79.6806829604419
75.3399990115149
81.3053918211527
82.8035549255071
81.1637179049344
85.3632256665115
86.7573961689029
79.4778738319496
80.9349084875013
81.5921214879792
80.7109217300939
76.3139725547807
82.1414651646484
75.8356344963212
78.0785409762354
81.7487710395530
82.8013055546603
81.0426679680293
85.9590818056062
76.2803153724397
78.6593661496306
80.2492762839822
80.4885216555969
73.6726193385933
85.7176440415797
85.1825156137993
87.8187958603501
79.5842202611197
81.9787940432406
91.0181477610718

Home vs Road Wins – Significance Test: For the NHL regular season, the Chicago Blackhawks won 27 out of 41 home games and won 18 out of 41 away games. Clearly the Blackhawks won a greater proportion of home games. Here we investigate whether or not they did significantly better at home than on the road. The table summarizes the relevant data. The p̂'s are actually population proportions but you should treat them as sample proportions. The standard error (SE) is given to save calculation time if you are not using software.

Data Summary

SE = 0.10990

The Test: Test the claim that the proportion of wins at home was significantly greater than on the road. Use a 0.05 significance level.

(a) Letting p̂₁ be the proportion of wins at home and p̂₂ be the proportion of wins on the road, calculate the test statistic using software or the formulaz =

where δ_p is the hypothesized difference in proportions from the null hypothesis and the standard error (SE) given with the data. Round your answer to 2 decimal places.
z =
To account for hand calculations -vs- software, your answer must be within 0.01 of the true answer.

(b) Use software or the z-table to get the P-value of the test statistic. Round to 4 decimal places.
P-value =

(c) What is the conclusion regarding the null hypothesis?

reject H₀

fail to reject H₀    

(d) Choose the appropriate concluding statement.

The data supports the claim that the proportion of wins at home was significantly greater than on the road.

While the proportion of wins at home was greater than on the road, the difference was not great enough to be considered significant.    

We have proven that the Blackhawks always do better at home games.

We have proven there was no difference in the proportion of wins at home than wins on the road.

DATA contains Part Quality data of three Suppliers. At α=0.05, style="color:rgb(34,34,34);">does Part Quality depend on Supplier, or should the cheapest Supplier be chosen? Part Quality Supplier Good Minor Defect Major Defect

A 100 5 8

B 160 10 4

C 150 7 11

Gun Murders - Texas vs New York - Significance Test
In 2011, New York had much stricter gun laws than Texas. For that year, the proportion of gun murders in Texas was greater than in New York. Here we test whether or not the proportion was significantly greater in Texas. The table below gives relevant information. Here, the p̂'s are population proportions but you should treat them as sample proportions. The standard error (SE) is given to save calculation time if you are not using software.

Data Summary

Standard Error (SE) = 0.02285

The Test: Test the claim that the proportion of gun murders was significantly greater in Texas than New York in 2011. Use a 0.01 significance level.

(a) Letting p̂₁ be the proportion of gun murders in Texas and p̂₂ be the proportion from New York, calculate the test statistic using software or the formulaz =

where δ_p is the hypothesized difference in proportions from the null hypothesis and the standard error (SE) is given with the data. Round your answer to 2 decimal places.
z =
To account for hand calculations -vs- software, your answer must be within 0.01 of the true answer.

(b) Use software or the z-table to get the P-value of the test statistic. Round to 4 decimal places.
P-value =

(c) What is the conclusion regarding the null hypothesis?

reject H₀

fail to reject H₀    

(d) Choose the appropriate concluding statement.

The data supports the claim that the proportion of gun murders was significantly greater in Texas than New York.

While the proportion of gun murders in Texas was greater than New York, the difference was not great enough to be considered significant.     

We have proven that the stricter gun laws in New York actually decreased the proportion of gun murders below the rate in Texas.

We have proven there was no difference in the proportion of gun murders between Texas and New York.

1. State and use the average payoff strategy to choose the best decision
2. State and use the aggressive strategy to choose the best decision
3. State and use the conservative strategy to choose the best decision
4. State and use the opportunity loss strategy to make the best decision
5. Suppose the decision maker obtains information that enables the following probabilities assessments: P(s_1) = 0.5; P(s_2) = 0.2; P(s_3) = 0.2; and P (s_4) = 0.1. use the expected value approach to determine the optimal strategy

. Refer to the accompanying table showing results from a Chembio test for Hepatitis C among HIV-infected patients.

a. Find the probability of selecting a subject with a positive test result, given that the subject does not have hepatitis C. Why is this case problematic for test subjects?

b. Find the probability of selecting a subject with a negative test result, given that the subject has hepatitis C. What would be an unfavorable consequence of this error?

c. Find the positive predictive value for the test. That is, find the probability that a subject has hepatitis C, given that the test yields a positive result. Does the result make the test appear to be effective?

d. Find the negative predictive value for the test. That is, find the probability that a subject does not have hepatitis C, given that the test yields a negative result. Does the result make the test appear to be effective?

Typing errors in a text are either nonword errors (as when "the" is typed as "teh") or word errors that result in a real but incorrect word. Spell‑checking software will catch nonword errors but not word errors. Human proofreaders catch 70%70% of word errors.

You ask a fellow student to proofread an essay in which you have deliberately made 1010 word errors.

(a) If ?X is the number of word errors missed, what is the distribution of ?X ? Select an answer choice.

?X is approximately Normal with ?=3μ=3 and ?=1.45σ=1.45

?X is binomial with ?=10n=10 and ?=0.3p=0.3

?X is binomial with ?=10n=10 and ?=0.7p=0.7

?X is Normal with ?=7μ=7 and ?=1.45σ=1.45

If ?Y is the number of word errors caught, what is the distribution of ?Y ? Select an answer choice.

?Y is binomial with ?=10n=10 and ?=0.7p=0.7

?Y is binomial with ?=10n=10 and ?=0.3p=0.3

?Y is approximately Normal with ?=3μ=3 and ?=1.45σ=1.45

?Y is Normal with ?=7μ=7 and ?=1.45σ=1.45

(b) What is the mean number of errors caught? (Enter your answer as a whole number.)

mean of errors caught =

What is the mean number of errors missed? (Enter your answer as a whole number.)

mean of errors missed =

(c) What is the standard deviation of the number of errors caught? (Enter your answer rounded to four decimal places.)

standard deviation of the number of errors caught =

What is the standard deviation of the number of errors missed? (Enter your answer rounded to four decimal places.)

standard deviation of the number of errors missed =

A chocolate chip cookie manufacturing company recorded the number of chocolate chips in a sample of 70 cookies. The mean is 22.24 and the standard deviation is 2.44. .Construct a 98​% confidence interval estimate of the standard deviation of the numbers of chocolate chips in all such cookies. How do I find the degree of freedom, which is 69, using the chi critical value table? 69 is not listed on there, just 60 and 70. Not sure how to use my TI-84 calculator to figure it out.

An article in Computers & Electrical Engineering, “Parallel simulation of cellular neural networks” (1996, Vol. 22, pp. 61–84) considered the speed-up of cellular neural networks (CNN) for a parallel general-purpose computing architecture based on six transputers in different areas. The data follow: 3.775302 3.350679 4.217981 4.030324 4.639692 4.139665 4.395575 4.824257 4.268119 4.584193 4.930027 4.315973 4.600101 Round your answers to 3 decimal places. Assume population is approximately normally distributed. (b) Construct a 99% two-sided confidence interval on the mean speed-up. Enter your answer; confidence interval, lower bound ≤μ≤ Enter your answer; confidence interval, upper bound (c) Construct a 99% lower confidence bound on the mean speed-up. Enter your answer in accordance to the item c) of the question statement ≤μ

All Greens is a franchise store that sells house plants and lawn and garden supplies. Although All Greens is a franchise, each store is owned and managed by private individuals. Some friends have asked you to go into business with them to open a new All Greens store in the suburbs of San Diego. The national franchise headquarters sent you the following information at your request. These data are about 27 All Greens stores in California. Each of the 27 stores has been doing very well, and you would like to use the information to help set up your own new store. The variables for which we have data are detailed below. x1 = annual net sales, in thousands of dollars x2 = number of square feet of floor display in store, in thousands of square feet x3 = value of store inventory, in thousands of dollars x4 = amount spent on local advertising, in thousands of dollars x5 = size of sales district, in thousands of families x6 = number of competing or similar stores in sales district A sales district was defined to be the region within a 5 mile radius of an All Greens store. I really need help with the bolded areas. Thank you.

x1 x2 x3 x4 x5 x6 231 3 294 8.2 8.2 11 156 2.2 232 6.9 4.1 12 10 0.5 149 3 4.3 15 519 5.5 600 12 16.1 1 437 4.4 567 10.6 14.1 5 487 4.8 571 11.8 12.7 4 299 3.1 512 8.1 10.1 10 195 2.5 347 7.7 8.4 12 20 1.2 212 3.3 2.1 15 68 0.6 102 4.9 4.7 8 570 5.4 788 17.4 12.3 1 428 4.2 577 10.5 14.0 7 464 4.7 535 11.3 15.0 3 15 0.6 163 2.5 2.5 14 65 1.2 168 4.7 3.3 11 98 1.6 151 4.6 2.7 10 398 4.3 342 5.5 16.0 4 161 2.6 196 7.2 6.3 13 397 3.8 453 10.4 13.9 7 497 5.3 518 11.5 16.3 1 528 5.6 615 12.3 16.0 0 99 0.8 278 2.8 6.5 14 0.5 1.1 142 3.1 1.6 12 347 3.6 461 9.6 11.3 6 341 3.5 382 9.8 11.5 5 507 5.1 590 12.0 15.7 0 400 8.6 517 7.0 12.0 8

(a) Generate summary statistics, including the mean and standard deviation of each variable. Compute the coefficient of variation for each variable. (Use 2 decimal places.) x s CV x1

(b) For each pair of variables, generate the correlation coefficient r. For all pairs involving x1, compute the corresponding coefficient of determination r2. (Use 3 decimal places.) r r2

x1, x2

(c) Perform a regression analysis with x1 as the response variable. Use x2, x3, x4, x5, and x6 as explanatory variables. Look at the coefficient of multiple determination. What percentage of the variation in x1 can be explained by the corresponding variations in x2, x3, x4, x5, and x6 taken together? (Use 1 decimal place.)

99.3%

(d) Write out the regression equation. (Use 2 decimal places for x3 and x6. Use 1 decimal place otherwise.)

x1=-18.86+16.2x2+.18x3+11.53x4+13.58x5+(-5.31x6)

If 12 new competing stores moved into the sales district but the other explanatory variables did not change, what would you expect for the corresponding change in annual net sales? (Use 2 decimal places.)

If you increased the local advertising by 9 thousand dollars but the other explanatory variables did not change, what would you expect for the corresponding change in annual net sales? (Use 2 decimal places.)

(f) Suppose you and your business associates rent a store, get a bank loan to start up your business, and do a little research on the size of your sales district and the number of competing stores in the district. If x2 = 2.8, x3 = 250, x4 = 3.1, x5 = 7.3, and x6 = 2, use a computer to forecast x1 = annual net sales and find an 80% confidence interval for your forecast (if your software produces prediction intervals). (Use 2 decimal places.)

forecast194.41

lower limit

upper limit

(g) Construct a new regression model with x4 as the response variable and x1, x2, x3, x5, and x6 as explanatory variables. (Use 2 decimal places for the intercept, 4 for x1, 5 for x3, and 3 otherwise.)

x4 = 4.14 + .0431 x1 + -.800 x2 + .00059x3 + -.661 x5 + Correct: .057 x6

Suppose an All Greens store in Sonoma, California, wants to estimate a range of advertising costs appropriate to its store. If it spends too little on advertising, it will not reach enough customers. However, it does not want to overspend on advertising for this type and size of store. At this store, x1 = 163, x2 = 2.4, x3 = 188, x5 = 6.6, and x6 = 10. Use these data to predict x4 (advertising costs) and find an 80% confidence interval for your prediction. (Use 2 decimal places.)

prediction

lower limit

upper limit

At the 80% confidence level, what range of advertising costs do you think is appropriate for this store? (Round to nearest integer.)

lower limit $

upper limit $

Older people often have a hard time finding work. AARP reported on the number of weeks it takes a worker aged plus to find a job. The data on number of weeks spent searching for a job contained in the table below.

1 31 48 3 8 30 34 13 15 35 8 6 44 29 16 20 4 27 22 14 18 17 14 40 17 7 48 45 9 24 9 11 39 11 5 51 16 28 1 40

a. Provide a point estimate of the population mean number of weeks it takes a worker aged plus to find a job. Round the answer to two decimal places. weeks

b. At confidence, what is the margin of error? Round the answer to four decimal places.

c. What is the confidence interval estimate of the mean? Round the answers to two decimal places. ,

d. Find the skewness. Round the answer to four decimal places

Consider the following hypothesis test:

H₀: p ≥ 0.75
H_a: p < 0.75

A sample of 400 items was selected. Compute the p-value and state your conclusion for each of the following sample results. Use α = .05.

Round your answers to four decimal places.

a. p = 0.66

p-value

Conclusion:
p-value Select greater than or equal to 0.05, reject, greater than 0.05, do not reject, less than or equal to 0.05, reject, less than 0.05, reject, equal to 0.05, do not reject, not equal to 0.05, do not reject H₀

b. p = 0.75

p-value

Conclusion:
p-value Select greater than or equal to 0.05, reject, greater than 0.05, do not reject, less than or equal to 0.05, reject, less than 0.05, reject, equal to 0.05, do not reject, not equal to 0.05, do not reject H₀

c. p = 0.71

p-value

Conclusion:
p-value Select greater than or equal to 0.05, reject, greater than 0.05, do not reject, less than or equal to 0.05, reject, less than 0.05, reject, equal to 0.05, do not reject, not equal to 0.05, do not reject H₀

d. p = 0.76

p-value

Conclusion:
p-value Select greater than or equal to 0.05, reject, greater than 0.05, do not reject, less than or equal to 0.05, reject, less than 0.05, reject, equal to 0.05, do not reject, not equal to 0.05, do not reject

Show all working.

Use the following data to:
a) draw a scatter plot
b) find the coefficient of correlation and test the significance at the 0.05 level
c) find the regression line
d) predict y' for x = 5

Number of alcoholic drinks - x            Score on a Dexterity Test - y
              2                                                    15
                1                                                    18
                3                                                   11
                4                                                     7
                2                                                    10
                1                                                    16
                5                                                    4
                6                                                     2

A company is developing a new high performance wax for cross country ski racing. In order to justify the price marketing​ wants, the wax needs to be very fast.​ Specifically, the mean time to finish their standard test course should be less than 55 seconds for a former Olympic champion. To test​ it, the champion will ski the course 8 times. The​ champion's times​ (selected at​ random) are

58.9​,

64.3​,

45.6​,

52.1​,

45.2​,

49.9​,

54.9​,

and

44.3

seconds to complete the test course. Should they market the​ wax? Assume the assumptions and conditions for appropriate hypothesis testing are met for the sample. Use 0.05 as the​ P-value cutoff level.

Calculate the​ P-value.

​P-value=