2.     The dataset QuizPulse10 contains pulse rates collected from 10 students in a class lecture and then from the same ten students during a quiz. We might expect the mean pulse rate to increase under the stress of a quiz. Use the dataset to test at the 5% significance level whether there is evidence to support this claim.

a.     Perform the hypothesis test in Minitab.

b.    Make a decision and mathematically justify your decision

c.     Interpret the results.

Do out-of-state motorists violate the speed limit more frequently than in-state motorists? This vital question was addressed by the highway patrol in a large eastern state. A random sample of the speeds of 2,500 selected cars was categorized according to whether the car was registered in the state or in some other state and whether or not the car was violating the speed limit. The data follow.

In state speeding cars: 521

Out of state speeding cars: 328

In state not speeding cars: 1141

Out of state not speeding cars: 510

a.) Do these data provide enough evidence to support the highway patrol's claim at the 5% significance level? Your conclusion must be in terms of the P-Value. Show all necessary work.

b). What type of error is possible and describe this error in terms of the problem?

c). Estimate the difference in the actual percentage of In State and Out of State speed limit violators using a 95% confidence interval. Show all necessary work. Using this interval estimation, is there sufficient evidence to support the highway patrol's claim? Explain Carefully.

d). Carefully interpret the confidence interval estimation.

1. A researcher who works for a national retail chain is interested in changes in employee satisfaction during the holiday season. The table below contains employee satisfaction scores taken at two time points (August 1^st and December 1^st) from a sample of 12 employees. Higher scores indicate more satisfaction. Conduct a hypothesis test to determine if there is an increase or decrease in employee satisfaction from Time 1 to Time 2. Use alpha .05.

1. State the null alternative hypothesis.
2. Conduct the hypothesis test.
3. State your findings.
4. Report an effect size measure. Please use the formula that we used in class

For a data set obtained from a sample of size n = 121 with x- = 44.25, it is known that σ = 5.4.

(a) What is the point estimate of µ?

(b) Find z score corresponding to a 95% confidence level, z_α/2. Recall that (1 − α)100% = 95%.

(c) Construct a 95% confidence interval for µ.

(d) What is the margin of error in part (c)?

3. The outside diameter of a part used in a gear assembly is known to be normally distributed with a mean of 39 mm and a standard deviation of 2.5 mm. The specifications on the diameter are (36,45), which means that part diameters between 36 mm and 45 mm are considered acceptable.

1. What is the proportion nonconforming?
2. What is Z bench?
3. If the mean of the process was centered between the specs, what would the proportion non-conforming be?

Chi-square tests are nonparametric tests that examine nominal categories as opposed to numerical values. Consider a situation in which you may want to transform numerical scores into categories. Provide a specific example of a situation in which categories are more informative than the actual values.

Suppose we had conducted an ANOVA, with individuals grouped by political affiliation (Republican, Democrat, and Other), and we were interested in how satisfied they were with the current administration. Satisfaction was measured on a scale of 1-10, so it was measured on a continuous scale. Explain what changes would be required so that you could analyze the hypothesis using a chi-square test. For instance, rather than looking at test scores as a range from 0 to 100, you could change the variable to low, medium, or high. What advantages and disadvantages do you see in using this approach? Which is the better option for this hypothesis, the parametric approach or nonparametric approach? Why?"

Sales personnel for Skillings Distributors submit weekly reports listing the customer contacts made during the week. A sample of 65 weekly reports showed a sample mean of 17.5 customer contacts per week. The sample standard deviation was 5.4. Provide 90% and 95% confidence intervals for the population mean number of weekly customer contacts for the sales personnel.

90% confidence interval, to 2 decimals:

( ,  )

95% confidence interval, to 2 decimals:

( , )

Three randomly selected households are surveyed as a pilot project for a larger survey to be conducted later. The numbers of people in the households are 2, 4, and 10. Consider the values of 2, 4, and 10 to be a population. Assume that samples of size n = 2 are randomly selected with replacement from the population of 2, 4, and 10. The nine different samples are as follows: (2, 2), (2, 4), (2, 10), (4, 2), (4, 4), (4, 10), (10, 2), (10, 4), and (10, 10). (i) Find the mean of each of the nine samples, then summarize the sampling distribution of the means in the format of a table representing the probability distribution. (ii) Compare the population mean to the mean of the sample means. (iii) Do the sample means target the value of the population mean? In general, do means make good estimators of population means? Why or why not?

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