The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that 77% of all fatally injured automobile drivers were intoxicated. A random sample of 54 records of automobile driver fatalities in Kit Carson County, Colorado, showed that 37 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than 77% in Kit Carson County? Use α = 0.05.

(a) What is the level of significance?

State the null and alternate hypotheses. Will you use a left-tailed, right-tailed, or two-tailed test?

H_o: p = 0.77; H₁: p > 0.77; right-tailed

H_o: p < 0.77; H₁: p = 0.77; left-tailed    

H_o: p = 0.77; H₁: p < 0.77; left-tailed

H_o: p = 0.77; H₁: p ≠ 0.77; two-tailed

(b) What sampling distribution will you use? Do you think the sample size is sufficiently large?

The normal distribution, since the sample size is large.

The t distribution, since the sample size is large.    

What is the value of the sample test statistic? (Use 2 decimal places.)


(c) Find the P-value of the test statistic. (Use 4 decimal places.)

Sketch the sampling distribution and show the area corresponding to the P-value.

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.    

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

(e) State your conclusion in the context of the application.

Reject the null hypothesis, there is sufficient evidence that the true proportion of driver fatalities related to alcohol is less than 0.77 in Kit Carson County.

Fail to reject the null hypothesis, there is insufficient evidence that the true proportion of driver fatalities related to alcohol is less than 0.77 in Kit Carson County.  

  Fail to reject the null hypothesis, there is sufficient evidence that the true proportion of driver fatalities related to alcohol is less than 0.77 in Kit Carson County.

Reject the null hypothesis, there is insufficient evidence that the true proportion of driver fatalities related to alcohol is less than 0.77 in Kit Carson County.

A large consumer electronics retailer is planning to expand its e-commerce operations. To this end, the management is considering establishing a strategic partnership with courier company XYZ for the delivery of customers’ orders.

Estimate the extend of the difference in delivery times between ABC and XYZ at 95% confidence level.

Write the relevant formulas in each case and show your calculations. Use appropriate tools from the Data Analysis Toolpak to verify your conclusions

The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. E refers to extroverted and I refers to introverted.

Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.

A. What is the level of significance?

B. State the null and alternate hypotheses. (from the following)

H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are independent.

H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are not independent.    

H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are not independent.

H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are independent.

C. Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)

D. Are all the expected frequencies greater than 5?

E. What sampling distribution will you use?

F. What are the degrees of freedom?

G. Find or estimate the P-value of the sample test statistic.

H. Based on your answers in parts (a) to (g), will you reject or fail to reject the null hypothesis of independence?

Ten randomly selected people took an IQ test A, and next day they took a very similar IQ test B. Their scores are shown in the table below. Person A B C D E F G H I J Test A 106 90 119 111 77 95 100 83 102 127

Test B 107 89 124 114 79 97 100 83 106 130

1. Consider (Test A - Test B). Use a 0.01 significance level to test the claim that people do better on the second test than they do on the first. (Note: You may wish to use software.)

(1)The test statistic is

(2) Construct a 99% confidence interval for the mean of the differences. Again, use (Test A - Test B).

<μ

Sally is tracking her spending. Every day for n days, Sally tracks the exact amount of dollars he spends that day, which is a nonnegative real number but not necessarily an integer. For positive integers i ≤ n, he spends xi dollars on day i. For example, on day 1, he spends x1 dollars. At the end of the n days, he calculates the average daily amount he spends on his food to be averagex = (x1 + x2 + · · · + xn)/n.

a) Prove that on at least one of the n days, Jarett paid at least averagex dollars.
(b) Prove that on fewer than half of the n days, Jarett paid strictly more than 2averagex dollars.

A study was conducted to determine if students in different college years differ on the number of hours of sleep they obtain. On a specified mid-semester Wednesday morning 25 Freshmen, 25 Sophomore, 25 Junior and 25 Senior college students reported the number of Hours of Sleep they had obtained on the previous evening. Answer the following question referring to the following tables. Descriptive Statistics Year Mean Std. Deviation N Freshmen 7.7080 1.86881 25 sophomore 7.1600 1.34412 25 Junior 7.4400 1.47422 25 Senior 5.8400 1.24766 25 Total 7.0370 1.64597 100 Tests of Between-Subjects Effects Source Sum of Squares df Mean Square F Sig. YEAR 51.515 3 17.172 7.607 .000 Error 216.698 96 2.257 Total 5220.150 100 Multiple Comparisons LSD Mean Difference (I-J) Std. Error Sig. (I) Year (J) Year Freshmen sophomore .5480 .42495 .200 Junior .2680 .42495 .530 Senior 1.8680* .42495 .000 sophomore Freshmen -.5480 .42495 .200 Junior -.2800 .42495 .512 Senior 1.3200* .42495 .002 Junior Freshmen -.2680 .42495 .530 sophomore .2800 .42495 .512 Senior 1.6000* .42495 .000 Senior Freshmen -1.8680* .42495 .000 sophomore -1.3200* .42495 .002 Junior -1.6000* .42495 .000

Based off of this data:

If we used multiple t-tests what would the overall alpha level be?

What is the Null Hypothesis for the one-way ANOVA?

Is the F test significant (give F and p)?

What do the multiple comparisons tell us? (i.e., what conclusions can be made about differences in attendance between classes?)

A recent study found that 64 children who watched a commercial for potato chips featuring a celebrity endorser ate a mean of 36 grams of potato chips as compared to a mean of 26 grams for 54 children who watched a commercial for an alternative food snack. Suppose that the sample standard deviation for the children who watched the​ celebrity-endorsed commercial was 21.1 grams and the sample standard deviation for the children who watched the alternative food snack commercial was 12.9 grams. Complete parts​ (a) through​ (c) below.

a. Assuming that the population variances are equal and alphaequals0.05​, is there evidence that the mean amount of potato chips eaten was significantly higher for the children who watched the​ celebrity-endorsed commercial? Let population 1 be the weights of potato chips eaten by children who watched the​ celebrity-endorsed commercial and let population 2 be the weights of potato chips eaten by children who watched the alternative food snack commercial. What are the correct null and alternative​ hypotheses? a)What is the test​ statistic? (Round to two decimal places)

what is the p value? (round to three decimal places)

b) Assuming that the population variances are​ equal, construct a 95​% confidence interval estimate of the difference mu 1 minus mu 2 between the mean amount of potato chips eaten by the children who watched the​ celebrity-endorsed commercial and children who watched the alternative food snack commercial.determine the 95​% confidence interval using​ technology, rounding to two decimal places.

c) Compare and discuss the results of​ (a) and​ (b). Check to make sure the results of​ (a) and​ (b) can be compared. Recall that confidence intervals are comparable only to the results of a​ two-tail hypothesis test. Since the test done in part​ (a) is a​ one-tail hypothesis​ test, these two results cannot be meaningfully compared.

the Jones family household includes Mr. and Mrs. Jones, 4 children, 2 cats, and 3 dogs. Every dat at 6 P.M. there is a Jones family stroll. The rules for a Jones family stroll are: Exactly 5 things (people + dogs + cats) go on each stroll. Each stroll must include at least one parent and at least one pet. There can never be a dog and a cat on the same stroll unless both parents go.

(a) How many different groupings are possible if exactly one parent and Rover, the oldest family dog, go on the stroll?

(b) How many different groupings are possible if exactly one parent goes on the stroll?

A social researcher is interested in understanding the effect of college education on wages. The workers in one group have earned an associate’s degree while members of the other group hold at least a bachelor’s degree. He would like to run a hypothesis test with α = .10 to see if those with a bachelor’s degree have significantly higher hourly wages than those with an associate’s degree.

k.Determine the critical t value(s) for this hypothesis test based on the degree of freedom, from (d), and the preset alpha level. (1 point total)

l.Compare the calculated t statistic with the critical t value by stating which is more “extreme”, and then make a decision about the hypothesis test by stating clearly “reject” or “fail to reject” the null hypothesis. (1 point total: .5 for comparison, .5 for decision)

m.Calculate the pooled standard deviation for the populations and then use it to calculate the standardized effect size of this test. (2 point total: 1 for pooled standard deviation, 1 for effect size. Deduct .5 if a result is wrong but the process is correct.)

A population of values has a normal distribution with μ=37.7μ=37.7 and σ=86.3σ=86.3. A random sample of size n=248n=248 is drawn.

1. Find the probability that a single randomly selected value is between 29.5 and 55.2. Round your answer to four decimal places.
   P(29.5<X<55.2)=P(29.5<X<55.2)=
   
2. Find the probability that a sample of size n=248n=248 is randomly selected with a mean between 29.5 and 55.2. Round your answer to four decimal places.
   P(29.5<M<55.2)=P(29.5<M<55.2)=  

An executive member of an ice cream company claimed that its product contained more than 500 calories per pint. To test this claim, 121 containers were randomly chosen and analyzed. The sample result showed that the mean was 506 calories with standard deviation of 33 calories. (a) At α= 5%, using the classical approach to test the claim. (2)Using P-value approach to conclude it.

1. The weight of 6-week-old poults (juvenile turkeys) are normally distributed with a mean 8.6 pounds and standard deviation 1.9 pounds. Find the first quartile of the weight.

A) 8.80 lb B) 8.06 lb C) 6.60 lb D) 7.33 lb

2. A gardener buys a package of seeds. Eighty-seven percent of seeds of this type germinate. The gardener plants 130 seeds. Approximate the probability that 81% or more seeds germinate.

A) 0.9761 B) 0.9884 C) 0.9904 D) 0.9842

3. A population has a standard deviation σ = 20.2. How large a sample must be drawn so that a 95% confidence interval for μ will have a margin of error equal to 4.2?

4. A college admissions officer takes a simple random sample of 120 entering freshmen and computes their mean mathematics SAT score to be 448. Assume the population standard deviation is 116. Based on a 98% confidence interval for the mean mathematics SAT score, is it likely that the mean mathematics SAT score for entering freshmen class is greater than 464? (Hint: you should first construct the confidence interval for the mean mathematics SAT score.). Answer Yes or No and give the reasoning for your answer.

Quick Start Company makes 12-volt car batteries. After many years of product testing, the company knows that the average life of a Quick Start battery is normally distributed, with a mean of 45.0 months and a standard deviation of 9.1 months.

(a) If Quick Start guarantees a full refund on any battery that fails within the 36-month period after purchase, what percentage of its batteries will the company expect to replace? (Round your answer to two decimal places.)
_ %

(b) If Quick Start does not want to make refunds for more than 13% of its batteries under the full-refund guarantee policy, for how long should the company guarantee the batteries (to the nearest month)?
_ months

Researchers conducted a study to determine whether magnets are effective in treating back pain. The results are shown in the table for the treatment​ (with magnets) group and the sham​ (or placebo) group. The results are a measure of reduction in back pain. Assume that the two samples are independent simple random samples selected from normally distributed​ populations, and do not assume that the population standard deviations are equal. Complete parts​ (a) and​ (b) below. Use a 0.05 significance level for both parts. Treatment Sham mu mu 1 mu 2 n 18 18 x overbar 0.55 0.45 s 0.52 1.15

The following table shows the Myers-Briggs personality preferences for a random sample of 406 people in the listed professions. E refers to extroverted and I refers to introverted.

Use the chi-square test to determine if the listed occupations and personality preferences are independent at the 0.05 level of significance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are independent.H₀: Myers-Briggs preference and profession are independent
H₁: Myers-Briggs preference and profession are not independent.    H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are independent.H₀: Myers-Briggs preference and profession are not independent
H₁: Myers-Briggs preference and profession are not independent.

(b) Find the value of the chi-square statistic for the sample. (Round the expected frequencies to at least three decimal places. Round the test statistic to three decimal places.)


Are all the expected frequencies greater than 5?

YesNo    

What sampling distribution will you use?

normalStudent's t    chi-squareuniformbinomial

What are the degrees of freedom?


(c) 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

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis of independence?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

(e) Interpret your conclusion in the context of the application.

At the 5% level of significance, there is insufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.At the 5% level of significance, there is sufficient evidence to conclude that Myers-Briggs preference and the profession are not independent.