Stats Chap 16-18
(Chap. 17: 9) What is the appropriate standard error to use for the hypothesis test? a.)What is the difference in the proportions of foreign born residents in both countries? b.) Whats the value of the z-statistic? c.) What do you conclude at alpha=0.05?
A.) 0.064 B.) 3.964 C.) Pvalue is less than 0.001. Very strong evidence, so reject the null hypothesis that the proportions are the same in the two countries.
(Chap. 17: 5)For the interval given in exercise three, explain what 95% confidence means.
If I were to take repeated samples of these sizes of Canadians and Americans, and compute two proportion confidence intervals, we would expect 95% of the intervals to contain the true difference in the proportions of foreign born citizens.
(Chap. 17: 13) Using the summary stats provided in exercise 11, researchers calculated a 95% confidence interval for the mean dif. between walmart and target purchase amounts. The interval was (-$14.15, -$1.85). Explain in context what this interval means.
We are 95% confident that the mean purchase amount at walmart is between $1.85 and $14.15 less than the mean purchase amount at target.
(Chap 16: 29)A clean air standard requires that vehicle exhaust emissions not exceed specified limits for various pollutants. Many states require that cars be tested annually to be sure they meet these standards... a) In this context, what is a Type I error? b) In this context, what is a Type II error? c) Which type of error would the shop's owner consider more serious? d) Which type of error might environmentalists consider more serious?
a.) Its decided that the shop isn't meeting standards when it is. b.) The shop is certified as meeting standards when its not. c.) Type 1 d.) Type 11
(Chap 16: 25) Sec. loan exercise describes the loan score method a bank uses to decide which applicants it will lend money... a.) In this context, what is meant by the power of the test. b.) What could the bank do to increase the power? c.) Whats the disadvantage of doing that?
a.) Power is the probability that the bank denies the loan that would not have been repaid. b.) Raise the cutoff score. c.) A larger number of trustworthy people would be denied credit, and the bank would miss the opportunity to collect interest on loans.
(Chap 16: 15)A researcher developing scanners to search for hidden weapons at airports has concluded that a new device is significantly better than the current scanner. He made this decision based on a test using α = 0.05. Would he have made the same decision at α = 0.10? How about α = 0.01? Explain.
Alpha= 0.10, Yes the Pvalue is less than 0.05, so is less than 0.10. But to reject Ho at alpha=0.01, the Pvalue must be below 0.01, which isn't necessairly the case.
(Chap 18: 25)Push-ups; Every year, the students at Gossett High School take a physical fitness test during their gym classes... create a 90% confidence interval for how many more push-ups boys can do than girls, on average, at that high school.
Based on these data, we are 90% confident that boys, on average, can do between 1.6 and 13.0 more push ups than girls (independent samples; not paired).
(Chap 18: 29)Is there a significant difference in calories between servings of strawberry and vanilla yogurt? Test an appropriate hypothesis and state your conclusion.
Ho: mu D = 0 vs. Ha: mu D=/= 0. data are paired by brand; brands are independent of each other; fewer than 10% of all yogurts (questionable); box-plot of differences shows an outlier (100) for great value: with the outlier included, the mean difference (strawberry-vanilla) is 12.5 calories with a T-stat of 1.332, with 11 df, for a P value of 0.2098. Deleting the outlier, the difference is even smaller, 4.55 calories with a T-stat of only 0.833 and a P value of 0.4241. With a P value so large, we do not reject Ho. We conclude that the data do not provide evidence of a difference in mean calories.
(Chap. 17: 15)The researchers in exercise 11 decide to test the hypothesis that the means are equal. The degrees of freedom formual gives 162.75 df. Test the null hypothesis at alpha= 0.05.
The difference is -$8 with an SE of 3.115, so the T stat is -2.569. With 162.75 or 161 df, the Pvalue is 0.011, which is less than 0.05. Reject the null hypothesis that the means are equal.
(Chap. 17: 63)Construct and interpret a 95% confidence interval for the mean additional amount that waterfront property is worth. (Df= 105.48).
These were random samples, both less than 10% of properties sold. Prices of houses should be independent, and random sampling makes the two groups independent. The box-plots make the price distributions appear to be reasonably symmetric, and with the large sample sizes the few outliers don't affect the means much. Based on this sample, we're 95% confident that, in New York, having a waterfront is worth, on average, about $59,121 to $140,898 more in sale price.
(Chap. 17: 3) The information used in exercise 1 was used to create a 95% two-proportion confidence interval for the differnece between canadian and us citizens who were born in foreign countries. Interpret this sent. in context. 95% confidence interval for P(canadians)- P(Americans.) is (3.24%, 9.56%)
We are 95% confident that, based on these data, the proportion of foreign born Canadians is between 3.24% and 9.56% higher than the proportion of foreign born Americans.
(Chap. 17: 7)If the info used in exercise 1 is used to make inferences about the proportion of all canadians and all americans born in other countries, what conditions must be met before proceeding? Are they met? explain.
We must assume that the data were collected randomly and that the Americans selected are independent of Canadians selected. Both assumptions should be met. Also, for both groups, we have at least 10 national born and foreign born citizens and the sample, sizes are less than 10% of the population sizes. All conditions for inferences are met.
(Chap. 17: 11) Do consumers spend more on a trip to Walmart (n=85, Y= $45, S= $21) or Target (n=80, Y= $53, S=$19)? To perform inference on these two samples, what conditions must be met? Are they? Explain.
We must assume the samples were random or otherwise independent of each other. We also assume that the distributions are roughly normal, so it would be a good idea to check a histogram to make sure there isn't strong skewness or outliers.
(Chap. 17: 51)In the July 2007 issue, consumer reports examined the calorie content of two kinds of hot dogs...would a 95% confidence interval for mu: meat- mu: beef include 0?
Yes, the high P value means that we lack evidence of difference, so 0 is plausible value for mu:meat - mu:beef.
(Chap. 17: 29)Researchers at the national cancer institute released the results of a study that investigated the effect of weed killing herbicides on house pets... A.) what's the standard error of the difference in the two proportions? B.) construct a 95% confidence interval for this difference? C.) state an appropriate conclusion.
a.) 0.035 b.) (0.356, 0.495) c.) We are 95% confident, based on these data, that the proportion of pets with a malignant lymphoma in homes where herbicides are used is between 35.6% and 49.5% higher than the proportion of pets with lymphoma in homes where no pesticides are used.
(Chap. 17: 57)The core plus mathematics project is an innovative approach to teaching math... A.) what's the margin of error for this confidence interval? B.) if we had created a 98% confidence interval, would the margin of error be larger or smaller? C.) explain what the calculated interval means In this context. D.) does this result suggest that students who learn math with CPMP will have significantly higher mean scores in algebra than those in traditional programs?
a.) 2.927 b.) larger. c.) based on this sample, we are 95% confident that students who learn math using CPMP method will score, on average, between 5.57 and 11.43 points higher on a test solving applied algebra problems with a calculator than students who learn by traditional methods. d.) yes, 0 isn't in the interval.
(Chap. 17: 55)Hot dogs, last one: explain why you think each of the following statements are true or false. A.) if I eat a meat hot dog instead of a beef hot dog, there's 90% chance I'll consume less fat. B.) 90% of meat hotdogs have between 1.4 and 6.5 grams less fat than beef hotdogs. C.) I'm 90% confident that meat hot dogs average between 1.4 and 6.5 grams less fat than beef hot dogs. D.)If I were to get more samples of both kinds of hotdogs, 90% of the time the meat hot dogs would average between 1.4 and 6.5 grams less fat than the beef hot dogs. E.) if I tested more samples, I'd expect about 90% of the resulting confidence intervals to include the true difference in mean fat content between the two kinds of hot dogs.
a.) False, the confidence interval is about means, not about individual hot dogs. b.) False, the confidence interval is about meas, not about individual hot dogs. c.) true. d.) False, Confidence intervals based on other samples will also try to estimate the true difference in population means; there's no reason to expect other samples to conform to this result. e.) true.
(Chap 16: 3)Which of the following are true? If false, explain briefly. a) A very high P-value is strong evidence that the null hypothesis is false. b) A very low P-value proves that the null hypothesis is false. c) A high P-value shows that the null hypothesis is true. d) A P-value below 0.05 is always considered sufficient evidence to reject a null hypothesis.
a.) False; A high Pvalue shows that the data are consistent with the null hypothesis, but provide no evidence for rejecting the null hypothesis. b.) False; It results in rejecting the null hypothesis, but doesn't prove that its false. c.) False; A high Pvalue shows that the data are consistent with the null hypothesis, but doesn't prove that the null hypothesis is true. d.) False; whether a Pvalue provides enough evidence to reject the null hypothesis depends on whether its smaller than the limit you set.
(Chap 16: 1) Which of the following are true? If false explain. a.) A Pvalue of 0.01 means that the null hypothesis is false. b.) A Pvalue of 0.01 means that the null hypothesis has a 0.01 chance of being true. c.) A Pvalue of 0.01 is evidence against the null hypothesis. d.) A Pvalue of 0.01 means we should definitely reject the null hypothesis.
a.) False; it shows evidence against it, but doesn't show its false. b.)False; the Pvalue isn't the probability that the null hypothesis is true. c.)True d.)False; whether a Pvalue provides enough evidence to reject the null hypothesis depends on the risk of a type 1 error that one is willing to assume (the alpha level).
(Chap 16: 27) In 2015, the U.S. census bureau reported that 62.2% of American families owned their homes-the lowest rate in 20 years... a.)In words what will their hypothesis be? b.)What would a type 1 error be? c.) What would a type 11 error be? d.)For each type of error, who would be harmed? e.)What would the power of the test rep. in this context.
a.) The null is that the level of home ownership remains the same. The alt. is that it rises. b.) The city concludes that home ownership is on the rise, but in fact the tax breaks don't help. c.) The city abandons the tax breaks, but they were helping. d.) A type 1 error causes the city to forgo tax revenue, while a type 11 error withdraws help from those who might have otherwise been able to buy a home. e.) The power of the test is the city's ability to detect actual increase in home ownership.
(Chap 16: 31)State regulators are checking up on repair shops to see if they are certifying vehicles that don't meet pollution standards. a.) In this context, whats meant by the power of the test the regulators are conducting? b.) Will the power be greater if they test 20 or 40 cars why? c.) Will the power be greater if they use a 5% or a 10% level of significacne? Why? d) Will the power be greater if they repair shop's inspectors are only a little out of compliance or a lot? why?
a.) The probability of detecting a shop thats not meeting standards. b.) 40 cars; larger n, which will reduce the standard error, making it easier to detect a failing shop. c.) 10%. Greater chance to reject Ho. d.) A lot. A larger effect size is easier to detect.
(Chap 16: 5)Which of the following are true? If false, explain briefly. a.) Using an alpha level of 0.05, a Pvalue of 0.04 results in rejecting the null hypothesis. b.) The alpha level depends on the sample size. c.) With an alpha level of 0.01, a Pvalue of 0.10 results in rejecting the null hypothesis. d.) Using an alpha level 0.05, a Pvalue of 0.06 means the null hypothesis is true.
a.) True b.)False; the alpha level isn't set independently and doesn't depend on sample size. c.) False; the Pvalue would have to be less than 0.01 to reject the null hypothesis. d.) False; it simply means we don't have enough evidence at that alpha level to reject the null hypothesis.
(Chap 16: 11)For each of the following situations, state whether a Type I or Type II, or neither error has been made. Explain briefly. (a) A bank wants to know if the enrollment on their website is above 30% based on a small sample of customers. They test H0:p= 0:3 versus HA:p >0:3 and reject the null hypothesis. Later they find out that actually 28% of all customers enrolled. (b) A student tests 100 students to determine whether other students on her campus prefer Coke or Pepsi and finds no evidence that preference for Coke is not 0.5.Later, a marketing company tests all students on campus and finds no difference. (c) A human resource analyst wants to know if the applicants this year score, on average, higher on their placement exam than the 52.5 points the candidates averaged last year. She samples 50 recent tests and finds the average to be 54.1 points. She fails to reject the null hypothesis that the mean is 52.5 points. At the end of the year, they find that the candidates this year had a mean of 55.3 points. (d) A pharmaceutical company tests whether a drug lifts the headache relief rate from the 25% achieved by the placebo. They fail to reject the null hypothesis because the p-value is 0.465. Further testing shows that the drug actually relieves headaches in 38% of people.
a.) Type 1 error. The actual value isn't greater than 0.3, but they rejected the null hypothesis. b.) No error; the acutal value is 0.50, which wasn't rejected. c.) Type 11 error. The actual value was 55.3 points, which is greater than 52.5. d.) Type 11 error. The null hypothesis wasn't rejected but it was false. The true relief rate was greater than 0.25.
(Chap 16: 23)Before lending someone money, banks must decide whether they believe the applicant will repay the loan. One strategy used is a point system. a.) When a person defaults on a loan, which type of error did the bank make? b.) Which kind of error is it when the bank misses an opportunity to make a loan to someone who would have repaid it? c.) Suppose the bank decides to lower the cutoff score from 250 points to 200. Is that analogous to choosing a higher or lower value of α for a hypothesis test. Explain. d.)What impact does this change in the cutoff value have on the chance of each type of error?
a.) Type 11 error b.) Type 1 error c.) By making it easier to get the loan, the bank has reduced the alpha level. d.) The risk of a type 1 error is decreased and the risk of a type 11 error is increased.
(Chap 18: 31) A tire manufacturer tested the braking performance of one of its tire models on a test track... a.) write a 95% confidence interval for the mean dry pavement stopping distance. Be sure to check the appropriate assumptions and conditions, and explain what your interval means. b.) Write a 95% confidence interval for the mean increase in stopping difference on wet pavement. Be sure to check the appropriate assumptions and conditions, and explain what your interval means.
a.) cars were probably not a simple random sample, but may be representative in terms of stopping distance; box-plot doesn't show outliers, but does indicate right skewness. A 95% confidence interval for the mean stopping distance on dry pavement is (131.8, 145.6) feet. b.) Data are paired by car; cars were probably not randomly chosen, but representative; box-plot shows an outlier (car4) with a difference of 12. With deletion of that car, a normal probability plot of the differences is relatively straight. Retaining the outlier, we estimate with 95% confidence that the average braking distance is between 38.8 and 62.6 feet more on wet pavement than on dry, based on this sample. without the outlier, the confidence interval is 47.2 to 62.8 feet.
(Chap 18: 13)Values for the labor force participation rate of women are published by the u.s bureau of labor stats... a.) which of these tests is appropriate for these data? b.) using the test you selected, state your conclusion.
a.) matched pairs; same cities in different periods. b.) there's a significant difference (P value = 0.0244) in the labor force participation rate for women in these cities; women's participation seems to have increased between 1968 and 1972.
(Chap 18: 27)Job satisfaction; A company institutes an exercise break for its workers to see if it will improve job satisfaction... a.) identify the procedure you would use to assess the effectiveness of the exercise program, and check to see if the conditions allow the use of that procedure. b.) test an appropriate hypothesis and state your conclusion. c.) if your conclusion turns out to be incorrect, what kind of error did you commit?
a.) paired sample test. Data are before/after for the same workers; workers randomly selected; assume fewer than 10% of all this company's workers; box-plot of differences shows them to be symmetric, with no outliers. b.) Ho: mu D = vs. Ha: mu D > 0, t= 3.60. P value = 0.0029. Because P< 0.01, reject Ho. These data show evidence that average job satisfaction has increased after implementation of the program. c.) Type 1.
(Chap 18: 1)Which of the following scenarios should be analyzed as paired data? a.) students take an MCAT prep course. Their before and after scores are compared. b.) 20 male and 20 female students in class take a midterm. We compare their scores. c.) A group of college freshmen are asked about the quality of the university cafeteria. A year later, the same students are asked about the cafeteria again. Do student's opinions change during their time at school?
a.) paired. b.) not paired. c.) paired.
(Chap. 17: 53)Hot dogs second helping...the resulting 90% confidence interval for mu:meat - mu: beef is (-6.5, -1.4). A.) the endpoints of this confidence interval are negative numbers, what does that indicate? B.) what does the fact that the confidence interval doesn't contain 0 indicate? C.) if we use this confidence interval to test the hypothesis that mu:meat - mu: beef =0, what's the corresponding alpha level?
a.) plausible values of mu:meat - mu:beef are all negative, so the mean fat content is probably higher for beef hot dogs. b.) the difference in sample means is significant. c.) 0.10
(Chap. 17: 33)A new vaccine was recently tested to see if it could prevent the painful and recurrent ear infections that many infants suffer from... A.) are the conditions for inference satisfied? B.) find a 95% confidence interval for the difference in rates of ear infection. C.) Use your confidence interval to explain whether think the vaccine is effective.
a.) yes, subjects are randomly divided into independent groups, and more than 10 successes and failures were observed in each group. b.) (4.7% , 8.9%) c.) We're 95% confident that the rate of infection is 5-9 percentage points lower. That's a meaningful reduction, considering the 20% infection rate among the unvaccinated kids.
(Chap. 17: 27)The centers for disease control and prevention reported a survey of randomly selected Americans... A.) are the assumptions and conditions necessary for inference satisfied? B.) create a 95% confidence interval for the difference in the proportions of senior men and women who have this disease. C.) interpret your interval in context. D.) does this confidence interval suggest that arthritis is more likely to afflict women than men?
a.) yes; random samples; less than 10% of the populations; samples are independent; more than 10 successes and failures in each sample. b.) (0.055, 0.140) c.) We are 95% confident, based on these samples, that the proportion of American women age 65 and older who suffer from arthritis is between 5.5% and 14.0% higher than the proportion of American men of the same age who suffer from arthritis. d.) yes, the entire interval lies above 0.