Reference no: EM131036895
1. T/F If there is not enough evidence to support the alternate hypothesis, we accept theNull hypothesis
2. Suppose that the mean time for a certain car to go from 0 to 60 miles per hour was 6.7 seconds. Suppose that you want to set up a statistical test to challenge the claim of 6.7 seconds. What would you use for the null hypothesis?
a) H0: mu > 6.7 seconds b) H0: mu = 6.7 seconds c) H0: mu < 6.7 seconds
d) H0: mu /= 6.7 seconds e) H0: mu <= 6.7 seconds
3) If the calculated zvalue for testing Ha: mu1 - mu2 /= 0 in a two-sampleztest of level .05 is 1.92, then
a. The P-value is .0548 and H0 should be rejected.
b. The P-value is .0548 and H0 should not be rejected.
c. The P-value is .0274 and H0should be rejected.
d. The P-value is .0274 and H0 should not be rejected
4) Suppose that the mean time for a certain car to go from 0 to 60 miles per hour was 7.7 seconds. Suppose that you want to test the claim that the average time to accelerate from 0 to 60 miles per hour is longer than 7.7 seconds. What would you use for the alternative hypothesis?
a) H0: mu <7.7 seconds b) H0: mu >7.7 seconds c) H0: mu =7.7 seconds
d) H0: mu >=7.7 seconds e) H0: mu <=7.7 seconds
5) Suppose that x has a distribution with mu= 15 and sigma = 5. If a random sample is taken of size n = 42, find mu.
a) 5.00 b) 0.77 c) 2.31 d) 14.00 e) 15.00
6) T/F Always make decision about the null hypothesis
7) T/F When actually practicing and estimating confidence intervals, it is more realistic to use a z interval than a t interval
8) A research firm wants to determine whether there's a difference in married couples between what the husband earns and what the wife earns. The firm takes a random sample of married couples and measures the annual salary of each husband and wife. What procedure should the firm use to analyze the data for the mean difference in salary within married couples?
a) One-sample t procedure
b) Two-sample t procedure
c) One-sample z procedure
d) Two-sample z procedure
e) Not enough information to determine which procedure should be used.
9)What is a parameter?
a) A conclusion about the value of a population parameter based on information about the corresponding sample statistic and probability
b) A numerical descriptive measure of a population
c) A set of measurements (or counts), either existing or conceptual
d) A probability distribution for a sample statistic
e) A numerical descriptive measure of a sample
10) Suppose a 95% confidence interval is computed for resulting in the interval (112.4, 121.6). Then,
a. 95% of the time,mu falls within the interval.
b. There is a 95% chance that mu falls within the interval.
c. 95% of all the possible values of mu fall with the interval.
d. 95% of all the possible samples produce intervals that do capture mu .
11)If sigma = 10, then the sample size required to estimate a population mean mu to within .5 with 95% confidence is
a. 40 b. 119 c. 1257 d. 1537
12) Suppose a certain species bird has an average weight of xbar = 3.20 grams. Based on previous studies, we can assume that the weights of these birds have a normal distribution with sigma = .29 grams. Find the sample size necessary for an 85% confidence level with a maximal error of estimate for the mean weights of the hummingbirds.
a a) 18 b) 12 c) 3 d ) 7 e) 5
13)Give an example of a population.
a) Seven cards chosen at random from a 52-card deck
b) A week of television shows watched by Americans as reported in a survey
c) The lengths of all trout in a lake
d) The automobiles bought by Americans polled in a telephone survey
e) Registered Oklahoma voters who voted in a U.S. presidential election
15) Suppose thirty-nine communities gave an average of xbar = 147.8 reported cases of larceny per year. Assume that sigma is known to be 39.9 cases per year. Find a 90%, 95%, and 98% confidence interval for the population mean annual number of reported larceny cases in such communities. Compare the margins of error. As the confidence level increase, do the margins of error increase?
a) The 90% confidence level has a margin of error of 65.6; the 95% confidence level has a margin of error of 78.2; and the 98% confidence level has a margin of error of 93.0. As the confidence level increases, the margins of error increases.
b) The 90% confidence level has a margin of error of 10.5; the 95% confidence level has a margin of error of 12.5; and the 98% confidence level has a margin of error of 14.9. As the confidence level increases, the margins of error increases.
c) The 90% confidence level has a margin of error of 1.7; the 95% confidence level has a margin of error of 2.0; and the 98% confidence level has a margin of error of 2.4. As the confidence level increases, the margins of error increases.
d) The 90% confidence level has a margin of error of 14.9; the 95% confidence level has a margin of error of 12.5; and the 98% confidence level has a margin of error of 10.5. As the confidence level increases, the margins of error decreases.
e) The 90% confidence level has a margin of error of 93.0; the 95% confidence level has a margin of error of 78.2; and the 98% confidence level has a margin of error of 65.6. As the confidence level increases, the margins of error decreases.
16)Suppose a certain species bird has an average weight of x bar = 3.85 grams. Based on previous studies, we can assume that the weights of these birds have a normal distribution with sigma = .31 grams. For a small group of 18 birds, find a 70% confidence interval for the average weights of these birds.
a 2.81 grams to 4.17 grams
b 2.81 grams to 3.93 grams
c 3.77 grams to 4.17 grams
d 3.77 grams to 3.93 grams
e 3.53 grams to 3.92 grams
17) T/F The higher the confidence level the wider the confidence interval.
18) Suppose the method of tree ring dating gave the following dates A.D. for an archaeological excavation site.
1234 1219 1266 1213 1252 1194 1272 1209 1227
Find a 80% confidence interval for the mean of all tree ring dates from this archaeological site.
a 1220.0 to 1244.0 b 1219.3 to 1243.5 c 1220.0 to 1244.2
d 1219.3 to 1244.2 e 1219.5 to 1243.5
19) Suppose a certain species bird has an average weight of xbar = 3.8 grams. Based on previous studies, we can assume that the weights of these birds have a normal distribution with sigma = .36 grams. For a small group of 18 birds, find the margin of error for a 70% confidence interval for the average weights of these birds.
a 0.19 grams
b 0.09 grams
c 0.52 grams
d 0.37 grams
e 0.04 grams
20) Letxbe a random variable representing dividend yield of Australian bank stocks. We may assume thatxhas a normal distribution with sigma = 2.8%. A random sample of 16 Australian bank stocks has a sample mean of xbar = 8.91%. For the entire Australian stock market, the mean dividend yield is mu = 6.4%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 6.4%? Use alpha = 0.05. What is the value of the test statistic?
a) -0 .224 b) -3 .586 c) 3.586 d) 0.896 e) 0.224
21) The heights of 18-year-old men are approximately normally distributed with mean 68 inches and standard deviation 3 inches. What is the probability that the average height of a sample of ten 18-year-old men will be between 66 and 69 inches?
a) 0.8357 b) 0.1643 c) 0.3357 d) 0.4179 e) 0.8817
22) T/F A type II error is called α
23) T/F Type I error is when you reject the null hypothesis that is really true.
24) The heights of 18-year-old men are approximately normally distributed with mean 68 inches and standard deviation 3 inches. What is the probability that an 18-year-old man selected at random is less than 71 inches tall?
a) 0.1587 b) 0.8413 c) 0.3413 d) 0.6826 e) 0.3174
25) Suppose thirty-eight communities gave an average of xbar = 125.2 reported cases of larceny per year. Assume that sigma is known to be 37.5 cases per year. Find a 85%, 90%, and 95% confidence interval for the population mean annual number of reported larceny cases in such communities. Compare the lengths of the confidence intervals. As the confidence levels increase, do the confidence intervals increase in length?
a) The 85% confidence level has a confidence interval length of 3.9; the 90% confidence level has a confidence interval length of 3.2; and the 95% confidence level has a confidence interval length of 2.8. As the confidence level increases, the confidence interval length decreases.
b) The 85% confidence level has a confidence interval length of 108.0; the 90% confidence level has a confidence interval length of 123.4; and the 95% confidence level has a confidence interval length of 147.0. As the confidence level increases, the confidence interval length increases.
c) The 85% confidence level has a confidence interval length of 2.8; the 90% confidence level has a confidence interval length of 3.2; and the 95% confidence level has a confidence interval length of 3.9. As the confidence level increases, the confidence interval lengths increases.
d) The 85% confidence level has a confidence interval length of 147.0; the 90% confidence level has a confidence interval length of 123.4; and the 95% confidence level has a confidence interval length of 108.0. As the confidence level increases, the confidence interval lengths decreases.
e) The 85% confidence level has a confidence interval length of 17.5; the 90% confidence level has a confidence interval length of 20.0; and the 95% confidence level has a confidence interval length of 23.8. As the confidence level increases, the confidence interval lengths increases.
26) Aseverestorm has an average peak wave height of 16.4 feet for waves hitting the shore. Suppose that a storm is in progress with a severe storm class rating. Let us say that we want to set up a statistical test to see if the wave action (i.e., height) is dying down or getting worse. If you wanted to test the hypothesis that the waves are dying down, what would you use for the alternate hypothesis? Is the P-value area on the left, right, or on both sides of the mean?
a) H1: mu is greater than 16.4 feet; the P-value area is on both sides of the mean
b) H1: mu is greater than 16.4 feet; the P-value area is on the left of the mean
c) H1: mu is not equal to 16.4 feet; the P-value area is on the right of the mean
d) H1: mu is not equal to 16.4 feet; the P-value area is on the left of the mean
e) H1: mu is less than 16.4 feet; the P-value area is on the left of the mean
27) T/F Large p-values shows support for the null hypothesis, so we accept that it is true
28)The heights of 18-year-old men are approximately normally distributed with mean 68 inches and standard deviation 3 inches. What is the probability that the average height of a sample of twenty 18-year-old men will be less than 69 inches? Round your answer to four decimal places.
a) 0.0968 b) 0.4032 c) 0.8064 d) 0.9320 e) 0.4516
29) T/F The width of a confidence interval for mu based on the z formula decreases as the sample size grows larger.
30) Letxbe a random variable representing dividend yield of Australian bank stocks. We may assume thatxhas a normal distribution with sigma = 2.4% . A random sample of 19 Australian bank stocks has a sample mean xbar = 8.71%. For the entire Australian stock market, the mean dividend yield is mu = 5.9% Do these data indicate that the dividend yield of all Australian bank stocks is higher than 5.9%? Use alpha = 0.05 Are the data statistically significant at the given level of significance? Based on your answers, will you reject or fail to reject the null hypothesis?
a) The P-value is less than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.
b) The P-value is less than the level of significance and so the data are statistically significant. Thus, we fail to reject the null hypothesis
c) The P-value is greater than the level of significance and so the data are statistically significant. Thus, we fail to reject the null hypothesis.
d) The P-value is greater than the level of significance and so the data are not statistically significant. Thus, we reject the null hypothesis.
e) The P-value is less than the level of significance and so the data are statistically significant. Thus, we reject the null hypothesis.
31) Letxbe a random variable representing dividend yield of Australian bank stocks. We may assume thatxhas a normal distribution with sigma = 2.6%. A random sample of 24 Australian bank stocks has a mean xbar = 7.63% For the entire Australian stock market, the mean dividend yield is mu = 5%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 5%? Use alpha = 0.05 What is the level of significance?
a) 0.900 b) 0.050 c) 0.950 d) 0.975 e) 0.100
32) T/F All other things being equal, choosing a smaller value ofαwill increases the probability of making a type II error.
33) T/F The level of significance of a test is the probability of making a type I error, given that the null hypothesis is true
34) Letxbe a random variable representing dividend yield of Australian bank stocks. We may assume thatxhas a normal distribution with sigma = 2.9%. A random sample of 9 Australian bank stocks has a sample mean of xbar = 9.79%. For the entire Australian stock market, the mean dividend yield is mu = 7.8%. Do these data indicate that the dividend yield of all Australian bank stocks is higher than 7.8%? Use alpha = 0.01 Find (or estimate) the P-value.
a) 0.010 b) 0.490 c) 0.020 d) 1.960 e) 0.980