front 1 Find the critical value z _{α} _{/2} that corresponds to the given confidence level. 92% | back 1 z _{α} _{/2 }= 1.75 8% ÷ 2 = 4% 1.0000 – 0.0400 = 0.9600 |
front 2 Find the critical value z _{α} _{/2} that corresponds to α = 0.04. | back 2 z _{α} _{/2 }= 2.05 0.04 ÷ 2 = 0.02 1.0000 – 0.0200 = 0.9800 |
front 3 Find the critical value z _{α} _{/2} that corresponds to a 98% confidence level. | back 3 2.33 2% ÷ 2 = 1% 1.0000 – 0.0100 = 0.9900 |
front 4 Find the critical value z _{α} _{/2} that corresponds to a 90% confidence level. | back 4 1.645 10% ÷ 2 = 5% 1.0000 – 0.0500 = 0.9500 |
front 5 Find the critical value z _{α} _{/2} that corresponds to a 91.28% confidence level. | back 5 1.71 8.72% ÷ 2 = 4.36% 1.0000 – 0.0436 = 0.9564 |
front 6 Find z _{α} _{/2} for α = 0.07 | back 6 1.81 0.07 ÷ 2 = 0.035 1.0000 – 0.035 = 0.9650 ≈ 0.9649 |
front 7 Confidence level 95%; n = 15; σ is known; population appears to be very skewed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 7 Neither normal nor t distribution applies. |
front 8 90%; n = 200; σ = 13.0; population appears to be skewed Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 8 z_{α/2} = 1.645 10% ÷ 2 = 5% 1.0000 – 0.0500 = 0.9500 |
front 9 Confidence level 95%; n = 19; σ is unknown; population appears to be normally distributed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 9 t _{α/2 } = 2.101 5% ÷ 2 = 2.5% 18 Degrees of Freedom; Area in One Tail; 0.025 |
front 10 Confidence level 99%; n = 28; σ = 31.6; population appears to be normally distributed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 10 z_{α/2} = 2.575 1% ÷ 2 = 0.5% 1.0000 – 0.0050 = 0.9950 |
front 11 A newspaper provided a "snapshot" illustrating poll results from 1910 professionals who interview job applicants. The illustration showed that 26% of them said the biggest interview turnoff is that the applicant did not make an effort to learn about the job or the company. The margin of error was given as ±3 percentage points. What important feature of the poll was omitted? | back 11 The confidence level In this poll, the sample size is given as 1910 professionals, the point estimate is given as 26%, and the confidence interval is given as ±3 percentage points around the point estimate. However, the confidence level is not provided. It is often 95%, but media reports often neglect to identify it. |
front 12 Find the critical value z _{α} _{/2} for α = 0.09. | back 12 z _{α} _{/2} = 1.70 |
front 13 Express the confidence interval (0.043, 0.095) in the form of p̂ – E < p < p̂ + E | back 13 0.043 < p < 0.095 |
front 14 A research institute poll asked respondents if they acted to annoy a bad driver. In the poll, n = 2356, and x = 909 who said that they honked. Use a 95% confidence level.
| back 14 a. 0.386 p̂ = x/n = 909/2356 = 0.3858234295 b. E = 0.0197 z_{α/2} = z_{0.05/2} = z_{0.025} = 1.96 E = z_{α/2} × √(p̂×q^ ÷ n) = 1.96 × √[0.386(1 – 0.386) ÷ 2356] = 0.0196583139 c. 0.366 < p < 0.406 First check that the requirements to construct a confidence interval used to estimate a population proportion are met. 1. The sample is a simple random sample. 2. The conditions for the binomial distribution are satisfied. 3. There are at least 5 successes and at least 5 failures. p̂ – E = 0.386 – 0.0197 = 0.3663 p̂ + E = 0.386 + 0.0197 = 0.4057 d. One has 95% confidence that the interval from the lower bound to the upper bound actually does contain the true value of the population proportion. |
front 15 In the week before and the week after a holiday, there were 10,000 total deaths, and 4939 of them occurred in the week before the holiday.
| back 15 a. 0.484 < p < 0.504 b. No, because the proportion could easily equal 0.5. The interval is not less than 0.5 the week before the holiday. |
front 16 An online site presented this question, 'Would the recent norovirus outbreak deter you from taking a cruise?' Among the 34,742 people who responded, 61% answered 'yes'. Use the sample data to construct a 90% confidence interval estimate for the proportion of the population of all people who would respond 'yes' to that question. Does the confidence interval provide a good estimate of the population proportion? | back 16 0.606 < p < 0.614 p̂ = 0.61 z _{α/2} _{} = z_{0.05} = 1.645 E = z_{α/2} × √(p̂×q^ ÷ n) = 1.645 × √[0.61(0.39) ÷ 34,742] = 0.0043 p = 0.61 ± 0.0043 No, the sample is a voluntary sample and might not be representative of the population. |
front 17 Use the data in the table above to answer the following questions. Find the sample proportion of candy that are red. Use that result to construct a 90% confidence interval estimate of the population percentage of candy that are red. Is the result consistent with the 30 % rate that is reported by the candy maker? | back 17 The proportion of red candy = 0.25 Number of red candy = 9 Pieces of candy in sample bag = 36 The proportion of red candy = 9/36 = 0.25 13.1% < p < 36.9 % z _{α/2} _{} = z_{0.05} = 1.645 E = z_{α/2} × √(p̂×q^ ÷ n) = 1.645 × √[0.25(0.75) ÷ 36] = 0.1187 p = 0.25 ± 0.1187 = 0.131, 0.369 Yes, because the confidence interval includes 30%. |
front 18 A study of 420,052 cell phone users found that 139 of them developed cancer of the brain or nervous system. Prior to this study of cell phone use, the rate of such cancer was found to be 0.0365% for those not using cell phones.
| back 18 a. 0.028% < p < 0.039 % p̂ = 139 ÷ 420,052 = 0.00033091 z_{0.025} = 1.96 E = 1.96 × √[0.00033091(0.99966909) ÷ 420,052] = 0.000055003 0.00033091 ± 0.000055 = 0.00028, 0.00039 b. No, because 0.0365% is included in the confidence interval. |
front 19 Many states are carefully considering steps that would help them collect sales taxes on items purchased through the Internet. How many randomly selected sales transactions must be surveyed to determine the percentage that transpired over the Internet? Assume that we want to be 95% confident that the sample percentage is within eight percentage points of the true population percentage for all sales transactions. | back 19 n = 151 z_{ α/2 } = z_{0.025} = 1.96 n = ((z_{ α/2} )^{2} × 0.25) ÷ E^{2} = (1.96^{2} × 0.25) ÷ 0.08^{2} = 150.0625 |
front 20 Find the sample size, n, needed to estimate the percentage of adults who have consulted fortune tellers. Use a 0.05 margin of error, use a confidence level of 98%, and use results from a prior poll suggesting that 13% of adults have consulted fortune tellers. | back 20 n = 245 z_{ α/2 } = z_{0.01} = 2.326
n = ((z_{ α/2}
)^{2} × p̂ × q^) ÷ E^{2}
= (2.326^{2} × 0.13 × 0.87) ÷ 0.05^{2} = 244.7608862 |
front 21 A programmer plans to develop a new software system. In planning for the operating system that he will use, he needs to estimate the percentage of computers that use a new operating system. How many computers must be surveyed in order to be 90% confident that his estimate is in error by no more than four percentage points?
| back 21 a. n = 423 b. n = 50 c. Yes, using the additional survey information from part (b) dramatically reduces the sample size. The required sample size decreases dramatically from 423 to 50, with the addition of the survey information. |
front 22 Which of the following groups has terms that can be used interchangeably with the others? | back 22 Percentage, Probability, and Proportion Percentage, probability, and proportion can be used interchangeably with each other. The critical value cannot be used interchangeably with these terms because the critical value is a number separating sample statistics that are likely to occur from those that are unlikely to occur. |
front 23 Which of the following is NOT true of the confidence level of a confidence interval?
| back 23 There is a 1 – α chance, where α is the complement of the confidence level, that the true value of p will fall in the confidence interval produced from our sample. The confidence level is the probability that the confidence interval actually does contain the true value of p, not the other way around. Saying that "there is a 1 – α chance, where α is the complement of the confidence level, that the true value of p will fall in the confidence interval produced from our sample" is a common misinterpretation of the confidence interval. |
front 24 Which of the following is NOT a requirement for constructing a confidence interval for estimating the population proportion?
| back 24 The trials are done without replacement. The trials are done without replacement is not a requirement because The 5% Guideline for Cumbersome Calculations states that if calculations are cumbersome and if a sample size is no more than 5% of the size of the population, treat the selections as being independent (even if the selections are made without replacement, so that they are technically dependent). |
front 25
| back 25 a. The confidence interval methods of this section are robust against departures from normality, meaning they work well with distributions that aren't normal, provided that departures from normality are not too extreme. Confidence interval methods are robust against departures from normality if either the sample size is greater than 30, the population is normally distributed, or the departure from normality is not too extreme, which can be checked by using a histogram or a dotplot. b. Yes, because the dotplot resembles a normal distribution and the sample size is greater than 30. Examine the dotplot and determine if it meets the conditions for a confidence interval robust against departures from normality. Confidence interval methods are robust against departures from normality if either the sample size is greater than 30, the population is normally distributed, or the departure from normality is not too extreme, which can be checked by using a histogram or a dotplot. |
front 26 Using the simple random sample of weights of women from a data set, we obtain these sample statistics: n = 40 and x̄ = 147.72 lb. Research from other sources suggests that the population of weights of women has a standard deviation given by σ = 32.99 lb.
| back 26 a. The best point estimate is 147.72 lb. b. The 99% confidence interval estimate is 134.29 lb < µ < 161.15 lb. z_{0.005 } = 2.575 E = z _{α/2} _{} × σ/√(n) = 2.575 × 32.99/√(40) = 13.43165578 x̄ – E < µ < x̄ + E 147.72 – 13.432 < µ < 147.72 + 13.432 134.288 < µ < 161.152 |
front 27 Randomly selected students participated in an experiment to test their ability to determine when one minute (or sixty seconds) has passed. Forty students yielded a sample mean of 59.3 seconds. Assuming that σ = 10.5 seconds, construct a 99% confidence interval estimate of the population mean of all students. Based on the result, is it likely that the students' estimates have a mean that is reasonably close to sixty seconds? | back 27 The 99% confidence interval for the population mean is 55.0 < µ < 63.6. z_{0.005 } = 2.575 E = z _{α/2} _{} × σ/√(n) = 2.575 × 10.5/√(40) = 4.275004112 x̄ – E < µ < x̄ + E 59.3 – 4.275 < µ < 59.3 + 4.275 55.025 < µ < 63.575 Yes, because the confidence interval includes sixty seconds. |
front 28 A study of the ages of motorcyclists killed in crashes involves the random selection of 145 drivers with a mean of 31.66 years. Assuming that σ = 10.7 years, construct a 95% confidence interval estimate of the mean age of all motorcyclists killed in crashes. Notice that the confidence interval limits do not include ages below 20 years. What does this mean? | back 28 The 95% confidence interval for the population mean is 29.92 < µ < 33.40. z_{0.025 } = 1.96 E = z _{α/2} _{} × σ/√(n) = 1.96 × 10.7/√(145) = 1.741629803 x̄ – E < µ < x̄ + E 31.66 – 1.7416 < µ < 31.66 + 1.7416 29.9184 < µ < 33.4016 The mean age of the population will most likely not be less than 20 years old. |
front 29 Salaries of 38 college graduates who took a statistics course in college have a mean, x̄, of $69,000. Assuming a standard deviation, σ, of $15,315, construct a 99% confidence interval for estimating the population mean mu. | back 29 $ 62,603 < µ < $ 75,397 |
front 30 Confidence level 99%; n = 24; σ is known; population appears to be very skewed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 30 Neither normal nor t distribution applies. |
front 31 Confidence level 98%; n = 27; σ is unknown; population appears to be normally distributed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 31 t_{ α/2} = 2.479 |
front 32 Confidence level 99%; n = 28; σ is known; population appears to be very skewed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 32 Neither normal nor t distribution applies. |
front 33 Confidence level 99%; n = 20; σ is unknown; population appears to be normally distributed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 33 t_{ α/2} = 2.861 |
front 34 Find the critical value z _{α/2} that corresponds to a 98% confidence level. | back 34 2.33 |
front 35 Find the critical value –z _{α/2} that corresponds to a confidence level of 97.8%. | back 35 –2.29 |
front 36 98%; n = 7; σ = 27; population appears to be normally distributed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 36 z_{ α/2} = 2.33 |
front 37 90%; n = 10; σ is unknown; population appears to be normally distributed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 37 t_{ α/2} = 1.83 |
front 38 90%; n = 9; σ = 4.2; population appears to be very skewed. Do one of the following, as appropriate. (a) Find the critical value z _{α} _{/2} (b) Find the critical value t _{α} _{/2} (c) State that neither the normal nor the t distribution applies. | back 38 Neither the normal nor the t distribution applies. |
front 39 Weight lost on a diet: 90% confidence n = 51 x̄ = 3.0 kg s = 5.1 kg Use technology and the given confidence level and sample data to find the confidence interval for the population mean µ. Assume that the population does not exhibit a normal distribution. Is the confidence interval affected by the fact that the data appear to be from a population that is not normally distributed? | back 39 1.8 kg < µ < 4.2 kg t _{α/2} ^{} = t_{0.05} = 1.676 E = t _{α/2} ^{} × s÷√(n) = 1.676 × 5.1÷√(51) = 1.20 x̄ – E < µ < x̄ + E 3.0 – 1.2 < µ < 3.0 + 1.2 1.8 < µ < 4.2 No, because the sample size is large enough. |
front 40 Listed below are measured amounts of lead (in micrograms per cubic meter, or µg/m^{3}) in the air. The EPA has established an air quality standard for lead of 1.5 µg/m^{3}. The measurements shown below were recorded at a building on different days. 5.40 1.10 0.47 0.75 0.71 1.30 Use the given values to construct a 95% confidence interval estimate of the mean amount of lead in the air. Is there anything about this data set suggesting that the confidence interval might not be very good? | back 40 –0.345 µg/m^{3} < µ < 3.589 µg/m^{3} (calc): s = 1.874; x̄ = 1.622 t_{df,} _{ α/2} _{} = t_{5, 0.025} = 2.571 E = t _{α/2} ^{} × s÷√(n) = 2.571 × 1.874÷√(6) = 1.967 x̄ – E < µ < x̄ + E 1.622 – 1.967 < µ < 1.622 + 1.967 -0.345 < µ < 3.589 Yes, the value of 5.40 appears to be an outlier. |
front 41 In a test of the effectiveness of garlic for lowering cholesterol, 42 subjects were treated with garlic in a processed tablet form. Cholesterol levels were measured before and after the treatment. The changes in their levels of LDL cholesterol (in mg/dL) have a mean of 3.4 and a standard deviation of 18.1.
| back 41 a. The best point estimate is 3.4 mg/dL. x̄ = 3.4 mg/dL b. –2.24 mg/dL < µ < 9.04 mg/dL t_{df, } _{α/2} _{} = t_{41, 0.025} = 2.0195 E = t _{α/2} ^{} × s÷√(n) = 2.020 × 18.1÷√(42) = 5.641639081 x̄ – E < µ < x̄ + E 3.4 – 5.642 < µ < 3.4 + 5.6442 -2.242 < µ < 9.042 c. The confidence interval limits contain 0, suggesting that the garlic treatment did not affect the LDL cholesterol levels. |
front 42 Twelve different video games showing substance use were observed and the duration times of game play (in seconds) are listed below. The design of the study justifies the assumption that the sample can be treated as a simple random sample. 4044 3877 3852 4017 4308 4803 4660 4028 5010 4817 4342 4313 Use the data to construct a 99% confidence interval estimate of µ, the mean duration of game play. | back 42 3983.0 < µ < 4695.5 (calc): s = 397.349; x̄ = 4339.25 t_{df, } _{α/2} _{} = t_{11, 0.005} = 3.106 E = t _{α/2} ^{} × s÷√(n) = 3.106 × 397.349÷√(12) = 356.2730344 x̄ – E < µ < x̄ + E 4339.25 – 356.273 < µ < 4339.25 + 356.273 3982.977 < µ < 4695.523 |
front 43 A physician wants to develop criteria for determining whether a patient's pulse rate is atypical, and she wants to determine whether there are significant differences between males and females. Use the sample pulse rates above.
| back 43 a. 64.0 < µ < 82.4 (calc): s = 12.795; x̄ = 73.2 t_{df, } _{α/2} _{} = t_{9, 0.025} = 2.262 E = t _{α/2} ^{} × s÷√(n) = 2.262 × 12.795÷√(10) = 9.15235571 x̄ – E < µ < x̄ + E 73.2 – 9.2 < µ < 73.2 + 9.2 64 < µ < 82.4 b. 64.7 < µ < 90.5 (calc): s = 18.007; x̄ = 77.6 t_{df, } _{α/2} _{} = t_{9, 0.025} = 2.262 E = t _{α/2} ^{} × s÷√(n) = 2.262 × 18.007÷√(10) =12.88053687 x̄ – E < µ < x̄ + E 77.6 – 12.9 < µ < 77.6 + 12.9 64.7 < µ < 90.5 c. No, because the two confidence intervals overlap, we cannot conclude that the two population means are different. |
front 44 An IQ test is designed so that the mean is 100 and the standard deviation is 17 for the population of normal adults. Find the sample size necessary to estimate the mean IQ score of statistics students such that it can be said with 99% confidence that the sample mean is within 7 IQ points of the true mean. Assume that σ = 17 and determine the required sample size using technology. Then determine if this is a reasonable sample size for a real world calculation. | back 44 The required sample size is 40. z _{α/2} = z_{0.005} = 2.575 n = [ (z _{α/2} × σ) ÷ E ] ^{2} = [ (2.575 × 17) ÷ 7 ] ^{2} = 39.10715561 Yes. This number of IQ test scores is a fairly small number. |
front 45 A student wants to estimate the mean score of all college students for a particular exam. First use the range rule of thumb to make a rough estimate of the standard deviation of those scores. Possible scores range from 300 to 2200. Use technology and the estimated standard deviation to determine the sample size corresponding to a 95% confidence level and a margin of error of 100 points. What isn't quite right with this exercise? | back 45 The range rule of thumb estimate for the standard deviation is 475. σ ≈ range ÷ 4 ≈ (2200 – 300) ÷ 4 ≈ 475 A confidence level of 95% requires a minimum sample size of 87. z _{α/2} = z_{0.025} = 1.96 n = [ (z _{α/2} × σ) ÷ E ] ^{2} = [ (1.96 × 475) ÷ 100 ] ^{2} = 86.6761 A margin of error of 100 points seems too high to provide a good estimate of the mean score. |
front 46 Which of the following is NOT a property of the Student t distribution?
| back 46 The standard deviation of the Student t distribution is s = 1. The standard deviation of the Student t distribution is greater than 1, unlike the standard normal distribution, which has a standard deviation of 1. |
front 47 Which of the following calculations is NOT derived from the confidence interval?
| back 47 The population mean, µ = (upper confidence limit) + (lower confidence limit) The population mean is not calculated from the confidence interval estimate; it is a fixed constant. |
front 48 Which of the following is NOT a requirement for constructing a confidence interval for estimating a population mean with σ known?
| back 48 The confidence level is 95%. A confidence level of 95% is not a requirement for constructing a confidence interval to estimate the mean with sigma known. The confidence level may vary. |
front 49 Which of the following would be a correct interpretation of a 99% confidence interval such as 4.1 < µ < 5.6? | back 49 We are 99% confident that the interval from 4.1 to 5.6 actually does contain the true value of µ. If we were to select many different samples of the same size and construct the corresponding confidence intervals, in the long run, 99% of them would actually contain the value of µ. |
front 50 Which of the following is NOT an equivalent expression for the confidence interval given by 161.7 < µ < 89.5?
| back 50 161.7 ±27.8 The interval 161.7 ±27.8 would actually produce 133.9 < µ < 189.5. |
front 51 Which of the following is NOT required to determine minimum sample size to estimate a population mean?
| back 51 The size of the population, N The minimum sample size does not depend on the size of the population of interest. |