Elementary Statistics: Active Learning Lecture Slides For use with Classroom Response Systems Chapter 4: Discrete Probability Distributions Elementary Statistics: Picturing the World Fifth Edition by Larson and Farber © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. True or false: The number of kittens in a litter is an example of a discrete random variable. True False © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. True or false: The number of kittens in a litter is an example of a discrete random variable. True False © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Determine the probability distribution’s missing probability value. 0.25 0.65 0.15 0.35 x 1 2 3 P(x) 0.25 0.30 ? 0.10 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Determine the probability distribution’s missing probability value. 0.25 0.65 0.15 0.35 x 1 2 3 P(x) 0.25 0.30 ? 0.10 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Let x represent the number of televisions in a household: Find the mean. 2 1.5 6 0.25 x 1 2 3 P(x) 0.05 0.20 0.45 0.30 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Let x represent the number of televisions in a household: Find the mean. 2 1.5 6 0.25 x 1 2 3 P(x) 0.05 0.20 0.45 0.30 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Let x represent the number of televisions in a household: Find the standard deviation. 1.29 0.837 0.146 1.12 x 1 2 3 P(x) 0.05 0.20 0.45 0.30 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Let x represent the number of televisions in a household: Find the standard deviation. 1.29 0.837 0.146 1.12 x 1 2 3 P(x) 0.05 0.20 0.45 0.30 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Forty-three percent of marriages end in divorce. You randomly select 15 married couples. Find the probability exactly 5 of the marriages will end in divorce. 0.160 0.015 0.039 0.333 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Forty-three percent of marriages end in divorce. You randomly select 15 married couples. Find the probability exactly 5 of the marriages will end in divorce. 0.160 0.015 0.039 0.333 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Forty-three percent of marriages end in divorce. You randomly select 15 married couples. Find the mean number of marriages that will end in divorce. 2.15 8.55 6.45 2.85 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Forty-three percent of marriages end in divorce. You randomly select 15 married couples. Find the mean number of marriages that will end in divorce. 2.15 8.55 6.45 2.85 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the probability using the standard normal distribution. P(z < 1.49) 0.9319 0.0681 0.6879 0.3121 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the probability using the standard normal distribution. P(z < 1.49) 0.9319 0.0681 0.6879 0.3121 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the probability using the standard normal distribution. P(z ≥ –2.31) 0.0104 0.0087 0.9896 0.9913 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the probability using the standard normal distribution. P(z ≥ –2.31) 0.0104 0.0087 0.9896 0.9913 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the probability using the standard normal distribution. P(–2.14 < z < 0.95) 0.1170 0.0681 0.1873 0.8127 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the probability using the standard normal distribution. P(–2.14 < z < 0.95) 0.1170 0.0681 0.1873 0.8127 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability a randomly selected person has an IQ score greater than 120. 0.9082 0.0918 0.6293 0.3707 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability a randomly selected person has an IQ score greater than 120. 0.9082 0.0918 0.6293 0.3707 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability a randomly selected person has an IQ score between 100 and 120. 0.9082 0.0918 0.4082 0.5918 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. Find the probability a randomly selected person has an IQ score between 100 and 120. 0.9082 0.0918 0.4082 0.5918 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the z-score that has 2.68% of the distribution’s area to its right. z = 0.9963 z = –1.93 z = –0.0037 z = 1.93 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the z-score that has 2.68% of the distribution’s area to its right. z = 0.9963 z = –1.93 z = –0.0037 z = 1.93 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What IQ score represents the 98th percentile? 131 69 113 145 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. IQ scores are normally distributed with a mean of 100 and a standard deviation of 15. What IQ score represents the 98th percentile? 131 69 113 145 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. American children watch an average of 25 hours of television per week with a standard deviation of 8 hours. A random sample of 40 children is selected. What is the probability the mean number of hours of television they watch per week is less than 22? 0.3520 0.0089 0.9911 0.6480 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. American children watch an average of 25 hours of television per week with a standard deviation of 8 hours. A random sample of 40 children is selected. What is the probability the mean number of hours of television they watch per week is less than 22? 0.3520 0.0089 0.9911 0.6480 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Use a correction for continuity to convert the following interval to a normal distribution interval. The probability of getting at least 80 successes x > 80.5 x > 79.5 x < 80.5 x < 79.5 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Use a correction for continuity to convert the following interval to a normal distribution interval. The probability of getting at least 80 successes x > 80.5 x > 79.5 x < 80.5 x < 79.5 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a point estimate for the mean price of all textbooks. $42 $12.30 $114.50 $2.73 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the critical value zc necessary to form a 98% confidence interval. 2.33 2.05 0.5040 0.8365 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the critical value zc necessary to form a 98% confidence interval. 2.33 2.05 0.5040 0.8365 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a 98% confidence interval for the mean price of all textbooks. (111.38, 117.62) (110.08, 118.92) (110.78, 118.22) (109.61, 119.39) © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. A random sample of 42 textbooks has a mean price of $114.50 and a standard deviation of $12.30. Find a 98% confidence interval for the mean price of all textbooks. (111.38, 117.62) (110.08, 118.92) (110.78, 118.22) (109.61, 119.39) © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Determine the minimum sample size needed to construct a 95% confidence interval for the mean age of employees at a company. The estimate must be accurate to within 0.5 year. Assume the standard deviation is 4.8 years. 18 19 354 355 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Determine the minimum sample size needed to construct a 95% confidence interval for the mean age of employees at a company. The estimate must be accurate to within 0.5 year. Assume the standard deviation is 4.8 years. 18 19 354 355 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the critical value tc necessary to form a 95% confidence interval with a sample size of 15. 1.960 2.145 2.131 2.120 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the critical value tc necessary to form a 95% confidence interval with a sample size of 15. 1.960 2.145 2.131 2.120 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. A random sample of 15 DVD players has a mean price of $64.30 and a standard deviation of $5.60. Find a 95% confidence interval for the mean price of all DVD players. (61.20, 67.40) (61.47, 67.13) (61.22, 67.38) (61.10, 67.51) © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. A random sample of 15 DVD players has a mean price of $64.30 and a standard deviation of $5.60. Find a 95% confidence interval for the mean price of all DVD players. (61.20, 67.40) (61.47, 67.13) (61.22, 67.38) (61.10, 67.51) © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a point estimate for the proportion of all Internet users who have high-speed Internet access at home. 1.28 0.78 0.22 195 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a point estimate for the proportion of all Internet users who have high-speed Internet access at home. 1.28 0.78 0.22 195 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a 90% confidence interval for the proportion of all Internet users who have high-speed Internet access at home. (1.19, 1.38) (0.728, 0.832) (0.731, 0.829) (0.737, 0.823) © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. In a survey of 250 Internet users, 195 have high-speed Internet access at home. Find a 90% confidence interval for the proportion of all Internet users who have high-speed Internet access at home. (1.19, 1.38) (0.728, 0.832) (0.731, 0.829) (0.737, 0.823) © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. You want to estimate, with 95% confidence, the proportion of households with pets. Your estimate must be accurate within 3% of the population proportion. No preliminary estimate is available. Find the minimum sample size needed. 1141 3267 1068 1067 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. You want to estimate, with 95% confidence, the proportion of households with pets. Your estimate must be accurate within 3% of the population proportion. No preliminary estimate is available. Find the minimum sample size needed. 1141 3267 1068 1067 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. State the null and alternative hypotheses. A company claims the mean lifetime of its AA batteries is more than 16 hours. H0: μ > 16 Ha: μ ≤ 16 H0: μ < 16 Ha: μ ≥ 16 H0: μ ≤ 16 Ha: μ > 16 H0: μ ≥ 16 Ha: μ < 16 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. State the null and alternative hypotheses. A company claims the mean lifetime of its AA batteries is more than 16 hours. H0: μ > 16 Ha: μ ≤ 16 H0: μ < 16 Ha: μ ≥ 16 H0: μ ≤ 16 Ha: μ > 16 H0: μ ≥ 16 Ha: μ < 16 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. State the null and alternative hypotheses. A student claims the mean cost of a textbook is at least $125. H0: μ > 125 Ha: μ ≤ 125 H0: μ < 125 Ha: μ ≥ 125 H0: μ ≤ 125 Ha: μ > 125 H0: μ ≥ 125 Ha: μ < 125 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. State the null and alternative hypotheses. A student claims the mean cost of a textbook is at least $125. H0: μ > 125 Ha: μ ≤ 125 H0: μ < 125 Ha: μ ≥ 125 H0: μ ≤ 125 Ha: μ > 125 H0: μ ≥ 125 Ha: μ < 125 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. True or false: Testing the claim that at least 88% of students have a cell phone would be a right-tail test. True False © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. True or false: Testing the claim that at least 88% of students have a cell phone would be a right-tail test. True False © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. You are testing the claim that the mean cost of a new car is more than $25,200. How should you interpret a decision that rejects the null hypothesis? There is enough evidence to reject the claim. There is enough evidence to support the claim. There is not enough evidence to reject the claim. There is not enough evidence to support the claim. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. You are testing the claim that the mean cost of a new car is more than $25,200. How should you interpret a decision that rejects the null hypothesis? There is enough evidence to reject the claim. There is enough evidence to support the claim. There is not enough evidence to reject the claim. There is not enough evidence to support the claim. © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. True or false: Given H0: μ = 40 Ha: μ ≠ 40 and P = 0.0436. You would reject the null hypothesis at the 0.05 level of significance. True False © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. True or false: Given H0: μ = 40 Ha: μ ≠ 40 and P = 0.0436. You would reject the null hypothesis at the 0.05 level of significance. True False © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the critical value, z0, for a left-tailed test at the 0.10 level of significance. z0 = –1.645 z0 = 1.645 z0 = –1.28 z0 = 1.28 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the critical value, z0, for a left-tailed test at the 0.10 level of significance. z0 = –1.645 z0 = 1.645 z0 = –1.28 z0 = 1.28 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the standardized test statistic z for the following situation: Claim: μ >15; s = 3.4 n = 40 z = 2.60 z = –2.60 z = –0.07 z = 12.90 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the standardized test statistic z for the following situation: Claim: μ >15; s = 3.4 n = 40 z = 2.60 z = –2.60 z = –0.07 z = 12.90 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the critical value(s), t0, for a two-tailed test, α = 0.05, and n = 8. –t0 = –1.96 and t0 = 1.96 –t0 = –2.306 and t0 = 2.306 –t0 = –1.895 and t0 = 1.895 –t0 = –2.365 and t0 = 2.365 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the critical value(s), t0, for a two-tailed test, α = 0.05, and n = 8. –t0 = –1.96 and t0 = 1.96 –t0 = –2.306 and t0 = 2.306 –t0 = –1.895 and t0 = 1.895 –t0 = –2.365 and t0 = 2.365 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. How many 4-letter television call signs are possible, if each sign must start with either a K or a W? 456,976 35,152 16 104 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. How many 4-letter television call signs are possible, if each sign must start with either a K or a W? 456,976 35,152 16 104 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The spinner shown is spun one time. Find the probability the spinner lands on blue. 0.375 0.5 0.125 0.25 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The spinner shown is spun one time. Find the probability the spinner lands on blue. 0.375 0.5 0.125 0.25 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The bar graph shows the cell phone provider for students in a class. One of these students is chosen at random. Find the probability that their provider is not AT&T. 0.3 0.6 0.125 0.4 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The bar graph shows the cell phone provider for students in a class. One of these students is chosen at random. Find the probability that their provider is not AT&T. 0.3 0.6 0.125 0.4 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. One card is selected at random from a standard deck, then replaced, and a second card is drawn. Find the probability of selecting two face cards. 0.050 0.053 0.038 0.462 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. One card is selected at random from a standard deck, then replaced, and a second card is drawn. Find the probability of selecting two face cards. 0.050 0.053 0.038 0.462 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. One card is selected at random from a standard deck, not replaced, and then a second card is drawn. Find the probability of selecting two face cards. 0.050 0.053 0.446 0.038 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. One card is selected at random from a standard deck, not replaced, and then a second card is drawn. Find the probability of selecting two face cards. 0.050 0.053 0.446 0.038 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is male, given that they prefer pepperoni. 0.333 0.6 0.208 0.556 Cheese Pepperoni Sausage Total Male 8 5 2 15 Female 4 3 9 10 24 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is male, given that they prefer pepperoni. 0.333 0.6 0.208 0.556 Cheese Pepperoni Sausage Total Male 8 5 2 15 Female 4 3 9 10 24 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. True or False: The following events are mutually exclusive. Event A: Being born in California Event B: Watching American Idol True False © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. True or False: The following events are mutually exclusive. Event A: Being born in California Event B: Watching American Idol True False © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is female or prefers sausage. 0.458 0.583 0.125 0.556 Cheese Pepperoni Sausage Total Male 8 5 2 15 Female 4 3 9 10 24 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is female or prefers sausage. 0.458 0.583 0.125 0.556 Cheese Pepperoni Sausage Total Male 8 5 2 15 Female 4 3 9 10 24 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. There are 15 dogs entered in a show. How many ways can first, second, and third place be awarded? 45 455 2,730 3,375 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. There are 15 dogs entered in a show. How many ways can first, second, and third place be awarded? 45 455 2,730 3,375 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. There are 13 students in a club. How many ways can four students be selected to attend a conference? 17,160 52 28,561 715 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. There are 13 students in a club. How many ways can four students be selected to attend a conference? 17,160 52 28,561 715 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the class width: 3 4 5 19 Class Frequency, f 1 – 5 21 6 – 10 16 11 – 15 28 16 – 20 13 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the class width: 3 4 5 19 Class Frequency, f 1 – 5 21 6 – 10 16 11 – 15 28 16 – 20 13 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Estimate the frequency of the class with the greatest frequency. 28 21 58 53 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Estimate the frequency of the class with the greatest frequency. 28 21 58 53 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. What is the maximum data entry? 96 38 9 41 Key: 3 | 8 = 38 3 8 9 4 0 2 7 5 1 1 4 8 6 3 3 3 8 9 9 7 0 0 1 1 2 4 7 8 8 8 8 9 8 2 2 3 4 7 7 8 9 9 9 1 1 4 5 6 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. What is the maximum data entry? 96 38 9 41 Key: 3 | 8 = 38 3 8 9 4 0 2 7 5 1 1 4 8 6 3 3 3 8 9 9 7 0 0 1 1 2 4 7 8 8 8 8 9 8 2 2 3 4 7 7 8 9 9 9 1 1 4 5 6 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mean. 78.5 79 474 78 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mean. 78.5 79 474 78 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the median. 78.5 79 79.5 78 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the median. 78.5 79 79.5 78 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mode. 78.5 79 79.5 78 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the mode. 78.5 79 79.5 78 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the standard deviation. 2.2 6 2 4.8 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The heights (in inches) of a sample of basketball players are shown: 76 79 81 78 82 78 Find the standard deviation. 2.2 6 2 4.8 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Determine the type of correlation between the variables. Positive linear correlation Negative linear correlation No linear correlation © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Determine the type of correlation between the variables. Positive linear correlation Negative linear correlation No linear correlation © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Calculate the correlation coefficient r, for temperature (x) and number of cups of coffee sold per hour (y). 0.946 –0.973 –2.469 81.760 x 65 60 55 50 45 40 35 30 25 y 8 10 11 13 12 16 19 22 23 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Calculate the correlation coefficient r, for temperature (x) and number of cups of coffee sold per hour (y). 0.946 –0.973 –2.469 81.760 x 65 60 55 50 45 40 35 30 25 y 8 10 11 13 12 16 19 22 23 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the t test statistic to determine if the correlation coefficient for temperature (x) and number of cups of coffee sold per hour (y) is significant. 10.8 –0.973 –11.1 –1.8 x 65 60 55 50 45 40 35 30 25 y 8 10 11 13 12 16 19 22 23 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the t test statistic to determine if the correlation coefficient for temperature (x) and number of cups of coffee sold per hour (y) is significant. 10.8 –0.973 –11.1 –1.8 x 65 60 55 50 45 40 35 30 25 y 8 10 11 13 12 16 19 22 23 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the equation of the regression line for temperature (x) and number of cups of coffee sold per hour (y). x 65 60 55 50 45 40 35 30 25 y 8 10 11 13 12 16 19 22 23 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. Find the equation of the regression line for temperature (x) and number of cups of coffee sold per hour (y). x 65 60 55 50 45 40 35 30 25 y 8 10 11 13 12 16 19 22 23 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The equation of the regression line for temperature (x) and number of cups of coffee sold per hour (y) is Predict the number of cups of coffee sold per hour when the temperature is 48º. 41.4 30.7 13.8 50.5 © 2012 Pearson Education, Inc.

© 2012 Pearson Education, Inc. The equation of the regression line for temperature (x) and number of cups of coffee sold per hour (y) is Predict the number of cups of coffee sold per hour when the temperature is 48º. 41.4 30.7 13.8 50.5 © 2012 Pearson Education, Inc.