2) Using the data in the table above, compute the sample mean. x 2 5 8 11 14 17 f 26 12 57 25 34 18 1) Using the data in the table above, compute the sample standard deviation. A) 4.582 B) 4.569 C) 5.128 D) 9.448 2) Using the data in the table above, compute the sample mean. 3) A research team has found that 39.7% of U.S. teen-aged drivers have access to a car that is exclusively theirs to use. If you conducted interviews with 3,458 U.S. teen-aged drivers, about how many would you expect to have exclusive access to a car? A) 40 B) 1,373 C) 2,085 D) 8,710

First step is to put the numbers into L1 and L2 by going to STAT-Edit. x 2 5 8 11 14 17 f 26 12 57 25 34 18 1) Using the data in the table above, compute the sample standard deviation. A) 4.582 B) 4.569 C) 5.128 D) 9.448 First step is to put the numbers into L1 and L2 by going to STAT-Edit. The quickest way from here is to go to STAT-Calc-1, designate L1 as your List, and L2 as the FreqList, and Calculate. The calculator will tell you that Sx = 4.582. The answer is A.

2) Using the data in the table above, compute the sample mean. x 2 5 8 11 14 17 f 26 12 57 25 34 18 2) Using the data in the table above, compute the sample mean. A) 4.582 B) 4.569 C) 5.128 D) 9.448 First step is to put the numbers into L1 and L2 by going to STAT-Edit. The quickest way from here is to go to STAT-Calc-1, designate L1 as your List, and L2 as the FreqList, and Calculate. The calculator will tell you that 𝑥 = 9.4476 The answer is D.

3). A research team has found that 39. 7% of U. S. teen-aged drivers 3) A research team has found that 39.7% of U.S. teen-aged drivers have access to a car that is exclusively theirs to use. If you conducted interviews with 3,458 U.S. teen-aged drivers, about how many would you expect to have exclusive access to a car? A) 40 B) 1,373 C) 2,085 D) 8,710 This is a binomial experiment. In a binomial experiment, as long as np and nq are both > 5, the mean (expected value) = np. (3458)(.397) = 1372.826 ≈ 1373. The answer is B. Remember that 𝜎 2 =𝑛𝑝𝑞 and 𝜎= 𝑛𝑝𝑞

Assume that this table is from a random sample and that the population it came from is normally distributed. 38 27 30 25 36 28 37 32 35 29 1) Construct a 95% confidence interval for the population mean using the table above. A) 29.417<𝜇<34.869 B) 29.67<𝜇<34.616 C) 29.76<𝜇<34.526 D) 29.167<𝜇<34.988 2) Construct a 90% confidence interval for the population standard deviation using the table above. A) 12.957<𝜎<49.175 B) 3.5996<𝜎<7.013 C) 5.892<𝜎<22.362 D) 3.468<𝜎<6.757 3) A study of 118 adults living in Norfolk, Virginia found that the average number of hours worked per week was 48.7 hours, with a standard deviation of 6.2 hours. Construct a 99% confidence interval for the population mean number of hours worked per week by adults living in Norfolk, Virginia. A) 47.205<𝜇<50.195 B) 45.5<𝜇<51.9 C) 42.5<𝜇<54.9 D) 47.23<𝜇<50.17

STAT-Edit, enter numbers into L1 Assume that this table is from a random sample and that the population it came from is normally distributed. 38 27 30 25 36 28 37 32 35 29 1) Construct a 95% confidence interval for the population mean using the table above. STAT-Edit, enter numbers into L1 STAT-Calc-1 on L1 to find the sample mean and standard deviation. 𝑥 =32.143 and 𝑠 𝑥 =4.721 To do this on the calculator, STAT-Test-8 (t-interval) because n < 30. Select Data and designate which list the numbers are in Enter the confidence level and calculate. 29.417<𝜇<34.869 The answer is A

We look in the columns for 0.95 and .05, and in the 13th row. Assume that this table is from a random sample and that the population it came from is normally distributed. 38 27 30 25 36 28 37 32 35 29 2) Construct a 90% confidence interval for the population standard deviation using the table above. To construct a confidence interval for variance and/or standard deviation, we use the Chi-Square distribution formula. (𝑛−1)( 𝑠 2 ) 𝑋 𝑅 2 < 𝜎 2 < (𝑛−1)( 𝑠 2 ) 𝑋 𝐿 2 ; 𝑋 𝐿 2 and 𝑋 𝑅 2 come from the 𝑋 2 chart. 1−𝑐 2 and 1+𝑐 2 tell us which columns to look in, and the Degrees of Freedom (n –1) tell us which row to look in. We look in the columns for 0.95 and .05, and in the 13th row. 𝑋 𝐿 2 = 5.892 and 𝑋 𝑅 2 = 22.362

This is the interval for the VARIANCE!! Assume that this table is from a random sample and that the population it came from is normally distributed. 38 27 30 25 36 28 37 32 35 29 2) Construct a 90% confidence interval for the population standard deviation using the table above. To construct a confidence interval for variance and/or standard deviation, we use the Chi-Square distribution formula. (𝑛−1)( 𝑠 2 ) 𝑋 𝑅 2 < 𝜎 2 < (𝑛−1)( 𝑠 2 ) 𝑋 𝐿 2 ; 𝑋 𝐿 2 and 𝑋 𝑅 2 come from the 𝑋 2 chart. (13)( 4.721 2 ) 22.362 < 𝜎 2 < (13)( 4.721 2 ) 5.892 12.957< 𝜎 2 <49.175 This is the interval for the VARIANCE!! To find the standard deviation interval, take the square roots of the variance interval endpoints. 3.6<𝜎<7.013 The answer is B

3). A study of 118 adults living in Norfolk, Virginia found that the 3) A study of 118 adults living in Norfolk, Virginia found that the average number of hours worked per week was 48.7 hours, with a standard deviation of 6.2 hours. Construct a 99% confidence interval for the population mean number of hours worked per week by adults living in Norfolk, Virginia. To do this on the calculator, STAT-Test-7 (z-interval) n > 30 Select Stats and enter 6.2, 48.7, 118, and .99 (in that order!!) Calculate. 47.23<𝜇<50.17 The answer is D