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Stat and Data Analysis One Sample t-interval and t-test Name: ____________________ 1. A company has set a goal of developing a battery that lasts over 5 hours (300 minutes) in continuous use. In a first test of a SRS of 12 of these batteries measured the following lifespans (in minutes): 321,295, 332, 351, 281, 336, 311, 253, 270, 326, 311, and 288. a. Is there evidence that the company has met its goal? Assume that the life-spans of the batteries is approximately Normal. Create an appropriate test. Conditions: 1. SRS 2. n > 30 3. pop > 10n Stated 12 < 30 but Normal Distribution 120 there are more than 120 batteries x = 306.25 s = 29.31 n = 12 df = 11 HO: μ = 300 HA: μ > 300 306.25 − 300 = 0.7387 29.31 12 P-Value = P(t11 > 0.7387) = 0.2378 Since the P-Value is greater than alpha(0.2378 > 0.05) we fail to reject the null hypothesis. There is not enough evidence that the mean life-span of the batteries is more than 300 minutes. t11 = b. Create a 98% confidence interval for the mean lifespan of this type of battery. df = 11 ∗ t11 = 2.718 29.31 306.25 ± 2.718 12 CI = 306.25 ± 22.30 ( 283.25,329.25 ) We are 98% confident that the true mean life-span of the batteries lies between 283.25 to 329.25 minutes. 2. Consumer Reports tested a SRS of 14 brands of vanilla yogurt and found the following numbers of calories per serving: 160 200 220 230 120 180 140 130 170 190 80 120 100 170 a. A diet guide claims that you will get 120 calories from a serving of vanilla yogurt. What does this evidence indicate? Use your confidence interval to test an appropriate hypothesis and state your conclusion. Conditions: 1. SRS 2. pop > 10n 3. n > 30 HO: μ = 120 HA: μ ≠ 120 Stated 10(14) = 140 There are more than 140 yogurts 14 < 30 but Normal Distribution x = 157.86 s = 44.75 n = 14 df = 13 157.86 − 120 = 3.1656 44.75 14 P-Value = 2•P(t13 > 3.1656) = 0.0074 Since the P-Value is less than alpha(0.0074 < 0.01) we reject the null hypothesis. There is significant evidence that the mean calories from a serving of vanilla yogurt is not 120 calories. t13 = b. Create a 95% confidence interval for the average calorie content of vanilla yogurt. (Assume the number of calories is unimodal and symmetric) x = 157.86 s = 44.75 n = 14 df = 13 ∗ = 2.160 t13 CI = 157.86 ± 2.160 44.75 14 CI = 157.86 ± 25.83 (132.03,183.69) We are 95% confident that the true mean calorie content of vanilla yogurt lies between 132.03 and 183.69 calories. 3. How much time do school-age children spend helping with housework? A random sample of 26 girls in two-parent families where both parents work full-time found a mean of 14.0 minutes per weekday with a standard deviation of 8.6 minutes. A histogram of the data showed a unimodal and approximately symmetric distribution. Construct and interpret a 98% confidence interval for the mean number of minutes spent doing housework. Conditions: 1. SRS 2. n > 30 3. pop > 10n - Stated - n = 26X Data is unimodal and symmetric - 10(26) = 260 ∗ t 25 = 2.485 x = 14.0 s = 8.6 8.6 n = 26 df = 25 CI : 14.0 ± 2.485 26 CI : 14.0 ± 4.19 CI : ( 9.81,18.19 ) We are 98% confident that the true mean number of minutes spent doing housework lies between 9.81 and 18.19 minutes.