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Statistics Review Sheet 2 OBJ: Review standard normal distribution, confidence intervals and regression

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I. Given a standard normal distribution, find the following areas. Draw and shade appropriate region.

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z st type of problem 1. Determine the area under the standard normal curve that lies between z = 0 and z = 1.83

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1 st type of problem – look up on table z

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4. Determine the area under the standard normal curve that lies to the left of z = z nd type of problem

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4. Determine the area under the standard normal curve that lies to the left of z = nd type of problem – look up on table z & subtract from.5.5 –.4970 =.003

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2. Determine the area under the standard normal curve that lies between -.38 and 1.2 z rd type of problem

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2. Determine the area under the standard normal curve that lies between -.38 and rd type of problem – look up on table z & add =.5329

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5. Determine the area under the standard normal curve that lies to the right of z th type of problem

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5. Determine the area under the standard normal curve that lies to the right of th type of problem – look up on table z & add to =.8643

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3. Determine the area under the standard normal curve that lies between.81 and 1.36 z th type of problem

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3. Determine the area under the standard normal curve that lies between.81 and th type of problem – look up on table z & subtract.4131 –.2910 =.1221

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z Given a normally distributed population with mean = 50 and standard deviation = 5. Each data item x can be converted to a standard value z by the formula: z = x – x s.

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Draw and shade area to find a value less than 57. z = x – x s z = 57 – 50 5 z = 1.4 Given a normally distributed population with mean = 50 and standard deviation = 5. Each data item x can be converted to a standard value z by the formula: z = x – x s z

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z Given a normally distributed population with mean = 50 and standard deviation = 5. Each data item x can be converted to a standard value z by the formula: z = x – x s

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z = x – x s z = 48 – 50 5 z = =.3087 z Draw and shade area to find a value between 41 and 48. z = x – x s z = 41 – 50 5 z = -1.8 Given a normally distributed population with mean = 50 and standard deviation = 5. Each data item x can be converted to a standard value z by the formula: z = x – x s

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Given a sample of 600 children where 486 preferred a P B & J sandwich for their lunch. s = p (1 – p) √ n 9. Compute p 10. Compute the standard deviation

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Given a sample of 600 children where 486 preferred a P B & J sandwich for their lunch. s = p (1 – p) √ n 9. Compute p Compute the standard error s = p (1 – p) √ n s =.81(.19) √ 600 s = √

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p – 2 s < p < p + 2s p – 3 s < p < p + 3s 11. Give a 95 % confidence interval for p. 12.Give a 99 % confidence interval for p.

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p – 2 s < p < p + 2s p – 3 s < p < p + 3s 11. Give a 95 % confidence interval for p. p – 2 s < p < p + 2s.81–2(.016) < p < (.016).778 < p < Give a 99 % confidence interval for p. p – 3 s < p < p + 3s.81 – 3(.016) < p < (.016).762 < p <.858

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Eleven used cars of the same make and model were randomly selected. The gas mileage and # of months since last tune up are noted below. 16. Find the regression equation. 15. Compute y. 14. Find r. 13. Make a scatter plot. TimeMPG

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Eleven used cars of the same make and model were randomly selected. The gas mileage and # of months since last tune up are noted below. 16. Find the regression equation. y 1 = -.33x Compute y Find r Make a scatter plot TimeMPG

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Eleven used cars of the same make and model were randomly selected. The gas mileage and # of months since last tune up are noted below Make a scatter plot TimeMPG

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17. At the 95 % confidence level, predict the mpg for a car whose last tune-up occurred 17 months ago. For 11 items, the r value found is bigger than the table value in both the 95% (.05) confidence level (.948>.602) and the 99% (.01)confidence level (.948>.735) 18. At the 99 % confidence level, predict the mpg for a car whose last tune-up occurred 30 months ago. That means the equation is good for both #17 and #18, so substitute the 17 months and the 30 months. y 1 = -.33(17) = and y 1 = -.33(30) = 9.93

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Describe the correlation between the following 21. # of days without rain and water level in a pond – (# of days without rain increase, water level decreases) 22.Money spent on groceries and number of people in a household + (more people in house more money is spent for groceries) 23. # of cats in a household and # of TV’s in house 0 ( no relationship between # of cats and #of TV’s)

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