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

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

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

z -3 -2 -1 0 1 2 3 1 st type of problem 1. Determine the area under the standard normal curve that lies between z = 0 and z = 1.83

1 st type of problem – look up on table..4664 z -3 -2 -1 0 1 2 3 1.83

4. Determine the area under the standard normal curve that lies to the left of z = - 2.75. z -3 -2 -1 0 1 2 3 2 nd type of problem

4. Determine the area under the standard normal curve that lies to the left of z = - 2.75. 2 nd type of problem – look up on table.003.4970.5 z -3 -2 -1 0 1 2 3 -2.75 & subtract from.5.5 –.4970 =.003

2. Determine the area under the standard normal curve that lies between -.38 and 1.2 z -3 -2 -1 0 1 2 3 3 rd type of problem

2. Determine the area under the standard normal curve that lies between -.38 and 1.2 3 rd type of problem – look up on table.1480.3849 z -3 -2 -1 0 1 2 3 -.38 1. 2 & add.1480 +.3489 =.5329

5. Determine the area under the standard normal curve that lies to the right of - 1.1. z -3 -2 -1 0 1 2 3 4 th type of problem

5. Determine the area under the standard normal curve that lies to the right of - 1.1. 4 th type of problem – look up on table.3643.5 z -3 -2 -1 0 1 2 3 -1.1 & add to.5.3643 +.5 =.8643

3. Determine the area under the standard normal curve that lies between.81 and 1.36 z -3 -2 -1 0 1 2 3 5th type of problem

3. Determine the area under the standard normal curve that lies between.81 and 1.36 5th type of problem – look up on table. 2910 z -3 -2 -1 0 1 2 3.81 1.36 & subtract.4131 –.2910 =.1221

z -3 -2 -1 0 1 2 3 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.

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.5.4192 z - 3 -2 -1 0 1 2 3 1.4.9192

z -3 -2 -1 0 1 2 3 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 = x – x s z = 48 – 50 5 z = -.4.4641 -.1554 =.3087 z -3 -2 -1 0 1 2 3 -1.8 -.4 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

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

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 486 600.81 10. Compute the standard error  s = p (1 – p) √ n s =.81(.19) √ 600 s = √.0002565.016

    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.

    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 <.81 + 2(.016).778 < p <.842 12.Give a 99 % confidence interval for p.  p – 3 s < p < p + 3s.81 – 3(.016) < p <.81 + 3(.016).762 < p <.858

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 816 1314 319 120 817 1016 15 2014 419 220 2313

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 + 19.83 15. Compute y. 16.64 14. Find r. -.948 20 13. Make a scatter plot. 15 10 5 0 5 10 15 20 25 TimeMPG 816 1314 319 120 817 1016 15 2014 419 220 2313

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. 20 13. Make a scatter plot. 15 10 5 0 5 10 15 20 25 TimeMPG 816 1314 319 120 817 1016 15 2014 419 220 2313

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 -.948 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) + 19.83 = 14.22 and y 1 = -.33(30)+19.83 = 9.93

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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