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

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Presentation on theme: "Statistics Review Sheet 2 OBJ: Review standard normal distribution, confidence intervals and regression."— Presentation transcript:

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

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

3 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

4 1 st type of problem – look up on table..4664 z -3 -2 -1 0 1 2 3 1.83

5 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

6 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

7 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

8 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

9 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

10 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

11 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

12 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

13 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.

14 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

15 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

16 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

17 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

18 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

19     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.

20     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

21 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

22 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

23 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

24 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

25 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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