Statistics Practice Quest 2

1. Statistics Practice Quest 2 OBJ: • Review standard normal distribution, confidence intervals and regression

2. 4th 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 1. Determine the area under the standard normal curve that lies to the right of - 1.1.

3. 2nd 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 to the left of z = - 2.75.

4. 3rd type of problem – look up on table .1480 .3849z -3 -2 -1 0 1 2 3 -.38 1. 2 & add .1480 + .3849 = .5329 3.Determine the area under the standard normal curve that lies between -.38 and 1.2

5. 2nd type of problem – look up on table .5 z -3 -2 -1 0 1 2 3 -2.75 & subtract from .5 .5 –.2642 = .2358 4. Determine the area under the standard normal curve that lies to the left of z = - .72.

6. 5th type of problem – look up on table .2910 z -3 -2 -1 0 1 2 3 .81 1.36 & subtract .4131 – .2910 = .1221 5. Determine the area under the standard normal curve that lies between .81 and 1.36

7. Draw and shade area to find a value greater than 55. z = x – x s z = 55 – 50 5 z =1 .3413 z - 3 -2 -1 0 1 2 3 1.4 .5 – .3413 .1587 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

8. Draw and shade area to find a value less than 42. z = x – x s z = 42 – 50 5 z = – 1.6 z - 3 -2 -1 0 1 2 3 – 1.6 .5 – .4452 = .0548 . 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

9. 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

10. 9. Compute p 486 600 .81 10. Compute the standard deviation   s = p (1 – p) √ n s = .81(.19) √ 600 .016 Given a sample of 600 children where 486 preferred a P B & J sandwich for their lunch.   s = p (1 – p)√ n

11. 11. Give a 95 % confidence interval for p. p – 2 s < p < p + 2s .81–2( ) < p < .81 + 2( ) .81–2(.016) < p < .81 + 2(.016) .778 < p < .842 Give a 99 % confidence interval for p.   p – 3 s < p < p + 3s .81 – 3( ) < p < .81 + 3( ) .81 – 3(.016) < p < .81 + 3(.016) .762 < p < .858    p – 2 s < p < p + 2s p – 3 s < p < p + 3s

12. 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. 14. Find the regression equation. y = -.33x + 19.83 16. Compute y. 16.64 15. Find r. -.948 13. Make a scatter plot.

13. 17. At the 95 % confidence level, predict the mpg for a car whose last tune-up occurred 17 months ago. n = 11 95% -.948 .602 Since r > table, significant so substitute 17 for x in equation y = -.33x + 19.83 y = -.33(17) + 19.83 14.25 mpg 18. At the 99 % confidence level, predict the mpg for a car whose last tune-up occurred 30 months ago. n = 11 99% -.948 .735 Since r > table, significant so substitute 30 for x in equation y =-.33x + 19.83 y = -.33(30) + 19.83 10 mpg