Statistics Homework Solutions 159

Statistics Homework Solutions 159 - 159 SECTION 6.13...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 159 SECTION 6.13 increase. If the level increases, the probability of rejecting H0 increases, so in particular, the probability of rejecting H0 when it is false increases. 50 000 versus H1 : µ 50 000. H1 is true, since the true value of µ is 49,500. ¡  § 5. (a) H0 : µ  (b) The level is the probability of rejecting H0 when it is true. Under H0 , X is approximately normally distributed with mean 50,000 and standard deviation σX 5000 100 500. ¡ ¤£ ¡ 49 400 .  The probability of rejecting H0 is P X ¥ ¤ ¨ 49 400 50 000 500 1 20. Under H0 , the z-score of 49,400 is z The level of the test is the area under the normal curve to the left of z 1 20. ¨ ¨ ¡ ¡ £ ©¥   ¡ ¤ ¡ ¡ ¡ Therefore the level is 0.1151. 49 500. ¡ The power is the probability of rejecting H0 when µ  X is approximately normally distributed with mean 49,500 and standard deviation σX 5000 100 500. ¡ 49 400 . ¨ ¡ ¡ ¨ ¤ 49 400 49 500 500 0 20. The z-score of 49,400 is z The power of the test is thus the area under the normal curve to the left of z 0 20. ¨  ¡ The probability of rejecting H0 is P X ¥ ¤£ ¡ £ ©¥ ¡   ¤ ¡ ¡ Therefore the power is 0.4207. µ0 , the 5% rejection region will be the region X  ¡¥ ¤ ¨ 49 177 5.  0 645. We will use z ¡ ¨ ¡ ¥¨ 49 500 500 0 65. £ ¥ 0 65. ¨  ¡ ¡ ¡  ¨  ¤ ¨ ¤ ¡ The power is therefore the area to the left of z ¨ ¡ 49 177 5 49 500. ¡ ¨ The z-score of 49,177.5 is z 49 177 5 when µ ¡  ¡ The power is therefore P X ¨  ¡ The rejection region is X 49 177 5. ¨ 1 645 500 ¡ 50 000 Therefore x5 1 645. ¡ The z-score corresponding to the 5th percentile is z x5 , where ¨ ¡ (c) Since the alternate hypothesis is of the form µ x5 is the 5th percentile of the null distribution. ¨ Thus the power is 0.2578. ¤ ¨ 49 920.  ¨ ¢  ¡  0 16. ¨ ¡ ¡ £ ©¥  ¡ ¡ ¥  ¤ ¡  ¨ ¤ The level is the area under the normal curve to the left of z ¡ 50 000 500 ¡ 49 920 ¨ The z-score of 49,920 is z 50 000. 0 16. ¨ ¤ 49 920 when µ 49 920.  ¡¥ ¡ Now compute the level of the test whose rejection region is X The level is P X ¡ 0 84 500 0 80 when µ 49 500. 49 500. The z-score corresponding to the  ¨ 49 500 ¨ Therefore x0 ¡¥ Therefore x0 is the 80th percentile of the normal curve when µ 80th percentile is z 0 84. x0 ¨ x0 where P X ¡ (d) For the power to be 0.80, the rejection region must be X  3. ...
View Full Document

This note was uploaded on 12/20/2011 for the course STA 3163 taught by Professor Mattgilg during the Fall '11 term at UNF.

Ask a homework question - tutors are online