Question

Q6. (16 points. In order to determine if, in high school basketball, right-handed players (group #1) are better free-throw shooters than left-handed players (group #2), two independent random samples of free shots were collected. Of 40 shots by right-handed players, 30 were successful and, of 60 shots by left-handed players, 30 were successful. Conduct the test of interest at the 10 level of significance. H, Vs. H Supporting work: Should H, be rejected? (Circle one): Yes No Q7. (16 points.) An investment banker believes that the average age of a millionaire is 50 years. To verify this, he will test H, u = 50 vs. H: 450 at the .05 level of significance based on a random sample 64 of millionaires. Suppose that o = 16. If, in reality, the average age of a millionaire is 51.5 years, what is the probability that the test will commit a type II error?

Answer 1

Q6)

In order to determine if,in high school baseball,right handed players(group #1) are better free-throw shooters than left-handed players (group #2). The two independent random sample of free shots were collected. of 40 shots by right-handed players, 30 were successful and,of 60 shots by left-handed players,30 were successful. Conduct the test of interest at the 0.10 level of significance.

Answer:

We have to test the null and alternative hypothesis,

Ho: There is no significant difference between the proportion of right handed players(group #1) and left handed players( group #2) regarding free throw shots. (i.e. p1=p2)   V/s

Ha: The proportion of right handed players(group #1) are better free-throw shooters than left-handed players (group #2). (i.e. p1>p2)

The value of test statistic for testing Ho V/s Ha is,

Z=2.5

and critical value of Z at α=0.10 for two tailed alternative is Zα=1.28

{ So, Yes,We should reject the null hypothesis }

Since Cal. Z>Zα, We reject the null hypothesis Ho and conclude that the proportion of the right handed players(group #1) are better free-throw shooters than left-handed players (group #2).

Explanation:

We have giver that, no = number of right-handed player = 40 H₂ = number of left-handed player=60 x = number of right-handed player who thoow shots free throw successfully number of left-handed player who shots free throw successfully, here x = 30 & X2 = 30 X2= л P - XI 30 ni 40 Let P = proportion of sight handed player who shoot a free-throw successfully - 3/4 -075 Pa = proportion of left handed player who shoot a free-throw successfully 22 = x2 = 630 ( The value of pooled proportion is ni P = p = = 0.5 itn2P2 30+30 n, tha 40+60 1- 0.6 0.4 - 0=1- have to test, Ho! There is no significant difference between the pooportion of right handed player and left handed player regarding the free throw shoots . v/s.

proportion of HA The Night handed player are better free throw shooters thon left handed player. je to test P = P2 v/s HA ΕΣΡ, use proportion To test the Ho and HA test, Under the Ho, test statistic is 2= (P1-P2) spoli nz + 40 0.75 -0.5) (06) (64) --) { 12=25 Since no, no are large, z~ N(0,1) The critical value of e calculated z at significance level x = 0.10 for right tailed alternative is 1.28. so since Zcal > 2x=128, we reject the null hypothesis Ho at a=0.10 significance level and conclude that the proportion of right handed player are better free throw shooter's than left handed players.