Question

Study E. A basketball player is practicing making free throws. On average, the probability of her making a free throw is 0.78. She finds that in the next 50 free throws, she makes 31. Is this outcome significantly more than what would be expected? Assume a = 0.05_2-tail and independence between shots.

Refer to Study E. What is the null hypothesis in terms of the variables in the study

Refer to Study E. What is the alternative hypothesis in terms of the variables in the study? (1 pt)

Refer to study E. What specific test-statistic will you use? Are you justified in using the appropriate statistic? Why or why not? (2 pts)

Refer to Study E. What is the obtained value of the statistic? (2 pts)

Refer to Study E. What the critical value of the statistic? Assume a = 0.05_2-tail. (1 pt)

Refer to Study E. What do you conclude regarding the null hypothesis using a = 0.05_2-tail? Justify your answer. (1 pt)

Answer 1

E) let p be the population proportion of free throws made by her

1)

null Hypothesis:               Ho:   p

=

0.780

2)

alternate Hypothesis:    Ha: p

>

0.780

3)

since np =50*0.78 =34 and n(1-p) =16 both are greater than 10 , we can use normal approximation of binomial distribution. test statistic to be used is z

4)

sample success x   =		31
sample size          n    =		50
std error   se =√(p*(1-p)/n) =		0.0586
sample proportion p̂ = x/n=		0.6200
test stat z =(p̂-p)/√(p(1-p)/n)=		-2.73

6)

for 0.052 level with right tailed test , critical z=

1.626

7)

since test statsitic is not higher than critical value ; we can not reject null hypothesis

we do not have sufficient evidence to conclude that  this outcome is significantly more than what would be expected