Question

Rachel claims that she can hit a bullseye 25% of the time at a distance of 50 feet to her target. Bob obnoxiously and loudly doubts her claim, so Rachel takes 20 shots from 50 feet and 4 of them hit the bullseye. Bob claims victory, but Rachel says that was actually really close to 25% and she usually does better.Luckily, a statistics student is there to settle the dispute. At a 0.05 significance level, is there enough evidence to show that Rachel's claim is wrong? (Hint: which claim can we show is false - the null or the alternative hypothesis?)

1.We have a sufficient sample size to conduct a valid test here

2.What is the value of the test statistic, written to two decimal places?

3.What is the p-value for the test, written to three decimal places?

Answer 1

1)

n• p and n(1 - p) ≥ 10 for the significance test

20 * 0.25 = 5

N0

2)

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.2 - 0.25)/sqrt(0.25*(1-0.25)/20)
z = -0.52

3)
P-value Approach
P-value = 0.603