Overall, it is known that about 30% of people tip taxi cab drivers. Suppose I ta

Question

Overall, it is known that about 30% of people tip taxi cab drivers. Suppose I take a sample of 25 taxi riders and find that only 6 out of the 25 gave a tip. I want to test to see if there is significant evidence that the amount of people who tip is significantly different than 30%.

(a) If I wanted to control my margin of error and set it to 5% with 90% confidence, what sample size would I need to take instead of the 25?

(b) Using my original sample size of 25, what would be the 99% confidence interval for the population proportion?

(c) What are the null and alternative hypotheses?

(d) What is the critical value at 99% confidence?

(e) Calculate the test statistic.

(f) Find the p-value.

(g) What conclusion would be made here at the 99% confidence level?

Explanation / Answer

a.
Compute Sample Size ( n ) = n=(Z/E)^2*p*(1-p)
Z a/2 at 0.1 is = 1.645
Sample Proportion = 0.24
ME = 0.05
n = ( 1.645 / 0.05 )^2 * 0.24*0.76
= 197.432 ~ 198

b.
Given that,
possibile chances (x)=6
sample size(n)=25
success rate ( p )= x/n = 0.24
success probability,( po )=0.3
failure probability,( qo) = 0.7
null, Ho:p=0.3
alternate, H1: p!=0.3
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.58
since our test is two-tailed
reject Ho, if zo < -2.58 OR if zo > 2.58
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.24-0.3/(sqrt(0.21)/25)
zo =-0.6547
| zo | =0.6547
critical value
the value of |z | at los 0.01% is 2.58
we got |zo| =0.655 & | z | =2.58
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -0.65465 ) = 0.51269
hence value of p0.01 < 0.5127,here we do not reject Ho

ANSWERS
---------------
null, Ho:p=0.3
alternate, H1: p!=0.3
test statistic: -0.6547
critical value: -2.58 , 2.58
decision: do not reject Ho
p-value: 0.51269

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