Question

A newspaper conducted a statewide survey concerning the 1998 race for state senator. The newspaper took a SRS of n=500 registered voters and found that 175 would vote for the Republican candidate. Let p represent the proportion of registered voters in the state who would vote for the Republican candidate.

We test

H0:p=.04

Ha:p>.04

What is the z-statistic for this test?

What is the P-value of the test?

A random sample of 1000 car owners in a particular city found 200 car owners who received a speeding ticket this year. Find a 95% confidence interval for the true percent of car owners in this city who received a speeding ticket this year. Express your results to the nearest hundredth of a percent.

Answer: ___________to ___________%

Answer 1

1)

Answer)

Null hypothesis Ho : P = 0.04

Alternate hypothesis Ha : P > 0.04

N = 500

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 20

N*(1-p) = 480

Both the conditions are met so we can use standard normal z table to estimate the P-Value

Test statistics z = (oberved p - claimed p)/standard error

Standard error = √{claimed p*(1-claimed p)/√n

Observed P = 175/500

Claimed P = 0.04

N = 500

After substitution

Test statistics z = 35.37

From z table, P(z>35.37) = 0

P-value = 0

2)

N = 1000

P = 200/1000

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 200

N*(1-p) = 800

Both the conditions are met so we can use standard normal z table to estimate the interval

Critical value z from z table for 95% confidence level is 1.96

Margin of error(MOE) = Z*√{P*(1-P)}/√N

MOE = 0.02479225685

Interval is given by

P-MOE to P+MOE

0.17520774314 to 0.22479225685

17.52% to 22.48%