Question

A publisher reports that 26% of their readers own a particular make of car. A marketing executive wants to test the claim that the percentage is actually below the reported percentage. A random sample of 280 found that 20% of the readers owned a particular make of car. Is there sufficient evidence at the 0.02 level to support the executive's claim? Step 4 of 7 : Determine the P-value of the test statistic. Round your answer to four decimal places.

H0: p=0.26 Ha: p<0.26 z=−2.29 One-Tailed Test

Answer 1

Solution:
Given in the question
Null hypothesis H0: p = 0.26
Alternate hypothesis Ha:p<0.26
Also
Number of sample = 280
sample proportion p^ = 0.2
Here we will use one proportion Z test as np and np(1-p) are both greater than 10
Z test stat = (p^ - p)/sqrt(p*(1-p)/n) = (0.20-0.26)/sqrt(0.26*(1-0.26)/280) = -0.06/0.026 = -2.29
So From Z table we found p-value as this is one tailed and Left tailed test
So P-value = 0.0110
At Level of significance () = 0.02, we can reject the null hypothesis as P-value is less than alpha value.
So we have significant evidence to support the claim that the percentage is actually below the reported percentage.