Test Statistic using Z-score

Question

From previous studies, it is concluded that 73% of people mind if others smoke near a building entrance. A researcher claims it has decreased and decides to survey 100 adults. Test the researcher's claim at the ∝=0.05 significance level. 

After surveying 100 adult Americans, the researcher finds that 10 people mind if others smoke near a building entrance. Compute the test statistic. Round to two decimal places.

Answer 1

Hi Bailey,

Formula for this test statistic is:

z=(p^-p₀₎/sd(p^) where:

p^=sample probability

p₀=hypothesized probability

sd(p^)=standard deviation of p^, which has its own formula

sd(p^)=sqrt(p₀(1-p₀))/sqrt(n) where:

p₀=hypothesized probability

n=sample size

Now, let’s get the standard deviation for this problem:

p₀=0.73

1-p₀=0.27

n=100

sd(p^)=sqrt((0.73*0.27)/100)

sd(p^)=0.0444

Now, returning to our initial test stat formula:

z=(p^-p₀₎/sd(p^)

p^=10/100=0.10

p₀=0.73

sd(p^)=0.0444 (computed above)

z=(0.10-0.73)/0.0444

z= -14.19

That is the answer to this question, but spoiler alert if you are ultimately asked to test this hypothesis: With a test stat like this, p-value will be essentially 0. You will be able to reject the null and conclude that true proportion of people who mind smoking near buildings is significantly less than 73%.