Question

1. a news report claims that 68% of dog owners have driven a car with their dog sitting in their lap. believing this percentage is too high, a veterinarian surveys a random sample of 300 dog owners and finds that 189 of these dog owners have let their dog sit in their lap while driving. if a hypothesis test is conducted at a significance level of 0.05, what should the veterinarian conclude?

a. because the difference between the sample proportion and the claimed population proportion is equal to exactly 0.05, there is evidence against the null hypothesis.
b. because the resulting p - value is larger than 0.05, there is evidence against the null hypothesis.
c. because the resulting p - value is smaller than 0.05, there is not enough evidence against the null hypothesis.
d. because the resulting p - value is smaller than 0.05, there is evidence against the null hypothesis.
e. because the resulting p - value is larger than 0.05, there is not enough evidence against the null hypothesis.

Explanation:

Step1: Calculate sample proportion

The sample proportion \( \hat{p}=\frac{189}{300}=0.63 \)

Step2: State hypotheses

\( H_0:p = 0.68 \), \( H_1:p<0.68 \)

Step3: Calculate test - statistic

The formula for the test - statistic \( z=\frac{\hat{p}-p}{\sqrt{\frac{p(1 - p)}{n}}}\)
Substitute \( \hat{p}=0.63\), \( p = 0.68\), \( n = 300\)
\( z=\frac{0.63-0.68}{\sqrt{\frac{0.68\times(1 - 0.68)}{300}}}\)
\( z=\frac{-0.05}{\sqrt{\frac{0.68\times0.32}{300}}}\)
\( z=\frac{-0.05}{\sqrt{\frac{0.2176}{300}}}\)
\( z=\frac{-0.05}{\sqrt{0.00072533}}\)
\( z=\frac{-0.05}{0.02693}\approx - 1.86\)

Step4: Find P - value

Since it is a left - tailed test, \( P - value=\Phi(z)\) where \( z=-1.86\)
From the standard normal table, \( \Phi(-1.86)=0.0314\)

Answer:

D. Because the resulting \( P \) - value is smaller than \( 0.05 \), there is evidence against the null hypothesis.