Question

a humane society claims that less than 69% of households in a certain country own a pet. in a random sample of 800 households in that country, 520 say they own a pet. at α = 0.05, is there enough evidence to support the society’s claim? complete parts (a) through (c) below. \\( h_0: p \geq 0.69 \\) \\( h_a: p < 0.69 \\) (b) use technology to find the p - value. identify the standardized test statistic. \\( z = \square \\) (round to two decimal places as needed )

Explanation:

Step1: Calculate sample proportion

The sample proportion $\hat{p}$ is calculated as the number of successes (households with pets) divided by the sample size $n$. So, $\hat{p}=\frac{520}{800}=0.65$.

Step2: Identify $p_0$, $n$

The hypothesized proportion $p_0 = 0.69$ and the sample size $n = 800$.

Step3: Calculate standard error

The standard error $SE$ for a proportion test is $\sqrt{\frac{p_0(1 - p_0)}{n}}=\sqrt{\frac{0.69\times(1 - 0.69)}{800}}=\sqrt{\frac{0.69\times0.31}{800}}\approx\sqrt{\frac{0.2139}{800}}\approx\sqrt{0.000267375}\approx0.01635$.

Step4: Calculate z - statistic

The z - statistic for a proportion test is given by $z=\frac{\hat{p}-p_0}{SE}=\frac{0.65 - 0.69}{0.01635}=\frac{- 0.04}{0.01635}\approx - 2.446$. Rounding to two decimal places, $z\approx - 2.45$.

Answer:

The standardized test statistic $z\approx\boxed{-2.45}$ (rounded to two decimal places). For the P - value, using a z - table or technology (like a calculator or statistical software) for a left - tailed test with $z=-2.45$, the P - value is approximately $P(Z < - 2.45)=0.0071$.