A city government claims that no more than $25\%$ of households have solar panels. A researcher suspects the rate is actually higher and surveys $200$ households, finding that $62$ have solar panels. Test if there is evidence that more than $25\%$ of households have solar panels using $\alpha =0.01$.

• A. Because $P$-value = $0.03$$>\alpha$, we REJECT $H_0$, there is ENOUGH evidence to suggest $p>0.25$.
• B. Because $P$-value = $0.01$ $>\alpha$, we FAIL TO REJECT $H_0$, there is NOT ENOUGH evidence to suggest $p>0.25$.
• C. Because $P$-value = $0.01$ $>\alpha$, we REJECT $H_0$, there is ENOUGH evidence to suggest $p>0.25$.
• D. Because $P$-value = $0.03$$>\alpha$, we FAIL TO REJECT $H_0$, there is NOT ENOUGH evidence to suggest $p>0.25$.

Answer: (D)

Verified step by step guidance

1. Step 1: Define the null and alternative hypotheses. Since the city claims no more than 25% have solar panels, the null hypothesis is \(H_0: p \leq 0.25\). The researcher suspects a higher rate, so the alternative hypothesis is \(H_a: p > 0.25\) (a right-tailed test).
2. Step 2: Identify the sample statistics. The sample size is \(n = 200\) households, and the number with solar panels is \(x = 62\). Calculate the sample proportion as \(\hat{p} = \frac{x}{n} = \frac{62}{200}\).
3. Step 3: Calculate the test statistic for a one-proportion z-test using the formula: \[z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}}\] where \(p_0 = 0.25\) is the hypothesized population proportion.
4. Step 4: Find the p-value corresponding to the calculated z-value. Since this is a right-tailed test, the p-value is the probability that the test statistic is greater than the observed z-value under the null hypothesis.
5. Step 5: Compare the p-value to the significance level \(\alpha = 0.01\). If the p-value is less than \(\alpha\), reject \(H_0\) and conclude there is enough evidence to support \(p > 0.25\). Otherwise, fail to reject \(H_0\) and conclude there is not enough evidence to support the claim.