A cereal company advertises that $50\%$ of households purchase its brand every month. A market research team wonders if the true proportion is different, so they survey $200$ households, and $114$ report that they purchase the brand at least once per month.
(A) Use $\alpha =0.10$ to test the claim.

• A. Because $P$-value > $\alpha$, we FAIL TO REJECT $H_0$, there is NOT ENOUGH evidence to suggest that p = 0.5.
• B. Because $P$-value > $\alpha$, we FAIL TO REJECT $H_0$, there is NOT ENOUGH evidence to suggest that p ≠ 0.5.
• C. Because $P$-value < $\alpha$, we REJECT $H_0$, there is ENOUGH evidence to suggest that p = 0.5.
• D. Because $P$-value < $\alpha$, we REJECT $H_0$, there is ENOUGH evidence to suggest that p ≠ 0.5.

Answer: (D)

Verified step by step guidance

1. Step 1: Define the null and alternative hypotheses. Since the company claims that 50% of households purchase the brand, the null hypothesis is \(H_0: p = 0.50\). The market research team suspects the true proportion is different, so the alternative hypothesis is two-tailed: \(H_a: p \neq 0.50\).
2. Step 2: Identify the sample statistics. The sample size is \(n = 200\), and the number of households reporting purchase is \(x = 114\). Calculate the sample proportion as \(\hat{p} = \frac{x}{n} = \frac{114}{200}\).
3. Step 3: Calculate the test statistic for a population proportion. Use the formula for the z-test: \[ z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0 (1 - p_0)}{n}}} \] where \(p_0 = 0.50\) is the hypothesized population proportion.
4. Step 4: Determine the \(P\)-value associated with the calculated \(z\)-value. Since this is a two-tailed test, the \(P\)-value is twice the area in the tail beyond the absolute value of \(z\) in the standard normal distribution.
5. Step 5: Compare the \(P\)-value to the significance level \(\alpha = 0.10\). If \(P\)-value \(<\) \(\alpha\), reject \(H_0\); otherwise, fail to reject \(H_0\). Then, interpret the result in the context of the problem, stating whether there is enough evidence to suggest the true proportion differs from 50%.