Question 9, 8.1.15, \( \begin{array}{l}\text { Part } 3 \text { of } 4\end{array} \) \( \begin{array}{l}\text { According to a survey, } 42 \% \text { of shoppers report buying organic food at every chance they get. A nutritionist thinks this rate has increased. The nutritionist surveys } 100 \text { shoppers and finds that } 45 \text { buy } \\ \text { organic food at every chance they get. Complete parts (a) through (c) below. } 89.17 \%, 89.17 \text { of } 100 \text { points }\end{array} \)

We are given that 42% (p₀ = 0.42) of shoppers report buying organic food whenever possible. A nutritionist believes that this rate has increased. To test this claim, a random sample of 100 shoppers is taken and 45 of them are found to buy organic food “at every chance they get” (sample proportion p̂ = 45/100 = 0.45). Below are the steps for parts (a)–(c). ───────────────────────────── Part (a): State the Hypotheses Let p represent the true proportion of shoppers who buy organic food at every opportunity. • Null hypothesis, H₀: p = 0.42 • Alternative hypothesis, Hₐ: p > 0.42 This is a right‐tailed test since the nutritionist claims that the rate has increased. ───────────────────────────── Part (b): Verify the Conditions for Using the Z-Test For a one‐sample proportion z-test, we check that: 1. The sample is a simple random sample. 2. The observations are independent. 3. The sample size is large enough so that the normal approximation holds. This requires that: • np₀ ≥ 10, and • n(1 – p₀) ≥ 10. Here, np₀ = 100 × 0.42 = 42, and n(1 – p₀) = 100 × 0.58 = 58. Both are well over 10, so the condition is met. ───────────────────────────── Part (c): Compute the Test Statistic and P-value 1. Compute the standard error using p₀:   SE = √[p₀(1 – p₀)/n] = √[(0.42 × 0.58)/100]     = √(0.2436/100)     = √(0.002436)     ≈ 0.04936 2. Compute the z test statistic:   z = (p̂ – p₀) / SE = (0.45 – 0.42) / 0.04936     = 0.03 / 0.04936     ≈ 0.607 3. Find the p-value. Since this is a right‐tailed test:   p-value = P(Z > 0.607). Using a standard normal table or calculator,   P(Z > 0.607) ≈ 0.27. ───────────────────────────── Conclusion Since the p-value (≈ 0.27) is much greater than a commonly used significance level (say, α = 0.05), we do not reject the null hypothesis. There is insufficient evidence to conclude that the proportion of shoppers who buy organic food “at every chance they get” has increased from 42%. ───────────────────────────── Final Answer Summary (a) H₀: p = 0.42  Hₐ: p > 0.42 (b) All conditions for using the z-test are satisfied. (c) The test statistic is z ≈ 0.61, and the p‑value is approximately 0.27. Thus, the evidence does not support the nutritionist’s claim that the proportion has increased.