Questions: A Food Marketing Institute found that 32% of households spend more than 125 a week on groceries. Assume the population proportion is 0.32 and a simple random sample of 368 households is selected from the population. What is the probability that the sample proportion of households spending more than 125 a week is less than 0.34?

Solution Steps

The standard error (SE) of the sample proportion is calculated using the formula:

$$SE = \sqrt{\frac{p(1 - p)}{n}}$$

where $p = 0.32$ (the population proportion) and $n = 368$ (the sample size). Substituting the values, we find:

$$SE = \sqrt{\frac{0.32 \times (1 - 0.32)}{368}} \approx 0.0243$$

Next, we calculate the Z-scores for the sample proportion. The Z-score for the upper bound (0.34) is calculated as follows:

$$Z_{end} = \frac{\hat{p} - p}{SE} = \frac{0.34 - 0.32}{0.0243} \approx 0.8255$$

For the lower bound, since we are considering the probability of the sample proportion being less than 0.34, the Z-score is:

$$Z_{start} = -\infty$$

Using the Z-scores, we can find the probability that the sample proportion is less than 0.34. The probability is given by:

$$P = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.8255) - \Phi(-\infty)$$

Since $\Phi(-\infty) = 0$ and $\Phi(0.8255)$ approaches 1, we have:

$$P \approx 1.0$$

Final Answer