A Food Marketing Institute found that 26% of households spend more than $125 a week on groceries. Assume the population proportion is 0.26 and a simple random sample of 191 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 0.27?

Assume that the conditions necessary to use the Normal curve have been satisfied. Round the answer to 4 decimal places.

There is a  probability that the sample proportion of households spending more than $125 a week is less than 0.27.


We are tasked with finding the probability that the sample proportion $\hat{p}$ of households spending more than $125 a week is less than 0.27, given that the population proportion $p = 0.26$ and the sample size $n = 191$. We are to assume that the sampling distribution of $\hat{p}$ follows a normal distribution because the conditions for the normal approximation have been met.

### Step 1: Determine the standard error of the sample proportion
The standard error (SE) of the sample proportion $\hat{p}$ is calculated using the formula:

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

Where:
- $p = 0.26$ is the population proportion,
- $n = 191$ is the sample size.

Substitute the values into the formula:

\[
SE = \sqrt{\frac{0.26(1 - 0.26)}{191}} = \sqrt{\frac{0.26 \times 0.74}{191}} = \sqrt{\frac{0.1924}{191}} \approx \sqrt{0.00100785} \approx 0.0317
\]

### Step 2: Find the Z-score
Next, we calculate the Z-score for the sample proportion $\hat{p} = 0.27$ using the formula:

\[
Z = \frac{\hat{p} - p}{SE}
\]

Substitute the known values:

\[
Z = \frac{0.27 - 0.26}{0.0317} = \frac{0.01}{0.0317} \approx 0.3155
\]

### Step 3: Find the probability
We now use the standard normal distribution to find the probability that the Z-score is less than 0.3155. Using a Z-table or standard normal distribution calculator, the probability corresponding to $Z = 0.3155$ is approximately:

\[
P(Z < 0.3155) \approx 0.6239
\]

Thus, the probability that the sample proportion is less than 0.27 is approximately 0.6239.

### Final Answer:
There is a **0.6239** probability that the sample proportion of households spending more than $125 a week is less than 0.27.

---

Do you need any further details or explanations? Here are 5 related questions to expand your understanding:

1. How would the probability change if the sample size was increased?
2. What if the population proportion was different, say 0.30? How would the calculation differ?
3. Why do we use the normal approximation in this case, and what are the conditions for it to be valid?
4. How can we calculate the probability that the sample proportion is greater than 0.27?
5. How would you interpret the result in the context of grocery spending?

**Tip:** The larger the sample size, the smaller the standard error, which results in a more accurate estimate of the population proportion.

A Food Marketing Institute found that 26% of households spend more than $125 a week on groceries. Assume the population proportion is 0.26 and a simple random sample of 191 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 0.27? Assume that the conditions necessary to use the Normal curve have been satisfied. Round the answer to 4 decimal places.

Find out the probability that the sample proportion of households spending more than $125 a week on groceries is less than 0.27. This problem explores statistical concepts, including the standard error and Z-score calculation, using a population proportion of 0.26 and a sample size of 191. Suitable for students in Grades 11-12.

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