Questions: Exercise 07.37 (Sampling Distribution of p) Question 9 The Food Marketing Institute shows that 17% of households spend more than 100 per week on groceries. Assume the population proportion is p=0.17 and a simple random sample of 800 households will be selected from the population. Use the z-table. a. Show the sampling distribution of p̄, the sample proportion of households spending more than 100 per week on groceries. E(p̄)= (to 2 decimals) σp̄= (to 4 decimals) b. What is the probability that the sample proportion will be within ± 0.02 of the population proportion (to 4 decimals)? c. Answer part (b) for a sample of 1600 households (to 4 decimals).

Exercise 07.37 (Sampling Distribution of p)
Question 9

The Food Marketing Institute shows that 17% of households spend more than 100 per week on groceries. Assume the population proportion is p=0.17 and a simple random sample of 800 households will be selected from the population. Use the z-table.
a. Show the sampling distribution of p̄, the sample proportion of households spending more than 100 per week on groceries.

E(p̄)= (to 2 decimals)

σp̄= (to 4 decimals)

b. What is the probability that the sample proportion will be within ± 0.02 of the population proportion (to 4 decimals)?

c. Answer part (b) for a sample of 1600 households (to 4 decimals).

Transcript text: Exercise 07.37 (Sampling Distribution of p ) Question 9 The Food Marketing Institute shows that $17 \%$ of households spend more than $\$ 100$ per week on groceries. Assume the population proportion is $p=0.17$ and a simple random sample of 800 households will be selected from the population. Use the $z$-table. a. Show the sampling distribution of $\bar{p}$, the sample proportion of households spending more than $\$ 100$ per week on groceries. \[ \begin{array}{l} E(\bar{p})=\square \text { (to } 2 \text { decimals) } \\ \sigma_{\bar{F}}=\square \text { (to } 4 \text { decimals) } \end{array} \] b. What is the probability that the sample proportion will be within $\pm 0.02$ of the population proportion (to 4 decimals)? $\square$ c. Answer part (b) for a sample of 1600 households (to $\mathbf{4}$ decimals). $\square$

Solution

Solution Steps

Step 1: Calculate the expected value of the sample proportion $ \bar{p} $

The expected value of the sample proportion $ \bar{p} $ is equal to the population proportion $ p $. Therefore: \[ E(\bar{p}) = p = 0.17 \]

Step 2: Calculate the standard deviation of the sample proportion $ \bar{p} $

The standard deviation of the sample proportion $ \bar{p} $ is given by: \[ \sigma_{\bar{p}} = \sqrt{\frac{p(1-p)}{n}} \] where $ p = 0.17 $ and $ n = 800 $. Substituting the values: \[ \sigma_{\bar{p}} = \sqrt{\frac{0.17 \times (1 - 0.17)}{800}} = \sqrt{\frac{0.17 \times 0.83}{800}} = \sqrt{\frac{0.1411}{800}} = \sqrt{0.000176375} \approx 0.0133 \]

Step 3: Calculate the probability that the sample proportion is within $ \pm 0.02 $ of the population proportion

To find the probability that the sample proportion $ \bar{p} $ is within $ \pm 0.02 $ of the population proportion $ p = 0.17 $, we use the $ z $-score formula: \[ z = \frac{\bar{p} - p}{\sigma_{\bar{p}}} \] For $ \bar{p} = 0.17 + 0.02 = 0.19 $: \[ z = \frac{0.19 - 0.17}{0.0133} \approx 1.50 \] For $ \bar{p} = 0.17 - 0.02 = 0.15 $: \[ z = \frac{0.15 - 0.17}{0.0133} \approx -1.50 \] Using the $ z $-table, the probability corresponding to $ z = 1.50 $ is approximately $ 0.9332 $, and for $ z = -1.50 $, it is approximately $ 0.0668 $. The probability that $ \bar{p} $ is within $ \pm 0.02 $ of $ p $ is: \[ P(-1.50 \leq z \leq 1.50) = 0.9332 - 0.0668 = 0.8664 \]

Step 4: Calculate the probability for a sample of 1600 households

For a sample size of $ n = 1600 $, the standard deviation of the sample proportion $ \bar{p} $ is: \[ \sigma_{\bar{p}} = \sqrt{\frac{0.17 \times 0.83}{1600}} = \sqrt{\frac{0.1411}{1600}} = \sqrt{0.0000881875} \approx 0.0094 \] Using the same $ z $-score formula: For $ \bar{p} = 0.17 + 0.02 = 0.19 $: \[ z = \frac{0.19 - 0.17}{0.0094} \approx 2.13 \] For $ \bar{p} = 0.17 - 0.02 = 0.15 $: \[ z = \frac{0.15 - 0.17}{0.0094} \approx -2.13 \] Using the $ z $-table, the probability corresponding to $ z = 2.13 $ is approximately $ 0.9834 $, and for $ z = -2.13 $, it is approximately $ 0.0166 $. The probability that $ \bar{p} $ is within $ \pm 0.02 $ of $ p $ is: \[ P(-2.13 \leq z \leq 2.13) = 0.9834 - 0.0166 = 0.9668 \]