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 sample of 900 households will be selected from the population. Use z-table.
a. Calculate σ(p̅), the standard error of the proportion of households spending more than $100 per week on groceries 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. What is the probability that the sample proportion will be within +/- 0.02 of the population proportion for a sample of 1,400 households (to 4 decimals)?
given : \(p=0.17, n=900\)
a. \(\sigma_{\hat{p}}=\sqrt{\frac{p *(1-p)}{n}}=\sqrt{\frac{0.17 *(1-0.17)}{900}}=0.0125\)
b. formula : \(z=\frac{\hat{p}-p}{\sqrt{\frac{p^{*(1-p)}{n}}{}}}\)
\(P(\) within \(\pm 0.02) \Rightarrow P(0.15 \(\Rightarrow P\left(\frac{0.15-0.17}{\sqrt{\frac{0.17 *(1-0.17)}{900}}}<\frac{\hat{p}-p}{\sqrt{\frac{p^{*(1-p)}}{n}}}<\frac{0.19-0.17}{\left.\sqrt{\frac{0.17 *(1-0.17)}{900}}\right)}\right.\) \(\Rightarrow P(-1.60 c. formula : \(z=\frac{\hat{p}-p}{\sqrt{\frac{p^{*(1-p)}{n}}{}}}\) \(P(\) within \(\pm 0.02) \Rightarrow P(0.15<\hat{p}<0.19)\) \(\Rightarrow P\left(\frac{0.15-0.17}{\sqrt{\frac{0.17 *(1-0.17)}{1400}}}<\frac{\hat{p}-p}{\sqrt{\frac{p^{*(1-p)}}{n}}}<\frac{0.19-0.17}{\sqrt{\frac{0.17 *(1-0.17)}{1400}}}\right)\) \(\Rightarrow P(-1.99
The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries
The Food Marketing Institute shows that of households spend more than per week on groceries. Assume the population proportion is and a simple random sample of households will be selected from the population. Use z-table. a. Calculate the sampling distribution of , the proportion of households spending more than per week on groceries. (to 2 decimals) (to 4 decimals) b. What is the probability that the sample proportion will be within of the population proportion (to 4 decimals)? eBook The Food Marketing Institute shows that...
The Food Marketing Institute shows that 15% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.15 and a sample of 800 households will be selected from the population. Use z-table.Calculate ( ), the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals).What is the probability that the sample proportion will be within +/- 0.02 of the population proportion (to 4 decimals)?What is...
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 sample of 700 households will be selected from the population. (a) Show the sampling distribution of p, the sample proportion of households spending more than $100 per week on groceries. (b) What is the probability that the sample proportion will be within ±0.02 of the population proportion? (c) Answer part (b) for a sample of...
The Food Marketing Institute shows that 16% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.16 and a sample of 600 households will be selected from the population. Use z-table. Calculate ( ), the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals). What is the probability that the sample proportion will be within +/- 0.02 of the population proportion (to 4...
The Food Marketing Institute shows that 16% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.16 and a sample of 900 households will be selected from the population. Use z-table. Calculate (), the standard error of the proportion of households spending more than $100 per week on groceries (to 4 decimals). What is the probability that the sample proportion will be within +/- 0.03 of the population proportion (to 4 decimals)?...
The Food Marketing Institute shows that 15% of households spend more than $100 per week on groceries. Assume the population proportion is p = 0.15 and a sample of 600 households will be selected from the population. What is the probability that the sample proportion will be within +/- 0.02 of the population proportion for a sample of 1,300 households (to 4 decimals)?
31. The Food Marketing Institute shows that 17% of households spend more than 105 weck on groceries. Assume the population proportion is p= .17 and a simple and Show the sampling distribution of the sample proportion of households spending sample of 800 households will be selected from the population, more than $100 per week on groceries. b. What is the probability that the sample proportion will be within 3.02 of the popular tion proportion? Answer part (b) for a sample...
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