41.
a. The sampling distribution of p-bar is normal since np = 800*0.17 = 136 > 10 and n(1-p) = 800*0.83 = 664 > 10.
The sampling distribution of p-bar has been represented below:
The sampling distribution of p-bar is done using STATKEY. We select Sampling Distribution for proportion then enter the proportion value as 0.17 and enter sample size as 800. Then we generate 1000 samples.
31. The Food Marketing Institute shows that 17% of households spend more than 105 weck on...
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 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 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....
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...
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 111 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.27 and 0.38?
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 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...
A Food Marketing Institute found that 53% of households spend more than $125 a week on groceries. Assume the population proportion is 0.53 and a simple random sample of 105 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is more than than 0.34? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer =
1/A Food Marketing Institute found that 27% of households spend more than $125 a week on groceries. Assume the population proportion is 0.27 and a simple random sample of 132 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 0.3? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer = 2/ A Food...
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 324 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? Answer = (Enter your answer as a number accurate to 4 decimal places.)