Question

In: Statistics and Probability

The Food Marketing Institute and Nielsen reported that 49% of U.S. consumers shop for consumer package...

The Food Marketing Institute and Nielsen reported that 49% of U.S. consumers shop for consumer package goods products online. Assume the population proportion is p=0.49 and a sample of 500 consumers will be selected from the population.

Calculate the expected value and the standard error for the sampling distribution of p ̅, the sample proportion of consumers who shop for consumer package goods product online.

Describe the sampling distribution of p ̅. Draw a graph of this probability distribution with its mean and standard deviation.

What is the probability that the sample proportion will be greater than 0.52?

What is the probability that the sample proportion will be within ±0.02 of the population proportion?

Answer part d for a sample of 1000 households.

Solutions

Expert Solution

The expected value and the standard error for the sampling distribution of

The expected value is obtained using the following formula,

The standard error is obtained using the following formula,

Sampling distribution of

The sample proportion is normally distributed with

The probability that the sample proportion will be greater than 0.52

The probability is obtained by calculating the z score,

From the z distribution table,

The probability that the sample proportion will be within 0.02 of the population proportion

The probability is obtained by calculating the z score,

From the z distribution table,

For a sample of 1000 households.

Now, the probability is,

From the z distribution table,


Related Solutions

The American Bankers Association reported that, in a sample of 100 consumer purchases in France, 49...
The American Bankers Association reported that, in a sample of 100 consumer purchases in France, 49 were made with cash, compared with 31 in a sample of 60 consumer purchases in the United States. Construct a 99 percent confidence interval for the difference in proportions. (Round your intermediate value and final answers to 4 decimal places.) The 99 percent confidence interval is from ____ to ____.
Marketing shapes consumer needs and wants versus marketing merely reflects the needs and wants of consumers...
Marketing shapes consumer needs and wants versus marketing merely reflects the needs and wants of consumers and take a position supporting your answer with concepts from the chapter. Note below Answer should be from this text. Text(s): Kotler, Phillip; Keller, Kevin Marketing Management 15 th Edition. Pearson/Prentice Hall ISBN: 978-0-13385646-
A Food Marketing Institute found that 45% of households spend more than $125 a week on...
A Food Marketing Institute found that 45% of households spend more than $125 a week on groceries. Assume the population proportion is 0.45 and a simple random sample of 113 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 0.43? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer =  (Enter your answer...
A Food Marketing Institute found that 34% of households spend more than $125 a week on...
A Food Marketing Institute found that 34% of households spend more than $125 a week on groceries. Assume the population proportion is 0.34 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 0.32? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer =  (Enter your answer as...
A Food Marketing Institute found that 47% of households spend more than $125 a week on...
A Food Marketing Institute found that 47% of households spend more than $125 a week on groceries. Assume the population proportion is 0.47 and a simple random sample of 89 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 0.35? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations. Answer =  (Enter your answer...
A Food Marketing Institute found that 34% of households spend more than $125 a week on...
A Food Marketing Institute found that 34% of households spend more than $125 a week on groceries. Assume the population proportion is 0.34 and a simple random sample of 101 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 0.35? There is a  probability that the sample proportion of households spending more than $125 a week is less than 0.35. Round the answer to 4...
A Food Marketing Institute found that 28% of households spend more than $125 a week on...
A Food Marketing Institute found that 28% of households spend more than $125 a week on groceries. Assume the population proportion is 0.28 and a simple random sample of 436 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? There is a  probability that the sample proportion of households spending more than $125 a week is less than 0.3. Round the answer to 4...
A Food Marketing Institute found that 35% of households spend more than $125 a week on...
A Food Marketing Institute found that 35% of households spend more than $125 a week on groceries. Assume the population proportion is 0.35 and a simple random sample of 143 households is selected from the population. What is the probability that the sample proportion of households spending more than $125 a week is between 0.32 and 0.46? Note: You should carefully round any z-values you calculate to 4 decimal places to match wamap's approach and calculations.
In a survery of 134 households, a Food Marketing Institute found that 68 households spend more...
In a survery of 134 households, a Food Marketing Institute found that 68 households spend more than $125 a week on groceries. Please find the 95% confidence interval for the true proportion of the households that spend more than $125 a week on groceries. A. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. Confidence interval = B. Express the same answer as a tri-linear inequality using decimals (not percents) accurate to...
In a survery of 352 households, a Food Marketing Institute found that 148 households spend more...
In a survery of 352 households, a Food Marketing Institute found that 148 households spend more than $125 a week on groceries. Please find the 98% confidence interval for the true proportion of the households that spend more than $125 a week on groceries. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. Confidence interval = Express the same answer as a tri-linear inequality using decimals (not percents) accurate to three decimal...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT