The 92 million Americans of age 50 and over control 50% of all discretionary income (AARP Bulletin, March 2008). AARP estimated that the average annual expenditure on restaurants and carry out food was $1873 for individuals in this age group. Suppose this estimate is based on a sample of 80 persons and that the sample standard deviation is $550.
a. At 95% confidence, what is the margin of error?
b. What is the 95% confidence interval for the population mean amount spent on restaurants and carryout food?
c. What is your estimate of the total amount spent by Americans of age 50 and over on restaurants and carryout food?
d. If the amount spent on restaurants and carryout food is skewed to the right, would you expect the median amount spent to be greater or less than $1873?
I'll help you with the first two:
For a):
The margin of error = 1.96(sd/√n)
sd = 550
n = 80
1.96 represents 95% using a z-table
For b):
CI95 = mean + or - margin of error
mean = 1873
Substitute the values into the formula and calculate.
Please help and explain the best you can. I am so lost.
Thank you in advance
To find the answers to these questions, we will use the formula for calculating the margin of error and confidence interval for a population mean based on a sample.
a. The margin of error can be calculated using the formula:
Margin of Error = z * (σ / √n)
where z is the z-score corresponding to the desired confidence level (95% in this case), σ is the standard deviation of the sample (550), and n is the sample size (80).
Plugging in the values, we get:
Margin of Error = 1.96 * (550 / √80) ≈ 1.96 * 61.27 ≈ 120.07
So, the margin of error is approximately $120.07.
b. The 95% confidence interval can be calculated using the formula:
Confidence Interval = sample mean ± Margin of Error
In this case, the sample mean is $1873 and the margin of error is $120.07.
Therefore, the 95% confidence interval is approximately $1873 ± $120.07, or $1752.93 to $1993.07.
c. To estimate the total amount spent by Americans of age 50 and over on restaurants and carryout food, we multiply the sample mean by the total population size. According to the information provided, there are 92 million Americans in this age group.
Estimated Total Amount Spent = sample mean * population size
= $1873 * 92,000,000 = $172,036,000,000
Therefore, the estimated total amount spent is approximately $172,036,000,000.
d. If the distribution of the amount spent on restaurants and carryout food is skewed to the right, then the median amount spent would be expected to be less than $1873.
To answer these questions, we will use the formula for calculating the margin of error and confidence intervals for a population mean:
Margin of Error = (Critical Value) * (Standard Deviation / Square Root of Sample Size)
Confidence Interval = Sample Mean ± (Margin of Error)
a. To find the margin of error, we need to identify the critical value for a 95% confidence level. Since the sample size is 80, we can assume that the population distribution is approximately normal. Looking up the critical value in a standard normal distribution table, we find that the critical value for a 95% confidence level is approximately 1.96.
Margin of Error = (1.96) * (550 / √80)
Margin of Error ≈ 12.26
Therefore, the margin of error is approximately $12.26.
b. To calculate the confidence interval, we use the sample mean, which is $1873, and the margin of error calculated in part (a).
Confidence Interval = $1873 ± $12.26
Confidence Interval ≈ ($1860.74, $1885.26)
So, the 95% confidence interval for the population mean amount spent on restaurants and carryout food is approximately $1860.74 to $1885.26.
c. To estimate the total amount spent by Americans aged 50 and over on restaurants and carryout food, we need to multiply the average annual expenditure per individual ($1873) by the total number of Americans aged 50 and over (92 million).
Total Amount Spent ≈ Average Expenditure per Individual * Total Population
Total Amount Spent ≈ $1873 * 92,000,000
Total Amount Spent ≈ $171,716,000,000
Therefore, the estimate of the total amount spent by Americans aged 50 and over on restaurants and carryout food is approximately $171,716,000,000.
d. If the amount spent on restaurants and carryout food is skewed to the right, it means that there are few individuals who spend significantly higher amounts than the average. In this scenario, the median amount spent would be expected to be less than the mean of $1873.