People end up tossing 12% of what they buy at the grocery store (Reader's Digest, March 2009). Assume this is the true population proportion and that you plan to take a sample survey of 540 grocery shoppers to further investigate their behavior.
a.Show the sampling distribution of ( ), the proportion of groceries thrown out by your sample respondents (to 4 decimals)?
To show the sampling distribution of the proportion of groceries thrown out by your sample respondents, you would need to calculate the standard error and use it to create a confidence interval.
1. Calculate the standard error:
The formula to calculate the standard error for a proportion is:
SE = √((p * (1 - p)) / n)
Where:
- SE is the standard error
- p is the true population proportion (given as 12% or 0.12)
- n is the sample size (given as 540)
SE = √((0.12 * (1 - 0.12)) / 540)
2. Calculate the margin of error:
The margin of error (MOE) is determined by multiplying the standard error by the critical value. For a 95% confidence level, the critical value is approximately 1.96.
MOE = 1.96 * SE
3. Calculate the lower and upper bounds of the confidence interval:
To determine the interval in which the true population proportion lies, you can subtract and add the margin of error from the sample proportion.
Lower bound = sample proportion - MOE
Upper bound = sample proportion + MOE
In this case, since you want to find the sampling distribution, the sample proportion would be the same as the true population proportion.
Sampling Distribution = [Lower bound, Upper bound]
For example, if the sample proportion (p-hat) is 0.12, the 95% confidence interval would be:
Sampling Distribution = [0.12 - MOE, 0.12 + MOE]
The sample distribution of p-bar
σp = √0.12*0.88/540) = 0.014