a. Show the sampling distribution of p, the proportion of groceries thrown out by your sample respondents. b. What is the probability that your survey will provide a sample proportion within 6.03 of the population proportion

The corresponding p - value will be  is < 0.00001 which is less than 0.05

Step-by-step explanation:

Suppose p= 0.12 and n= 540 then the standard deviation can be taken out as √p(1-p)/n= √0.12(1-0.12)/540= 0.014

For the probability that the survey will provide a sample proportion within 6.03 of the population proportion

we will find the z statistic

z= p^-p/√p(1-p)/n

But √p(1-p)/n= 0.014 and p^-p= 6.03

Putting the values

z= 6.03/0.014= 430.71

The corresponding p - value will be  is < 0.00001 which is less than 0.05

you just switch the numbers basically so it would be 3/17

3.07*10^-4

move the decimal 4 places to the left

.000307