Question

According to a 2009 Reader's Digest article, people throw away approximately 14% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 176 grocery shoppers to investigate their behavior. What is the probability that the sample proportion exceeds 0.17?

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 152 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?

According to a 2009 Reader's Digest article, people throw away approximately 11% of what they buy at the grocery store. Assume this is the true proportion and you plan to randomly survey 123 grocery shoppers to investigate their behavior. What is the probability that the sample proportion is between 0.05 and 0.14?

Answer 1

Solution(a)
Given in the question
Population proportion = 0.14
Sample Proportion = 0.17
No. of sample = 176
we need to find P(sample proportion>0.17) = 1-P(Sample Proportion<0.17)
We will use Z test
Z = (0.17-0.14)/sqrt(0.14*0.86/176) = 0.03/0.0261= 1.15
So from Z table we found P-value
P(sample proportion>0.17) = 1- 0.8749 = 0.1251

Solution(b)
population proportion = 0.28
No. of sample = 152
We need to find P(0.27<smaple proportion<0.38) = P(sample proportion<0.38) - P(sample proportion<0.27)
Z = (0.27-0.28) / sqrt(0.28*0.72/152) = -0.01/0.03641 = -0.27
Z = (0.38-0.28)/sqrt(0.28*0.72/152) = 2.75
From Z table we found
P(0.27<smaple proportion<0.38) = P(sample proportion<0.38) - P(sample proportion<0.27) = 0.9970 - 0.3936 = 0.6034

Solution(c)

P = 0.11
No. of sample = 123
P(0.05<sample proportion<0.14) = P(sample proportion<0.14) - P(sample proportion<0.05)
Z = (0.05-0.11)/sqrt(0.14*0.86/123) = -1.92
Z = (0.14-0.11)/sqrt(0.14*0.86/123) = 0.96
from Z table we found p-value
P(0.05<sample proportion<0.14) = 0.8315 - 0.0274 = 0.8041