Question

A Food Marketing Institute found that 56% of households spend more than $125 a week on groceries. Assume the population proportion is 0.56 and a simple random sample of 91 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 0.4?

Answer 1

Soln,

Given that p= 0.56 of spends more than $ 125

and n= 91 of saimple random sample is selected

hence according to question that what is the probability of spending more than $125 is or 40 %

Since it is from a large population and n > 30 hence Z test is applied

Hence P value to Z> -3.08 is 0.999 which is almost equal to 1.

Table is shown below one thing is noted that the graph always shows the value to the left of the graph hence for z>-3.08 we will use 1-0.0010=0.999 as the p - value