This is practice exam that is not graded. To practice for real exam coming up. **Exercise 07.37**The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population proportion is $ p = 0.17 $ and a simple random sample of 800 households will be selected from the population. Use the z-table.a. Calculate the sampling distribution of $\bar{p}$, the proportion of households spending more than $100 per week on groceries.- $ E(\bar{p}) = $ _____ (to 2 decimals)- $ \sigma_{\bar{p}} = $ _____ (to 4 decimals)b. What is the probability that the sample proportion will be within ±0.02 of the population proportion (to 4 decimals)?c. What is the probability that the sample proportion will be within ±0.02 of the population proportion for a sample of 1600 households (to 4 decimals)?[**Icon Key**]- Save- Submit Assignment for Grading- Check My Work**Question 8 of 8**

Given p=0.17 n=800E(p¯)=p=0.17 σp¯=(p*(1-p)n) =0.17*(1-0.17)800 =0.0133 The sampling distribution of the sample mean p^ has mean 0.17 and standard deviation is 0.0133 .b) We need first to find test statisticz=p^-pp*(1-p)nz=±0.020.17*(1-0.17)800z=±1.51 ThereforeP(p^-p<0.02)=P(-1.51<z<1.51) =1-2*P(z<-1.51) =1-2*(0.0655) (by using z table) =0.8690The probability that the sample proportion will be within ±0.02 of the population proportion is equal to 0.8690 c) Given : n=1600 The test statistic is:z=p^-pp*(1-p)nz=±0.020.17*(1-0.17)1600z=±2.13 P(p^-p<0.02)=P(-2.13<z<2.13) =1-2*P(z<-2.13) =1-2*(0.0166) (by using z table) =0.9668The probability that the sample proportion will be within ±0.02 of the population proportion for a sample of 1600 households is equal to 0.9668