Question

A recent opinion poll found that 245 out of 250 people are opposed to a new tax. Calculate the sample proportion, and the standard error of the population, SEP. Round to the nearest whole percent and the SEP to the nearest hundredth.

Marine biologists catching tuna for research found that 16 out of 28 tuna had elevated mercury levels. Calculate the sample proportion, and the standard error of the population, SEP. Round to the nearest whole percent and the SEP to the nearest hundredth.

Can someone help with both problems?

Answer 1

Solution-1: Sample size -n -250 Sample number of people that are opposed to a new tax-x Sample proportion of people that are opposed to a new tax 245 x 245 n 250 0.980 = 98% Standard error of the population p(1 -p) 0.980 x (1 0.980) -0.0089 250 Solution-2: Sample size = n 28 Sample number of tuna that had elevated mercury levelsx Sample proportion of tuna that had elevated mercury levels p =-=-= 0.571 57.1% 16 x 16 n 28 Standard error of the population p(1 -p) 0.571 X (1 0.571) 0.0935 28

Answer 2

• use the given information to calculate the sample proportion, SEP, for each of the described sample populations. Round p to the nearest whole percent and round the SEP to the nearest hundreths a new window screen was found to block 1,400 out of 1,540 types of insects from getting through the window