Question

Thirty-two percent of all Americans drink bottled water more than once a week (Natural resources Defense Council, December 4, 2015). Suppose you have been hired by the Natural Resources Defence Council to investigate bottled water consumption in St. Paul. You plan to select a sample of St. Paulites to estimate the proportion who drink bottled water more than once a week. Assume the sample proportion of St. Paulites to estimate the proportion who drink bottled water more than once a week is 0.32, the same as the overall proportion of americans who drink bottled water more than once a week. Use z-table. a. Suppose you select a sample of 540 St.Paulites. Show the sampling distribution of p (to 4 decimals) ク b. Based upon a sample of 540 St. Paulites, what is the probability that the sample proportion will be within 0.08 of the population proportion (to 4 decimals). probability= .9999 C. Suppose you select a sample of 230 St.Paulites. Show the sampling distribution of (to 4 decimals) =| .0201 0 d. Based upon a smaller sample of only 230 St. Paulites, what is the probability that the sample proportion will be within 0.08 of the population proportion (to 4 decimals) probability = .9907 e. As measured by the increase in probability, how much do you gain in precision by taking the larger sample in parts (a) and (b) rather than the smaller sample in parts (c) and (d)? Reduced by Have gain in precision by increasing the sample.

Answer 1

Answer:

Step-by-step explanation:

given E(p) = 0.32

random sample size n =540

a.) sample distribution=

standad deviation of p$\sqrt{\frac{p(1-p}{n} } =\sqrt{\frac{0.32(1-0.32}{540} } =0.0201$

b.)   $z=\frac{p^{i}-p_{0} }{\sqrt{\frac{p_{0}(1-p_{0})}{n} } } \\\\\\\frac{0.32-0.08}{\sqrt{\frac{0.08*0.92}{540} } }=20.51$

probability = 0.9999

c.) n = 230

E(p) = 0.32

$\sqrt{\frac{p(1-p}{n} } =\sqrt{\frac{0.32*0.68}{230} } =0.0308$

d.)  $z=\frac{p^{i}-p_{0} }{\sqrt{\frac{p_{0}(1-p_{0})}{n} } } \\\\\\\frac{0.32-0.08}{\sqrt{\frac{0.08*0.92}{230} } }=13.41$

probability = 0.9999

e.)  20.51-13.41=7.1

reduced by 7.1 and have gained in precision by increasing the sample.