Question

1. In 2009 an industry spokesperson said that over 82% of Americans owned a cellular phone. Test this claim using a significance level of a = 0.05, given that in a random sample of 1234 Americans, 1023 owned a cell phone. What is P? (2 dec. places) e) Draw a picture of the sampling distribution of P below and shade in the area that corresponds to the p-value you found. Label axis with the appropriate values. a) Write the null and alternative hypotheses. Ho: Ha: b) To determine if the conditions are met to perform a z-test, complete the following. f) State your conclusion and interpret its meaning in context of the original claim (Use a = 0.05.) PE 유 Is np = 210? Is ng = 2 10? c) Write the rejection rule by filling in the blank: Reject He if p-value d) Use STAT-TESTS-1-PropZTest to find the test statistic z and the p-value za p-value =

Answer 1

sal papulation propartion =P=0.82 an 82% ( propastion of A American owned celletas phone in 2009) 1023 = 0.829 1234 sample peropartion a) Ho! P = 0.82 P 7 0.82 H, P = 0.82 q 1-P = 0.18 np 1234 (0.82) - 1011.88 > 10 (TRUE) ng - 1234 (0.18) = 222.12 >10 (TRUE) c) Reject to it p-nalue (x= lave of significance) d) z = P(1-P) Pa n n Z = 0.829 -0.800 1.02 0.163 (082) (0.18) 1234 L = 6.258 : scare is too high it means p-value i's approximately xero. P-nalue = 4x1010 an zero