According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, assume some state had 447 complaints of identity theft out of 1740 consumer complaints. Do these data provide enough evidence to show that that state had a higher proportion of identity theft than 23% ? Test at the 4% level.How can I resolve the parts that are blanks ?

Check those assumptions: 1. np = 400.2 of which is 2 v of 10 2. n(1 - p) = 1339.8 of which is 2 vy of 10 3. N = 1000000 of which is 2 v of If no N is given in the problem, use 1000000 N: NAME THE PROCEDURE The conditions are met to use a 1-Proportion Z-Test 1 T: TEST STATISTIC ne The symbol and value of the random variable on this problem are as follows: Leave this answer as a fraction. pvox The formula set up of the test statistic is as follows. : (Leave any values that were given as fractions as fractions) p- p p(1-p) 0.23 OF )/ V( 0.23 v o . ( 1 0.23 1740 of ) Final answer for the test statistic from technology. Round to 2 decimal places: Z= O: OBTAIN THE P-VALUE Report to 4 decimal places. It is possible when rounded that a p-value is 0.0000 P-value = M: MAKE A DECISION Since the p-value s v v we reject H