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In a random sample of 1000 homes in a certain city, it is found that 229 are heated by oil. Find 99% confidence intervals for the proportion of homes in this city that are heated by oil using both of the accompanying methods for computing large-sample confidence intervals for p. Click here to view the methods for large-sample confidence intervals for p. Click here to view page 1 of the normal probability table. Click here to view page 2 of the normal probability table.
If p is the proportion of successes in a random sample of size n and q=1-p, an approximate 100(1-x)% confidence interval for the binomial parameter p is given by the two methods below, where Zx/2 is the z-value leaving an area of a/2 to the right. Method 1: p-Zα/2√ n Method 2: 2 p+ 2 Zα/2 2n 4359439 <p< 4n 1+ 1+ 2n 2 <p<p+Zα/2. n Zα/2 1 + 2 α/2 n n n + 2 2 Za12 n + 2012 1+ 2 n 2 Z/2 2 4n
(a) A random sample of 200 voters in a town is selected, and 123 are found to support an annexation suit. Find the 96% confidence interval for the fraction of the voting population favoring the suit. (b) What can be asserted with 96% confidence about the possible size of the error if the fraction of voters favoring the annexation suit is estimated to be 0.77? Click here to view page 1 of the normal probability table. Click here to view page 2 of the normal probability table.