In a random sample of 1000 homes in a certain city, it is found that 229 are heated by oil. Find 99% confidence interval

Post by **answerhappygod** » Mon Jul 11, 2022 11:41 am

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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. The confidence interval using method 1 is 0.195 <p< 0.263 (Round to three decimal places as needed.) The confidence interval using method 2 is (Round to four decimal places as needed.) <p< (...)
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
Ten engineering schools in a country were surveyed. The sample contained 225 electrical engineers, 70 being women; 200 chemical engineers, 20 being women. Compute a 99% confidence interval for the difference between the proportions of women in these two fields of engineering. Is there a significant difference between the two proportions? Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. C Let p, be the population proportion of electrical engineers that are women in the schools that were surveyed and let p₂ be the population proportion of chemical engineers that are women in the schools that were surveyed. The 99% confidence interval is <P₁-P₂ <. (Round to three decimal places as needed.)
A survey of 1000 students found that 282 chose professional baseball team A as their favorite team. In a similar survey involving 760 students, 240 of them chose team A as their favorite. Compute a 95% confidence interval for the difference between the proportions of students favoring team A in the two surveys. Is there a significant difference? Click here to view page 1 of the normal probability table. Click here to view page 2 of the normal probability table. C Let p, be the population proportion studied in the first survey and let p₂ be the population proportion studied in the second survey. The 95% confidence interval is ]<P₁-P₂ <- (Round to four decimal places as needed.)