Find the confidence interval for estimating the population proportion for parts (a) through ) Think carotary...
need help with this one Find the confidence interval for estimating the population proportion for parts (a) through (c). a. 92.5% confidence level; n = 600; P=0.20 b. 99% confidence level; n = 140; p = 0.04 C. Q=0.09; n = 375; p=0.90 Click the icon to view the standard normal table of the cumulative distribution function. a. The confidence interval is <P (Round to four decimal places as needed.) b. The confidence interval is <p<. (Round to four decimal...
confidence level and sample data to find confidence interval for estimating a population round your answer to the same number of decimal places as a sample Question 2 2 pts Assume that a sample is used to estimate a population proportion p. Find the margin of error E that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. 90% confidence; n 390, x-146
Use the confidence level and sample data to find a confidence interval for estimating the population . Round your answer to the same number of decimal places as the sample mean. Test scores: n = 104, 7 = 95.3, 0 = 6.5; 99% confidence 93.7 < < 96.9 O94.1 << 96.5 93.8 < p < 96.8 94.2 << 96.4
A business student is interested in estimating the 95% confidence interval for the proportion of students who bring laptops to campus. He wishes a precise estimate and is willing to draw a large sample that will keep the sample proportion within six percentage points of the population proportion. What is the minimum sample size required by this student, given that no prior estimate of the population proportion is available? Use Table 1. (Round intermediate calculations to 4 decimal places and...
Use the confidence level and sample data to find a confidence interval for estimating the population. Round your answer to the same number of decimal places as the sample mean A random sample of 87 light bulbs had a mean life of 479 hours with a standard deviation of a = 30 hours. Construct a 90% confidence interval for the mean life, ja, of all light bulbs of this type O A 472 hr < < 486 he B. 474...
Use the confidence level and sample data to find a confidence interval for estimating the population u. Round your answer to the same number of decimal places as the sample mean. 20) A group of 59 randomly selected students have a mean score of 29.5 with a standard deviation of 20) 5.2 on a placement test. What is the 90% confidence interval for the mean score,, of all students taking the test? A) 27.8 <u <31. 2 B ) 28.2<u<30....
Find zα/2 for each of the following confidence levels used in estimating the population proportion. Round your answers to 3 decimal places.) zα/2 a. 90% b. 98% c. 85% d. 95% e. 99%
a. You wish to compute the 95% confidence interval for the population proportion. How large a sample should you draw to ensure that the sample proportion does not deviate from the population proportion by more than 0.12? No prior estimate for the population proportion is available. Round intermediate calculations to at least 4 decimal places and "z" value to 3 decimal places. Round up your answer to the nearest whole number.) Sample Size - b. A business student is interested...
Use the confidence level and sample data to find a confidence interval for estimating the population. Round your answer to the same number of decimal places as the sample mean 19) A group of 5 randomly selected students have a mean score of 295 with a standard deviation of 19 5.2 on a placement test. What is the 90% confidence interval for the mean score of all students taking the test? A) 27.8 < <312 ) 27.9<<311 C)282 << 308...
1.) Assume that a sample is used to estimate a population proportion p. Find the margin of error m that corresponds to the given statistics and confidence level. Round the margin of error to four decimal places. 95% confidence; n = 380, x = 50 Group of answer choices 0.0340 0.0408 0.0306 0.0357 2.) Formulate the indicated conclusion in nontechnical terms. Be sure to address the original claim. A researcher claims that 62% of voters favor gun control. Assuming that...