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1. Use the given degree of confidence and sample data to construct a confidence interval for the point) population proportion p. A survey of 865 voters in one state reveals that 408 favor approval of an issue before the legislature. Construct the 95% confidence interval for the true proportion of all voters in the state who favor approval. 0 0.438<p0.505 0 0.444 p0.500 0 0.435<p<0.508 O 0.471 p0.472 2. Use the given data to find the minimum sample size required to estimate the populatiorn (i point) proportion. Margin of error: 0.09; confidence level 90%; from a prior study, p is estimated by 0.23. 060 O 53 180 0 5 3. Find the indicated critical z value. point) Find the critical value ze2 that corresponds to a 98% confidence level. 2.575 2.05 2.33 1.75 4. Use the given degree of confidence and sample data to construct a confidence interval for the s paint population proportion p. Of 140 adults selected randomly from one town, 35 of them smoke. Construct a S38 confidence interval for the true percentage of all adults in the town that smoke 15.654 K p < 34,4 25.0% < p < 31.0%
5. Solve the problem. Round the point estimate to the nearest thousandth. 1 point When 430 randomly selected light bulbs were tested in a laboratory, 224 lasted more than 500 hours. Find a point estimate of the proportion of all light bulbs that last more than 500 hours 0:0.521 0.479 0.519 0.343 6. Use the confidence level and sample data to find a confidence interval for estimating the 1 point) population u. Round your answer to the same number of decimal places as the sample mean. Test scores: n = 75, x :46.1, σ 58; 98% confidence O 44.8 < μ < 47.4 45.0 < μ < 47.2 0 44.5 < μ < 47.7 7. Use the confidence level and sample data to find a confidence interval for estimating the 1 point) population u. Round your answer to the same number of decimal places as the sample mean. A group of 66 randomly selected students have a mean score of 22 4 with a standard deviation of 2.8 on a placement test. What is the 90% confidence interval for the mean score. u, of all students taking the test? 21.6 < 23.2 w 21.7 < < 23.1 21.8 <u* 23.0 21.523.3 8. 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. 4 point) A rancom ssmple of 130 full-grown obsters had a mean weight of 21 ounces and a standard deviation of 3.0 ounces. Construct a 98% confidence in tral for the popelation mean. 0z or
9. Use the confidence level and sample data to find the margin of error E. Round your answer to the (point) same number of decimal places as the sample mean unless otherwise noted. College students annual earnings: 99% confidence; n-76,--$3016, ơ . S872 $233 $891 $258 $196 10. Use the confidence level and sample data to find the margin of error E. Round your answer to the point) same number of decimal places as the sample mean unless otherwise noted. 45, x-1 1.9 years ơ : 2.0 years Replacement times for washing machines: 90% confidence: n 0.5 yr 0.1 yr 0.4 yr 2.9 yr
0 0
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Answer #1

1) Sample proportion is 4080.472

The (1-a)100% confidence interval is given by

(1-P

For 95% confidence interval 21.96

So,

20-2 = (1.96)V/0 472(1-0.172) ~ 0.033 865

Hence the interval is

(0.472-0.033 : 0.472+0.033)=(0.439·0.505)

Option 1st is correct

2) The margin of error is given by

p(1 - )

Using the given values we get

0.09 1.645,/ η (18.2778)20.1771 0.23(1 - 0.23) 60

So sample size required is 60

Option first is correct

3) The critical value for 98% confidence is 2.33

Option 3rd is correct

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