Question 211330:
Section 6.1: Confidence Intervals for the Mean (Large Samples)
1. Find the critical value zc necessary to form a confidence interval at the given level of confidence. (References: definition for level of confidence page 311, end of section exercises 5 – 8 page 317)
a. 95%
b. 75%
2. The new Twinkle bulb has a mean life of hours with a standard deviation hours. A random sample of 50 light bulbs is selected from inventory. The sample mean was found to be 500 hours.
a. Find the margin of error E for a 95% confidence interval.
Round your answer to the nearest hundredths.
b. Construct a 95% confidence interval for the mean life, of all Twinkle bulbs. (References: example 5 page 315, end of section exercises 51 - 56 pages 319 - 320)
Answer: a.
b.
3. A standard placement test has a mean of 115 and a standard deviation of = 10. Determine the minimum sample size if we want to be 95% certain that we are within 3 points of the true mean. (References: example 6 page 316, end of section exercises 58 - 62 pages 321 - 322)
Section 6.2: Confidence Intervals for the Mean (Small Samples)
4. An experimental egg farm is raising chickens to produce low cholesterol eggs. A lab tested 16 randomly selected eggs and found that the mean amount of cholesterol was 190 mg. The sample standard deviation was found to be s = 18.0 mg on this group. Assume that the population is normally distributed. (References: example 2 and 3 pages 327 - 328, end of section exercises 5 - 16 pages 330 - 331)
a. Find the margin of error for a 95% confidence interval. Round your answer to the nearest tenths.
b. Find a 95% confidence interval for the mean cholesterol content for all experimental eggs. Assume that the population is normally distributed.
Section 6.3: Confidence Intervals for Population Proportions
5. The new Twinkle bulb is being developed to last more than 1000 hours. A random sample of 100 of these new bulbs is selected from the production line. It was found that 64 lasted more than 1000 hours. Find the point estimate for the population proportion, the margin of error for a 95% confidence interval and then construct to the 95% confidence interval for the population proportion.
a. Find the margin of error E.
b. Construct a 95% confidence interval for the population proportion p of all Twinkle bulbs.
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! 1. Find the critical value zc necessary to form a confidence interval at the given level of confidence. (References: definition for level of confidence page 311, end of section exercises 5 – 8 page 317)
a. 95%
Ans: z = +/-1.96
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b. 75%
Ans: z = +/-1.150..
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2. The new Twinkle bulb has a mean life ??? of hours with a standard deviation hours. A random sample of 50 light bulbs is selected from inventory. The sample mean was found to be 500 hours.
Comment. I put ??? where some word(s) are missing.
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a. Find the margin of error E for a 95% confidence interval.
Round your answer to the nearest hundredths.
E = 1.96[s/sqrt(100)]
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b. Construct a 95% confidence interval for the mean life, ??? of all Twinkle bulbs. (References: example 5 page 315, end of section exercises 51 - 56 pages 319 - 320)
Answer:
a. ???
b. ???
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3. A standard placement test has a mean of 115 and a standard deviation of s = 10. Determine the minimum sample size if we want to be 95% certain that we are within 3 points of the true mean. (References: example 6 page 316, end of section exercises 58 - 62 pages 321 - 322)
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n = [1.96*10/3]^2 = 42.68
Round up to n = 49
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Section 6.2: Confidence Intervals for the Mean (Small Samples)
4. An experimental egg farm is raising chickens to produce low cholesterol eggs. A lab tested 16 randomly selected eggs and found that the mean amount of cholesterol was 190 mg. The sample standard deviation was found to be s = 18.0 mg on this group. Assume that the population is normally distributed. (References: example 2 and 3 pages 327 - 328, end of section exercises 5 - 16 pages 330 - 331)
a. Find the margin of error for a 95% confidence interval. Round your answer to the nearest tenths.
E = 1.96[18/sqrt(16) = 1.96*18/4 = 8.82
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b. Find a 95% confidence interval for the mean cholesterol content for all experimental eggs. Assume that the population is normally distributed.
95%: 190-8.82 < u < 190+8.82
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Section 6.3: Confidence Intervals for Population Proportions
5. The new Twinkle bulb is being developed to last more than 1000 hours. A random sample of 100 of these new bulbs is selected from the production line. It was found that 64 lasted more than 1000 hours. Find the point estimate for the population proportion, the margin of error for a 95% confidence interval and then construct to the 95% confidence interval for the population proportion.
point estimate = (64/100) = 0.64
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a. Find the margin of error E.
E = 1.96*sqrt[0.64*0.36/100] = 0.0941..
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b. Construct a 95% confidence interval for the population proportion p of all Twinkle bulbs.
0.64 - 0.0941 < p < 0.64 + 0.0941
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Cheers,
Stan H.
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