Assignment 5B 1) The Gallop Poll folks found that 27% of adult’s surveyed (national survey) said that they had been in a tornado. How many adults would need to be surveyed to estimate the true...

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Assignment 5B
1) The Gallop Poll folks found that 27% of adult’s surveyed (national survey) said that they had been in a tornado. How many adults would need to be surveyed to estimate the true proportion with a 95% confidence that is 5% wide?
2) A recent study indicated that 29% of 100 women over age 55 in the study were widows.
a) How large a sample must one take to be 90% confident that the estimate is within 0.05 of the true proportion?
b) If no estimate of the sample proportion is available, how large a sample is needed?
3) An old study of 1000 people in Phoenix found that 27 were obese. Find the 99% confidence interval of the population proportion of persons living in Phoenix who are obese.
Assignment 5A
1) A study of 35 golfers showed that their average score was 92. The standard deviation of the sample was 5.
a) Find the best point estimate of the mean.
b) Find the 95% confidence interval of the mean score for all golfers.
c) Find the 95% confidence interval of the mean score if the sample of 60 golfers is used.
d) Which interval is smaller? Explain why.
2) Pizza Hut wishes to find the 90% confidence interval of the true mean cost of a medium “veggie” pizza. How large a sample is needed if the store manager wants to be accurate to within $0.10? A previous study gave a standard deviation of $0.57.
3) A sample of 17 states has the cigarette taxes (in cents) listed:
112 120 98 55 71 35 99 124 62
150 150 55 100 132 20 70 93
Find the 98% confidence interval for this tax in all 50 states, the District of Columbia, and Puerto Rico.



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Assignment 5B 1) The Gallop Poll folks found that 27% of adult’s surveyed (national survey) said that they had been in a tornado. How many adults would need to be surveyed to estimate the true proportion with a 95% confidence that is 5% wide? 2) A recent study indicated that 29% of 100 women over age 55 in the study were widows. a) How large a sample must one take to be 90% confident that the estimate is within 0.05 of the true proportion? b) If no estimate of the sample proportion is available, how large a sample is needed? 3) An old study of 1000 people in Phoenix found that 27 were obese. Find the 99% confidence interval of the population proportion of persons living in Phoenix who are obese. Assignment 5A 1) A study of 35 golfers showed that their average score was 92. The standard deviation of the sample was 5. a) Find the best point estimate of the mean. b) Find the 95% confidence interval of the mean score for all golfers. c) Find the 95% confidence interval of the mean score if the sample of 60 golfers is used. d) Which interval is smaller? Explain why. 2) Pizza Hut wishes to find the 90% confidence interval of the true mean cost of a medium “veggie” pizza. How large a sample is needed if the store manager wants to be accurate to within $0.10? A previous study gave a standard deviation of $0.57. 3) A sample of 17 states has the cigarette taxes (in cents) listed: 112 120 98 55 71 35 99 124 62 150 150 55 100 132 20 70 93 Find the 98% confidence interval for this tax in all 50 states, the District of Columbia, and Puerto Rico.



Answered Same DayDec 20, 2021

Answer To: Assignment 5B 1) The Gallop Poll folks found that 27% of adult’s surveyed (national survey) said...

David answered on Dec 20 2021
128 Votes
Assignment 5B
1) The Gallop Poll folks found that 27% of adult’s surveyed (national survey) said that they had been
in a
tornado. How many adults would need to be surveyed to estimate the true proportion with a 95%
confidence that is 5% wide?
Answer:
Here the sample estimate of the proportion is, p = 0.27. Here, α = 0.05. Therefore, zα/2 = z0.025 = 1.96.
Here the width of the interval is 0.05 therefore the margin of error, E = 0.025.
Let n = required sample size
Therefore, n = (z0.025)
2*p*(1-p)/E2 = 1211.49 ≈ 1212
1212 adults would need to be surveyed
2) A recent study indicated that 29% of 100 women over age 55 in the study were widows.
a) How large a sample must one take to be 90% confident that the estimate is within 0.05 of the true
proportion?
Answer:
Here the sample estimate of the proportion is, p = 0.29. Here, α = 0.1. Therefore, zα/2 = z0.05 = 1.645. Here
the margin of error, E = 0.05.
Let n = required sample size
Therefore, n = (z0.05)
2*p*(1-p)/E2 = 222.87 ≈ 223
b) If no estimate of the sample proportion is available, how large a sample is needed?
Since we do not know any estimate we take 0.5 (conservative estimate)
Therefore, the required sample size, n = (z0.05)
2*p*(1-p)/E2 = 270.6 ≈ 271
This...
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