7. Standard deviation, σ = 500
Confidence coefficient = 0.99
Margin of Error, E = (Upper limit - Lower limit)/2 = (9600-9400)/2 = 100
Significance level, α = 1-CL = 0.01
Critical value, z = NORM.S.INV(0.01/2) = 2.5758
Sample size, n = (z * σ / E)² = (2.5758 * 500 / 100)² = 165.87 = 166
Answer a. 166
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8. Answer b, I and II are both false.
7. in an area. A sample survey is to be conducted to determine the mean family...
A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $79.50 Devo a 95% confidence interval estimate of the mean amount spent per day by a family of four visiting Niagara Falls (to 2 decimals). . Based on the...
A sample survey of 62 discount brokers showed that the mean price charged for a trade of 100 shares at $50 per share was $31.22. The survey is conducted annually. With the historical data available, assume a known population standard deviation of $19. (a) Using the sample data, what is the margin of error in dollars associated with a 95% confidence interval? (Round your answer to the nearest cant) $ (b) Develop a 95% confidence interval for the mean price...
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QUESTION 4 Question 4-5 are based on the following information: A survey conducted by the American Automobile Association showed that a family of four spends an average of $215.60 per day while on vacation. Suppose a sample of 64 families of four vacationing at Niagara Falls resulted in a sample mean of $252.45 per day and a sample standard deviation of $70.00. Develop a 95% confidence interval estimate of the mean amount spent per day by a family of four...
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A survey is to be conducted to determine the average driving in miles by MS dents. The investigator wants to know how are the sample should be taken to confidence interval on the true average within 1.5 miles. A similar study on the past found that the population standard deviation was 82 miles Bonus(5 points): A recent study of 750 Internet users in Europe found that 35% of Internet users were women. What is the 96% confidence...