Question

1.A doctor wants to estimate the HDL cholesterol of all​ 20- to​ 29-year-old females. How many subjects are needed to estimate the HDL cholesterol within 3 points with 99 % confidence assuming sigma equals 11.5 question mark Suppose the doctor would be content with 90 % confidence. How does the decrease in confidence affect the sample size​ required? A 90 % confidence level requires nothing subjects. ​(Round up to the nearest whole number as​ needed.)

2.People were polled on how many books they read the previous year. How many subjects are needed to estimate the number of books read the previous year within one book with 95​% ​confidence? Initial survey results indicate that sigmaequals11.1 books. A 95 % confidence level requires nothing subjects.

3.You are given the sample mean and the sample standard deviation. Use this information to construct the​ 90% and​ 95% confidence intervals for the population mean. Which interval is​ wider? If​ convenient, use technology to construct the confidence intervals. A random sample of 33 ​eight-ounce servings of different juice drinks has a mean of 98.8 calories and a standard deviation of 48.2 calories. The​ 90% confidence interval is left parenthesis nothing comma nothing right parenthesis . ​(Round to one decimal place as​ needed.)

Answer 1

1)

The following information is provided,
Significance Level, α = 0.01, Margin or Error, E = 3, σ = 11.5

The critical value for significance level, α = 0.01 is 2.58.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (2.58 * 11.5/3)^2
n = 97.81

Therefore, the sample size needed to satisfy the condition n >= 97.81 and it must be an integer number, we conclude that the minimum required sample size is n = 98
Ans : Sample size, n = 98

2)

The following information is provided,
Significance Level, α = 0.05, Margin or Error, E = 1, σ = 11.1

The critical value for significance level, α = 0.05 is 1.96.

The following formula is used to compute the minimum sample size required to estimate the population mean μ within the required margin of error:
n >= (zc *σ/E)^2
n = (1.96 * 11.1/1)^2
n = 473.32

Therefore, the sample size needed to satisfy the condition n >= 473.32 and it must be an integer number, we conclude that the minimum required sample size is n = 474
Ans : Sample size, n = 474

3)

sample mean, xbar = 98.8
sample standard deviation, s = 48.2
sample size, n = 33
degrees of freedom, df = n - 1 = 32

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.69

ME = tc * s/sqrt(n)
ME = 1.69 * 48.2/sqrt(33)
ME = 14.18

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (98.8 - 1.69 * 48.2/sqrt(33) , 98.8 + 1.69 * 48.2/sqrt(33))
CI = (84.6 , 113.0)

The 90% confidenc einterval is (84.6 , 113.0)

sample mean, xbar = 98.8
sample standard deviation, s = 48.2
sample size, n = 33
degrees of freedom, df = n - 1 = 32

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.04

ME = tc * s/sqrt(n)
ME = 2.04 * 48.2/sqrt(33)
ME = 17.117

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (98.8 - 2.04 * 48.2/sqrt(33) , 98.8 + 2.04 * 48.2/sqrt(33))
CI = (81.7 , 115.9)

The 95% Confidence interval is (81.7 , 115.9)

The 95% interval is wider