Question

1.Quon wants to advertise how many chocolate chips are in each Big Chip cookie at his bakery. He randomly selects a sample of 73 cookies and finds that the number of chocolate chips per cookie in the sample has a mean of 6.6 and a standard deviation of 2.8. What is the 95% confidence interval for the number of chocolate chips per cookie for Big Chip cookies? Assume the data is from a normally distributed population. Round answers to 3 decimal places where possible.

2.A botanist wishes to estimate the typical number of seeds for a certain fruit. She samples 45 specimens and counts the number of seeds in each. Use her sample results (mean = 33.3, standard deviation = 5.1) to find the 90% confidence interval for the number of seeds for the species. Enter your answer as an open-interval (i.e., parentheses) accurate to 3 decimal places.

3.You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=28.3. You would like to be 95% confident that your estimate is within 4 of the true population mean. How large of a sample size is required?

Answer 1

1)

sample mean, xbar = 6.6
sample standard deviation, s = 2.8
sample size, n = 73
degrees of freedom, df = n - 1 = 72

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 1.99

ME = tc * s/sqrt(n)
ME = 1.99 * 2.8/sqrt(73)
ME = 0.652

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (6.6 - 1.99 * 2.8/sqrt(73) , 6.6 + 1.99 * 2.8/sqrt(73))
CI = (5.948 , 7.252)

2)

sample mean, xbar = 33.3
sample standard deviation, s = 5.1
sample size, n = 45
degrees of freedom, df = n - 1 = 44

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, tc = t(α/2, df) = 1.68

ME = tc * s/sqrt(n)
ME = 1.68 * 5.1/sqrt(45)
ME = 1.277

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (33.3 - 1.68 * 5.1/sqrt(45) , 33.3 + 1.68 * 5.1/sqrt(45))
CI = (32.023 , 34.577)

3)

The following information is provided,
Significance Level, α = 0.05, Margin or Error, E = 4, σ = 28.3

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 * 28.3/4)^2
n = 192.29

Therefore, the sample size needed to satisfy the condition n >= 192.29 and it must be an integer number, we conclude that the minimum required sample size is n = 193
Ans : Sample size, n = 193

Rounding to whole number answer would be 193 otherwise 192