Answer:
To construct a 90% confidence interval for the mean amount of juice in all such bottles, we first need to find the sample mean and sample standard deviation:
Sample mean, x = (15.2 + 15.5 + 15.9 + 15.5 + 15.0 + 15.7 + 15.0 + 15.7)/8 = 15.4375
Sample standard deviation, s = s = sqrt[((15.2-15.4375)^2 + (15.5-15.4375)^2 + (15.9-15.4375)^2 + (15.5-15.4375)^2 + (15.0-15.4375)^2 + (15.7-15.4375)^2 + (15.0-15.4375)^2 + (15.7-15.4375)^2)/7] = 0.339
Using a t-distribution with degrees of freedom (n-1) = 7 and a 90% confidence level, we can find the t-value as 1.895.
The 90% confidence interval can then be calculated as:
x plus or minus (t-value)*(s/sqrt(n))
= 15.4375 plus or minus (1.895)*(0.339/sqrt(8))
= (15.16, 15.72)
Therefore, the answer is A.
To find a 99% confidence interval for the mean monthly income for all workers at the plant, we use the formula:
x plus or minus (z-value)*(σ/sqrt(n))
where x is the sample mean, σ is the population standard deviation, n is the sample size, and z-value is the critical value from the standard normal distribution for a 99% confidence level, which is 2.576.
Plugging in the given values, we get:
x plus or minus (z-value)*(σ/sqrt(n))
= 2150 plus or minus (2.576)*(250/sqrt(18))
= (2096, 2204)
Therefore, the answer is A.