Question

1) A statistician randomly selects 27 dates to examine the occupancy rate of a hotel. While reviewing the hotel's occupancy records for these dates, the statistician calculates the standard deviation of the number of rented rooms to be 5.86. Assuming that the distribution of the number of rented rooms follows a normal distribution, provide an estimate for the population standard deviation at a 95% confidence level.

2) Assume the distribution of the volumes of bottles produced in a glass bottle factory is N(μ, 25). A random sample is taken from the production line. Calculate the number of bottles that need to be selected for the length of the 95% confidence interval for the μ parameter to be 1.

3) A factory operates with two type A machines and one type B machine. The random variable Y, representing the weekly maintenance cost of the type A machines, follows a Normal distribution with mean μ1 and variance 2σ2. The random variable X, representing the weekly maintenance cost of the type B machines, also follows a Normal distribution but with mean μ2 and variance 2σ3. The expected weekly maintenance cost for the factory is given as μ1 + 2μ2. Given a random sample of weekly maintenance costs drawn from type A machines, Y1, Y2, ..., Yn, and an independent random sample drawn from the weekly maintenance costs of type B machines, X1, X2, ..., Xm, calculate a 95% confidence interval for μ1 + 2μ2.

Answer 1

1. To estimate the population standard deviation at a 95% confidence level, we can use the following formula:

s/√n * tα/2,n-1 = margin of error

where s is the sample standard deviation (5.86), n is the sample size (27), tα/2,n-1 is the t-value for the 95% confidence level with n-1 degrees of freedom (26 degrees of freedom in this case), and the margin of error is the range in which we expect the true population standard deviation to lie.

First, we need to find the t-value for the 95% confidence level with 26 degrees of freedom. We can do this using a t-distribution table or a calculator and find that tα/2,n-1 = t0.025,26 = 2.056.

Next, we can plug in the values we have:

5.86/√27 * 2.056 = 2.696

This means that we can be 95% confident that the true population standard deviation lies within a range of ±2.696. Therefore, we can estimate the population standard deviation to be between 3.164 (5.86 - 2.696) and 8.556 (5.86 + 2.696) at a 95% confidence level.

2. If the distribution of the volumes of bottles produced in a glass bottle factory is N(μ, 25), then the standard deviation of the population is σ = 5 (since σ^2 = 25).

The length of the 95% confidence interval for the mean (μ) is given by:

length = 2 * (z-value) * (standard error)

where the z-value is the critical value from the standard normal distribution for a 95% confidence level (1.96) and the standard error is the standard deviation of the sample mean, which is given by:

standard error = σ / sqrt(n)

where n is the sample size.

To find the sample size (n) needed for the length of the 95% confidence interval to be 1, we can set up the following equation:

1 = 2 * 1.96 * (5 / sqrt(n))

Solving for n, we get:

n = (2 * 1.96 * 5)^2 / 1^2

n = 96.04

Since n must be a whole number, we round up to get:

n = 97

Therefore, we need to select at least 97 bottles from the production line to estimate the population mean (μ) with a 95% confidence interval whose length is no more than 1 unit.