Question

A research company desires to know the mean consumption of milk per week among males over age 25. A sample of 650 males over age 25 was drawn and the mean milk consumption was 3.4 liters. Assume that the population standard deviation is known to be 1.3 liters. Construct the 95% confidence interval for the mean consumption of milk among males over age 25. Round your answers to one decimal place.

Answer 1

Explanation:

Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Since the sample size is large and the population standard deviation is known, we would use the following formula and determine the z score from the normal distribution table.

Margin of error = z × σ/√n

Where

σ = population standard Deviation

n = number of samples

From the information given

x = 3.4

σ = 1.3

n = 650

To determine the z score, we subtract the confidence level from 100% to get α

α = 1 - 0.95 = 0.05

α/2 = 0.05/2 = 0.025

This is the area in each tail. Since we want the area in the middle, it becomes

1 - 0.025 = 0.975

The z score corresponding to the area on the z table is 1.96. Thus, confidence level of 95% is 1.96

Margin of error = 1.96 × 1.3/√650 = 0.1

Confidence interval = 3.4 ± 0.1

The lower end of the confidence interval is

3.4 - 0.1 = 3.3 litres

The upper end of the confidence interval is

3.4 + 0.1 = 3.5 litres

Answer 2

`3.3\leq x\leq 3.5`

Explanation:

The 95% confidence interval can be calculated as:

`x'-z_(\alpha/2)(s)/(√(n)) \leq x\leq x'+z_(\alpha/2)(s)/(√(n))`

Where
`x` is the mean consumption of milk,
`x'` is the sample mean,
`s` is the population standard deviation,
`n` is the size of the sample,
`\alpha` is 5% and
`z_(\alpha /2)` is the z-value that let a probability of
`\alpha /2` on the right tail.

So, replacing
`x'` by 3.4 liters,
`s` by 1.3 liters, n by 650 and
`z_(\alpha /2)` by 1.96, we get that the 95% confidence interval is equal to:

`3.4-(1.96*(1.3)/(√(650)))\leq x\leq 3.4+(1.96*(1.3)/(√(650))) \\3.4-0.1\leq x\leq 3.4+0.1\\3.3\leq x\leq 3.5`