Question

The following are the heat-producing capacities of coal from two mines (in millions of calories per ton): Mine A: 8,500, 8,330, 8,480, 7,960, 8,030 Mine B: 7,710, 7,890, 7,920, 8,270, 7,860. Assuming that the data constitute independent random samples from normal populations with equal variances, construct a 99% confidence interval for the difference between the true average heat-producing capacities of coal from the two mines.

Answer 1

Answer: The confidence interval will be -91.87 < μ₁-μ₂ <751.87

Step-by-step explanation: Confidence Interval is the interval of certainty in which the true population mean is in.

In a confidence interval for the difference between two averages with a sample size less than 30, the calculations are:

$x_{1}-x_{2}$ ± $t.S_{p}\sqrt{\frac{1}{n_{1}} +\frac{1}{n_{2}} }$

in which

x₁ and x₂ are the sample mean of each data set

t is the probability found in the t-table whose df = n₁ + n₂ - 2

$S_{p}$ is the pooled estimate of the common standard deviation, assuming variances are similar, and can be calculated as:

$S_{p}=\sqrt{\frac{(n_{1}-1)(s_{1})^{2}+(n_{2}-1)(s_{2})^{2}}{n_{1}+n_{2}-2} }$

n₁ and n₂ are sample sizes

The data to construct the interval are:

Mine A:

x₁ = 8260

s₁ = 251.9

n₁ = 5

Mine B:

x₂ = 7930

s₂ = 206.52

n₂ = 5

Then:

$S_{p}=\sqrt{\frac{(5-1)(251.9)^{2}+(5-1)(206.52)^{2}}{8}$

$S_{p}=\sqrt{\frac{253814.44+170602.04}{8}$

$S_{p}=$ 230.33

Using a table, for a 99% confidence and df = 8, t-score = 2.896.

Doing calculations:

(8260-7930) ± (2.896)(230.33)($\sqrt{\frac{1}{5} +\frac{1}{5} }$)

330 ± (2.896)(230.33)($\sqrt{0.4}$)

330 ± 421.87

Interval will be

$-91.87<\mu_{1}-\mu_{2}<751.87$

A 99% confidence interval for the difference between the true average heat-producing capacities of coal from the mines is between -91.87 and 751.87