Question

A chemical company has two factories, one in London and the other in Windsor. We wish to determine if there is any difference between the two factories with respect to how much methane is being put into the air. Readings of methane expulsion are taken at randomly spaced intervals in the two cities. The 8 readings in London show an average of 0.23 parts per million (ppm) with a standard deviation of 0.07 ppm. The 11 readings in Windsor show an average of 0.32 ppm with a standard deviation of 0.12 ppm.

a) Is the methane expulsion different in the two cities? First, you need to determine whether you should use the pooled or non-pooled test. Once you have determined that, you can test your hypothesis at the 5% level of significance by calculating the p-value. Make sure you state your null and alternative hypothesis (in symbols and words), make your decision and draw your conclusion.

(b) Calculate a 99% confidence interval for the difference in actual mean methane levels for the two cities and provide a clear interpret the resulting interval.

Answer 1

Solution:

Part a

Here, we have to use two sample t test for the difference between two population means assuming equal population variances. Here, we use pooled test, because sample sizes are small and less than 30.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: The methane expulsion is not different in the two cities.

Alternative hypothesis: H_a: The methane expulsion is different in the two cities.

H₀: µ₁ = µ₂ versus H_a: µ₁ ≠ µ₂

Test statistic formula for pooled variance t test is given as below:

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

We are given

X1bar = 0.23

X2bar = 0.32

S1 = 0.07

S2 = 0.12

n1 = 8

n2 = 11

df = n1 + n2 – 2 = 8 + 11 – 2 = 17

α = 0.05

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = [(8 – 1)* 0.07^2 + (11 – 1)* 0.12^2]/(8 + 11 – 2)

S_p² = 0.0105

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (X1bar – X2bar) / sqrt[S_p²*((1/n1)+(1/n2))]

t = (0.23 – 0.32) / sqrt[0.0105*((1/8)+(1/11))]

t = -0.09/0.0476

t = -1.8913

P-value = 0.0758

(by using t-table)

P-value > α = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that the methane expulsion is different in the two cities.

Part b

Here, we have to find 99% confidence interval for the difference between two population means.

Confidence interval for difference between two population means is given as below:

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Where S_p² is pooled variance

S_p² = [(n1 – 1)*S1^2 + (n2 – 1)*S2^2]/(n1 + n2 – 2)

S_p² = 0.0105

Standard error = sqrt[S_p²*((1/n1)+(1/n2))] = 0.0476

Confidence level = 99%

df = 17

Critical value = 2.8982

(by using t-table)

(X1bar – X2bar) = 0.23 – 0.32 = -0.09

Confidence interval = (X1bar – X2bar) ± t*sqrt[S_p²*((1/n1)+(1/n2))]

Confidence interval = -0.09 ± 2.8982*0.0476

Confidence interval = -0.09 ± 0.137954

Lower limit = -0.09 - 0.137954 = -0.2279

Upper limit = -0.09 + 0.137954 = 0.0479

Confidence interval = (-0.2279, 0.0479)

We are 99% confident that the population mean difference between methane expulsion for two cities will lies within -0.2279 and 0.0479.