Question

The mean daily evaporation for the 10 days on which Pan A was used was 19.10 mm, and the mean evaporation on the 10 days on which Pan B was used was 17.24mm. The variance estimated from the sample from Pan A is 7.72 mm2 , and the variance estimated from the sample from Pan B is 5.36 mm2 . Assuming that these two samples come from normal distribution with equal variances,

(a) Does the experimental evaporation pan, Pan A, give significantly higher evaporation rates than the standard pan, Pan B, at the 1% level of significance?

(b) What is the p-value for the test?

Answer 1

Solution:

Part a)

Here, we have to use pooled variance two sample t test for population means.

H₀: µ₁ = µ₂ versus H_a: µ₁ > µ₂

(One tailed / upper tailed test)

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 = 19.1, X2bar = 17.24, S1 = 2.778489, S2 = 2.315167, n1 = 10, n2 = 10

α = 0.01, df = n1 + n2 – 2 = 10+10-2 = 18

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

S_p² = [(10 – 1)* 2.778489^2 + (10 – 1)* 2.315167^2]/(10 + 10 – 2)

S_p² = 6.5400

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

t = (19.1 – 17.24) / sqrt[6.5400*((1/10)+(1/10))]

t = 1.6263

Critical t value = 2.5524

Test statistic value t is less than critical t value.

So, we do not reject the null hypothesis.

There is insufficient evidence to conclude that Pan A give significantly higher evaporation rates than the standard pan B.

Part b)

df = 18

t = 1.6263

P-value = 0.0606

(By using t-table or excel)