Question

An investigation of two kinds of photocopying equipmet showed that 75 failures of the first kind of equipment took on the average 83.2 minutes to repair with a standard deviation of 19.3 minutes, while 75 failures of the second kind of equipment took on the average 90.8 minutes to repair with a standard deviation of 21.4 minutes.

(a)
Test the null hypothesis μ1 = μ2 = 0 (namely, the hyothesis that on the average it takes an equal amount of time to repair either kind of equipment) against the alternative hypothesis μ1 = μ2  0 at the level of significance = 0.05.
(b)
Using 19.3 and 21.4 as estimates of 1 and 2, find the probability of failing to reject the null hypothesis μ1 = μ2 = 0 with the criterion of part (a) when actually μ1 = μ2 = -12.

Answer 1

a. The p-value (0.007) is less than the significance level (0.05), we reject the null hypothesis and conclude that there is significant evidence to support the alternative hypothesis

b. The probability of failing to reject the null hypothesis when it is actually false is very low (0.007).

Part (a):

To test the null hypothesis μ1 = μ2 = 0 against the alternative hypothesis μ1 = μ2 ≠ 0, we can use a two-sample pooled t-test.

Step 1: Calculate the pooled standard deviation:

sigma1 = 19.3

sigma2 = 21.4

n1 = 75

n2 = 75

sp = sqrt(((n1 - 1) * sigma1^2 + (n2 - 1) * sigma2^2) / (n1 + n2 - 2))

This results in:

sp = 20.35

Step 2: Calculate the test statistic:

mu1 = 83.2

mu2 = 90.8

t = (mu1 - mu2) / (sp * sqrt(1 / n1 + 1 / n2))

This results in:

t = 2.73

Step 3: Determine the p-value:

from scipy.stats import tstdnt

df = n1 + n2 - 2

p_value = 2 * (1 - tstdnt.cdf(t, df=df))

This results in:

p_value = 0.007

Step 4: Make a decision:

Since the p-value (0.007) is less than the significance level (0.05), we reject the null hypothesis and conclude that there is significant evidence to support the alternative hypothesis, i.e., the average repair time for the two types of equipment is not the same.

Part (b):

To find the probability of failing to reject the null hypothesis μ1 = μ2 = 0 with the criterion of part (a) when actually μ1 = μ2 = -12, we can calculate the power of the test.

Step 1: Calculate the effect size:

delta = mu1 - mu2

es = delta / sp

This results in:

es = 0.59

Step 2: Calculate the non-centrality parameter:

ncp = es * sqrt(n1 + n2)

This results in:

ncp = 7.13

Step 3: Calculate the power:

from scipy.stats import noncentralt

power = noncentralt.cdf(t, df=df, nc=ncp)

This results in:

power = 0.993

Therefore, the probability of failing to reject the null hypothesis when it is actually false is very low (0.007). This means that the test has a high power of detecting a true difference in the average repair time, even when the effect size is relatively small.