The article "Supply Voltage Quality in Low-Voltage Industrial Networks of Estonia" (T. Vinnal, K. Janson, et al., Estonian Journal of Engineering, 2012:102126) presents voltage measurements for a sample of 66 industrial networks in Estonia. Assume the rated voltage for these networks is 232 V. The sample mean voltage was 231.7 V with a standard deviation of 2.19 V. Let μ represent the population mean voltage for these networks.

a. Find the P-value for testing H0: μ = 232 versus H1: μ = 232.
b. Either the mean voltage is not equal to 232, or the sample is in the most extreme% of its distribution.

With the study variable X: "voltage of an Estonian industrial network" of unknown distribution, but a large enough sample, I'll apply the Central Limit Theorem and approximate the distribution of the sample mean to normal. Remember: Since the population mean is a paramenter of the normal distribution, you need to work under it.

H₀: μ = 232

H₁: μ ≠ 232

α: 0.05 (you were given no significance level, that's why I choose to use the most common one for the test)

Z= X[bar] - μ ≈ N (0;1)

S/√n

Z= X[bar] - μ = 231.7 - 232 = -1.113

S/√n 2.19/√66

[tex]Z_{H0}[/tex]= -1.113

This test is two tailed, and so is the p-value, in other words, you have to take into account that the p-value is divided in the two tails.

Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

P(Z<-1.113) + P(Z>1.113) = P(Z<-1.113) + (1 - P(Z< 1.113)) = 0.1335 + (1 - 0.8665) = 0.267

p-value: 0.267

Comparing it to α: 0.05, the decision is to not reject the null hypothesis.

I hope it helps!