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.

Step-by-step explanation:

Hello!

You need to test if the true mean of the voltage of Estonian networks is 232.

You are given a sample of n= 66 industrial networks, with sample mean X[bar]= 231.7 V and standard deviation S= 2.19 V.

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

= -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!

n

step-by-step explanation:

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