Hypothesis Test for the Mean


This entry is part 3 of 4 in the series Statistics Hypothesis

57. Let’s do a hypothesis test on a single population. Either we either know the population variance or we don’t know the population variance. Suppose we know the variance. We have a sample dataset in the form of a column of numbers. Suppose they are salaries of data analysts. They could be just about any set of numbers.

  • single population
  • known population variance
  • Sample dataset of size n

We can easily calculate the sample mean because we have the data and we know how many observations we have.

What is our hypothesis? A hypothesis is a statement. Suppose someone is asserting that the average data analyst salary is $70,000 per year. Is that reasonable? We needed a two-sided test, as we are interested in knowing both if the salary is significantly less than that or significantly more than that. The null hypothesis is: the population mean salary is $70,000. We denote it as mu zero equals $70,000. The alternative hypothesis is that the population mean salary is different than $70,000.

H_0 : \mu_0 = 70000
H_1 : \mu_0 \neq 70000

Testing is done by standardizing the variable at hand and comparing it to the lowercase z, which follows a standard normal distribution. We standardize a variable by subtracting the mean and dividing by the standard deviation.

the formula for standardization becomes: Capital Z is equal to the sample mean, minus the value of interest from the null hypothesis, divided by the standard error. In this way, we obtain a distribution with a mean of zero and a standard deviation of one.

Z = \left( \dfrac{\bar{x} - \mu_0}{\dfrac{\sigma}{\sqrt{n}}} \right) \hspace{0.3cm} \texttildelow  \hspace{0.3cm} N(0,1)

Critical Value

What is the critical value? It is the lowercase z from the table.

Z Score

This uppercase Z should not be mistaken with lowercase z. The uppercase Z is the standardized variable associated with the test and will be called the z-score from now on.

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