All language subtitles for 005 Test for the Mean. Population Variance Known_en

-: All right.

Now that we've covered the necessary theory,

it is time for some testing.

We're going to explore two types of tests

drawn from a single population

and drawn from multiple populations.

This is very similar to confidence intervals

for a single population,

and confidence intervals for two populations

that we covered previously.

In the next few videos, we will run tests for a single mean

with both known variance and unknown variance.

Let's start with a test

in which the variance is known, shall we?

For this test,

we will use our good old data scientist salary example.

Here's the data set one more time.

By now, I hope you are able to calculate the sample mean,

it is $100,200.

The population variance is known

and its standard deviation is equal to $15,000.

Moreover, the sample size is 30.

However, you saw that according to Glassdoor,

the popular salary information website,

the mean data scientist salary is $113,000.

The sample that is available on Glassdoor

is based on self-reported numbers,

and you would like to see if its value is correct.

We needed a two-sided test,

as we are interested in knowing

both that the salary is significantly less than that

or significantly more than that.

The null hypothesis is, the population means salary

is $113,000.

We denoted as mu zero equals $113,000.

The alternative hypothesis

is that the population mean salary

is different than $113,000.

All right.

Formula time, almost.

Testing is done by standardizing the variable at hand

and comparing it to the z,

which follows a standard normal distribution.

Remember standardization?

We learned about it in the previous section.

Back then, I told you it was very important

and you will now see why.

For those that don't remember,

I suggest watching the video

on standardization once again.

For the others, I will quickly go through it.

We standardize a variable by subtracting the mean

and dividing by the standard deviation.

Since it is a sample, we use the standard error.

Thus, the formula for standardization becomes

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 0 and a standard deviation of 1.

This Z should not be mistaken with z.

The Z is the standardized variable

associated with the test

and will be called the Z-score from now on.

The z is the one from the table

that we've talked about before,

and henceforth will be referred to as the critical value.

All right, how does testing work?

Think about this.

The z is normally distributed

with a mean and standard deviation of 1.

The Z is normally distributed

with a mean of X bar minus mu zero

and a standard deviation of 1.

Standardization lets us compare the means.

The closer the difference of X bar and mu zero to zero,

the closer Z-score itself to zero.

This implies a higher chance to accept the null hypothesis.

Let's go back to the example.

So what is the value of our standardized variable?

We plug in the numbers that we have

from the beginning of the lesson.

What we get is a Z-score of minus 4.67.

Now, we will compare the absolute value

of minus 4.67 with a z of alpha divided by 2,

where alpha is the significance level.

Note that we use the absolute value

as it is much easier to always compare positive Z's

with positive z's.

Moreover, some z tables don't include negative values.

You should be aware that the two statements,

minus 4.67 is lower than the negative critical value,

is the same as 4.67 is higher

than the positive critical value.

Thus, our decision rule becomes,

absolute value of the Z-score

should be higher than the absolute value

of the critical value.

Using 5% significance,

our alpha is 0.05.

Since it is a two-sided test,

we check the table for z of 0.025.

A corresponding value is 1.96.

The last thing we need to do is compare

our standardized variable to the critical value.

If the Z-score is higher than 1.96,

we would reject the null hypothesis.

If it is lower, we will accept it.

4.67 is higher than 1.96,

therefore, we reject the null hypothesis.

The answer is that at the 5% significance level,

we have rejected the null hypothesis,

or at 5% significance,

there is no statistical evidence

that the mean salary is $113,000.

There are many other ways to express this

and you'll probably hear more about this

later on in the course.

What if we had a different significance level?

Using 1% significance,

we have an alpha of 0.01.

So z of alpha divided by 2 is 2.58.

Once again, our Z-score of 4.67 is higher than 2.58,

so we would reject the null hypothesis

even at the 1% significance.

But how much further can we go

before we could not reject the null hypothesis anymore?

0.5%?

0.1%?

There is a special technique that allows us

to see what the significance level is,

after which we will be unable to reject the null hypothesis.

We will see it in our next video.

Stay tuned.