All language subtitles for 008 Test for the Mean. Population Variance Unknown_en

-: Alright, great.

Now that we know what the P-value is and how to use it,

we will get back to hypothesis testing.

We saw only one of two possible cases. Remember?

We haven't covered the more commonly observed case

when the population variance is unknown.

Alright. Imagine you are the marketing analyst of a company

and that you've been asked to estimate

if the email open rate of one of the firm's competitors

is above your company's.

Your company has an open rate of 40%.

An email open rate is a measure

of how many people on the email list actually open

the emails they have received.

At first, you struggle to figure out

how to get such specific information

about a competitor company,

but then you see that an employee of that competitor company

posted a selfie on Facebook saying "LOL.

The email management software we are using drives me nuts."

In the background, you can see her screen

and it shows clearly the summaries

of the last 10 email campaigns that were sent

and their corresponding open rates. Bingo!

With your statistical skills, that's all you need.

A little help from Facebook.

Let's state the hypotheses.

Null hypothesis mean open rate is lower or equal to 40%.

Alternative hypothesis mean open rate is higher than 40%.

Note that in hypothesis testing,

we are aiming to reject the null hypothesis.

When we wanna test if the open rate is higher than 40%,

the null hypothesis actually states the opposite statement.

Also, pay attention that this time,

we are dealing with a one-sided test.

Alright. Your boss told you

that 0.05 is an adequate significance level for this test,

so that's what you'll use.

Here's the data set.

You calculate the sample mean and get 37.7%.

The sample standard deviation is 13.74%,

thus the standard error is 4.34%.

You assume that the population of open rates of sent emails

is normally distributed.

Like confidence intervals with variance unknown

in a small sample,

the correct statistic to use is the t-statistic.

Remember, you do not know the variance

and the sample is not big enough.

This means that the variable follows

the student's T distribution

and you must employ the t-statistic.

Let's calculate it then.

We calculate the T score the same way as the Z score.

The T score is equal to the sample mean,

minus the hypothesized mean value,

divided by the standard error.

The result that we get is -0.53.

-: As we said earlier,

it is easier to work with positive numbers.

So, we should compare the absolute value of -0.53

with the appropriate T with N minus one degrees of freedom

at 0.05 one-sided significance.

We quickly navigate through the table

and get 1.83 at the 5% significance critical value.

Okay, 0.53 is lower than 1.83.

Remember the Decision rule?

If the absolute value of the T score

is lower than the statistic from the table,

we cannot reject the null hypothesis.

Therefore, we must accept it.

What you do next is you go and tell your boss

that at this level of significance, statistically,

we cannot say that the email open rate of our competitors

is higher than 40%.

Okay. What about the second measurement we saw?

What was that?

Ah, yes.

The P-value.

The P-value of this statistic is 0.304.

As the P-value is greater than the significance level

of 0.05, we come to the same conclusion.

We cannot reject the null hypothesis.

Let's do a quick check.

If the significance level was 0.01,

the P-value would still be higher

and we wouldn't reject the null hypothesis.

This is an important observation

that we haven't noted before.

If we cannot reject a test at 0.05 significance,

we could not reject it at smaller levels either.

Alright. That's all for now.

Make sure you learn the material by doing the exercises

after this lesson.

Thanks for watching.