Welcome to an example
of a one sample hypothesis test for a proportion.
Joon believes that 50% of first-time brides
in the United States are younger than their grooms.
She performs a hypothesis test to determine
if the percentage is the same or different from 50%.
Joon samples 100 first-time brides and 53 reply
that they are younger than their grooms.
For the hypothesis test
she uses a 10% level of significance.
Let's begin by listing all of the given information.
First, the 10% level of significance indicates that alpha
is equal to 0.1 is a decimal.
Next, Joon samples 100 first time brides
and therefore the sample size n is 100,
53 reply that they are younger than their grooms
and therefore x, the number of successes is 53
which indicates p prime, the sample proportion
is equal to 0.53.
Because she is testing to determine if the percentage
is the same or different from 50%,
the alternative hypothesis is p doesn't equal 0.50
and therefore the null hypothesis is p equals 0.50.
Notice because of the not equals here,
we will have a two-tailed test.
And then finally notice how the sample proportion
is 0.03 units from the population proportion
which we can obtain by subtracting 0.50 from 0.53.
Notice there is no mention of a mean,
the information is given in percentages
which is why we know the problem involves a proportion.
Because we have a proportion, we know the test statistic
is a z-score.
Now looking at the graph here on the left,
notice how the population proportion is right in the middle,
the sample proportion of 0.53 is to the right
because we have a two tail test though
to find the value on the left,
we have to subtract 0.03 from 0.50
which does give us a 0.47 on the left.
And therefore the combined area on the left and right
is equal to the p-value and each area
is equal to half of the p-value.
This also indicates that the p-value is equal
to the probability that p prime
is less than equal to 0.47 on the left
or p prime is greater than or equal to 0.53 on the right.
And now let's go to the calculator determine the z-score
and the p-value.
And because we have a proportion,
we will be performing a 1-PropZTest.
To do this press Stat, right arrow to tests,
select option five for 1PropZTest.
I've already entered the key information,
the population proportion is 0.5 enter,
x equals 53 enter and equals 100 enter
the null hypothesis is p doesn't equal 0.50,
we select doesn't equal arrow down
to calculate, press Enter.
We have a z-score of 0.6 and a p-value
of approximately 0.5485.
Let's go ahead and record this
and now let's go back to the calculator.
Let's run the test again and select draw.
So stat, right arrow to test, option five,
everything stays the same, except now we go down to draw,
press Enter.
Notice how this gives us a graph that resembles
our graph here on the left and below we do have the z-score
as well as the p-value.
Before we interpret the p-value and then compare to alpha,
let's take a look at the z-score of 0.6.
For a two tailed test when alpha is equal to 0.1,
notice how the z-scores are 1.645
on the right and negative 1.645 on the left.
We'll notice how the z-score of 0.6
is done in the rejection region which indicates
we are not going to reject the null hypothesis
but before we formally conclude this,
let's interpret the p-value and then compare to alpha.
The interpretation of the p-value is
if the null hypothesis is true
then there is a 0.5485 tolerability or 54.85% chance
that the sample or estimated proportion is 0.53 or more
or 0.47 or less, meaning this is not a rare event.
So if we compare alpha to the p-value,
notice how the p-value is high compared to alpha.
Looking at our notes below, if the p-value is high,
the null must fly or more formerly because the p-value
is greater than alpha or because alpha
is less than the p-value,
we do not reject the null hypothesis.
So again, in conclusion, at a 10% level of significance,
the sample data do not show sufficient evidence
that the percentage of first-time brides
who are younger than their grooms is different from 50%.
Again, this is because we are not rejecting
the null hypothesis which is p equals 0.50.
I hope you found this helpful.