Statistic 6

This week will continue to discuss hypothesis testing and confidence interval, but

now we will discuss 2 samples.

Just like with 1 – sample hypothesis testing there are 4 steps we will follow. To

review those 4 steps please review the Week 6 Hypothesis Testing PDF.

But the conclusion will still be the same.

If the p-value is < alpha, you Reject Ho and state this test is significant.

If the p-value is > alpha, you Do Not Reject Ho and state this test is not significant.

In this document we will discuss 2 – sample proportion hypothesis testing and

confidence intervals.

There are still 3 different hypothesis scenarios with a 2 – sample proportion

hypothesis test.

Lower Tail Test (1 tail):

Ho: 𝑝1 − 𝑝2 = 0

Ha: 𝑝1 − 𝑝2 < 0

Upper Tailed Test (1 tail):

Ho: 𝑝1 − 𝑝2 = 0

Ha: 𝑝1 − 𝑝2 > 0

Two Tailed Test:

Ho: 𝑝1 − 𝑝2 = 0

Ha: 𝑝1 − 𝑝2 ≠ 0

The hypothesized value is 0 and the same key words apply from a 1 – sample

hypothesis test to determine which scenario to use.

The Z – Test Statistic = 𝑝1− 𝑝2

√𝑝∗𝑞( 1

𝑛1 +

1

𝑛2 )

Where 𝑝 = 𝑥1+𝑥2

𝑛1+𝑛2

𝑞 = 1 − 𝑝

We will then use =NORM.S.DIST function in Excel to find the p-value. This Excel

function should look familiar from Week 4.

Example:

In a developing section of a district 50 people were surveyed and 38 were in favor

of the new proposal. For the rest of the district 100 people were surveyed and

only 65 people were in favor of the new proposal. Is there evidence that the

number of people favoring the new proposal is greater in the developing section

than the rest of the district? Use alpha = .05

First step is to state the hypothesis scenario. Because the key word says greater

this means it is an upper tailed test.

Ho: 𝑝1 − 𝑝2 = 0

Ha: 𝑝1 − 𝑝2 > 0

The first proportion is favoring the new proposal in the developing district and the

second proportion is favoring the new proposal in the rest of the district.

𝑝1 = 38

50 = .76

𝑞1 = 1 − .76 = .24

𝑝2 = 65

100 = .65

𝑞2 = 1 − .65 = .35

𝑝 = 38 + 65

50 + 100 = .68667

𝑞 = 1 − .68667 = .31333

Now that we have these values we can plug them in to find the Test Statistic.

Z – Test Statistic = .76−.65

√.68667∗.31333( 1

50 +

1

100 )

= 1.369

Now that we have the Z-Test Statistics we can use the =NORM.S.DIST function to

find the p-value.

And yes, we can have a negative Z- Test Statistic, if we do that is fine. You DO

NOT have to take the absolute value of anything. Use the Test Stat. as is in the

Excel function.

In Excel input =NORM.S.DIST(1.369,TRUE)

We will write out TRUE because this test is cumulative.

We see this p-value = .9145 BUT remember when we use this function in Excel,

this function is in the less than form. This means if we were running a Lower

Tailed test, this would be our p-value. BUT since we are running an Upper Tailed

Test we need to take 1 - .9145 to get the p-value for our test.

P-value = 1 - .9145 = .0855.

We see the p-value for our upper tailed test is .0855. If we compare this to .05,

we see that:

.0855 > .05. Since the p-value is greater than alpha, We Do Not Reject Ho. This

test is not significant and No, there is no evidence that the proportion of people

favoring the new proposal is greater in the developing section than the rest of the

district at alpha = .05.

What if we were running a two tailed test? To find this p-value we would take

whichever p-value is smaller and multiple it by 2.

.0855*2 = .171. The p-value for a two tailed test would be .171.

Now that we ran a hypothesis test, let calculate a confidence interval and draw

the same conclusion.

The equation for a 2 – sample proportion is:

𝑝1 − 𝑝2 ± 𝑍𝛼 2

∗ √

𝑝1𝑞1 𝑛1

+ 𝑝2𝑞2

𝑛2

Where Standard Error (SE) = √ 𝑝1𝑞1

𝑛1 +

𝑝2𝑞2

𝑛2

Margin of Error (ME) = 𝑍𝛼 2

∗ √

𝑝1𝑞1

𝑛1 +

𝑝2𝑞2

𝑛2

Plugging in what we know:

. 76 − .65 ± 𝑍𝛼 2

∗√ . 76 ∗ .24

50 +

. 65 ∗ .35

100

The last thing we need to find is the Z- Critical Value. We will use the

=NORM.S.INV function to find this. This function should look familiar from Week

4.

If we want to find a 95% confidence interval, then alpha = 1 - .95 = .05. But

because this is a confidence interval and we need to take into account the plus

AND minus on both sides if the bell-shaped curve we will divide alpha be 2. .05/2

= .025. Then we take 1 - .025 = .975. We will use this value in our Excel function.

=NORM.S.INV(.975)

We see the Z – Critical Value is 1.96. We will plug this into the equation and

solve.

. 76 − .65 ± 𝑍𝛼 2

∗√ . 76 ∗ .24

50 +

. 65 ∗ .35

100

. 76 − .65 ± 1.96√ . 76 ∗ .24

50 +

. 65 ∗ .35

100

. 76 − .65 ± 𝑍𝛼 2

∗√ . 76 ∗ .24

50 +

. 65 ∗ .35

100

. 76 − .65 ± 1.98(.076961)

. 11 ± .1508

(-.0408, .2608)

The confidence interval goes from -4.08% to 26.08%. This interval goes from a

negative value to a positive value. This means that 0 is in fact in this interval.

Because 0 is in the interval it is Not Significant, and we Do Not Reject Ho. This is

the same conclusion that we got with the hypothesis test.