Significance Tests for Comparing Proportions

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Significance Tests for Comparing Proportions

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Question 1

106

Question 1.

If H_o: p₁ = p₂, what is the hypothesized value of p₁ - p₂?

One

Five

Zero

One hundred

Incorrect. If p_{1 =} p₂, then p₁ - p₂= 0.

Correct. If p_{1 =} p₂, then p₁ - p₂= 0.

Incorrect. Try again.

Question 2

208

Question 2.

Why do we use $\widehat{p}$ as our estimates of both p₁ and p₂?

Because it is the average of

{\hat{p}}_{1}

$\widehat{p}_{1}$ and

{\hat{p}}_{1}

$\widehat{p}_{1}$

Because it is easier than using

{\hat{p}}_{1}

$\widehat{p}_{1}$ and

{\hat{p}}_{2}

$\widehat{p}_{2}$ to estimate p₁ and p₂, respectively.

Because we assume H_o: p₁ = p₂ is true

Incorrect. Because we assume H_o: p₁ = p₂ is true, we can use the marginal,

\hat{p}

$\widehat{p}$ , as our estimates of both p₁ and p₂.

Correct. Because we assume H_o: p₁ = p₂ is true, we can use the marginal,

\hat{p}

$\widehat{p}$ , as our estimates of both p₁ and p₂.

Incorrect. Try again.

Question 3

234

Question 3.

For the data in this two way table, what is the value of $\widehat{p}$ , the marginal proportion of success?

0.35

0.45

0.30

0.40

Incorrect.

\frac{40 + 30}{100 + 100}

$\frac{40+30}{100+100}$ =

\frac{70}{200}

$\frac{70}{200}$ = 0.35

Correct.

\frac{40 + 30}{100 + 100}

$\frac{40+30}{100+100}$ =

\frac{70}{200}

$\frac{70}{200}$ = 0.35

Incorrect. Try again.

Question 4

313

Question 4.

In addition to appropriately collected data, what condition is necessary for using the following formula to find P-value using the standard Normal table?

z = $\frac{(\widehat{p}_{1} - \widehat{p}_{2})-0}{ \sqrt{\widehat{p}(1-\widehat{p})( \frac{1}{n_{1}}+ \frac{1}{n_{2} } )} }$

n₁

\geq

$\geq$ 30 and n₂

\geq

$\geq$ 30

(n₁ + n₂)p₁

\geq

$\geq$ 10 and (n₁ + n₂)p₂

\geq

$\geq$ 10

n_{1} \hat{p} \geq 5

$n_{1}\widehat{p} \geq 5$ ,

n_{1} (1 - \hat{p}) \geq 5

$n_{1}(1 - \widehat{p}) \geq 5$ ,

n_{2} \hat{p} \geq 5

$n_{2}\widehat{p} \geq 5$ , and

n_{2} (1 - \hat{p}) \geq 5

$n_{2}(1 - \widehat{p}) \geq 5$

Incorrect.

n_{1} \hat{p} \geq 5

$n_{1}\widehat{p} \geq 5$ ,

n_{1} (1 - \hat{p}) \geq 5

$n_{1}(1 - \widehat{p}) \geq 5$ ,

n_{2} \hat{p} \geq 5

$n_{2}\widehat{p} \geq 5$ , and

n_{2} (1 - \hat{p}) \geq 5

$n_{2}(1 - \widehat{p}) \geq 5$ is the other necessary condition. These requirements for sample size must be met in order to use the z-score to find P-value using the standard Normal table.

Correct.

n_{1} \hat{p} \geq 5

$n_{1}\widehat{p} \geq 5$ ,

n_{1} (1 - \hat{p}) \geq 5

$n_{1}(1 - \widehat{p}) \geq 5$ ,

n_{2} \hat{p} \geq 5

$n_{2}\widehat{p} \geq 5$ , and

n_{2} (1 - \hat{p}) \geq 5

$n_{2}(1 - \widehat{p}) \geq 5$ is the other necessary condition. These requirements for sample size must be met in order to use the z-score to find P-value using the standard Normal table.

Incorrect. Try again.

Question 5

360

Question 5.

What is the appropriate alternative hypothesis for testing whether aspirin reduces the risk of a heart attack? Use p₁ = the heart attack rate for doctors taking aspirin and p₂ = the heart attack rate for doctors taking a placebo.

H_a: p₁ $\neq$ p₂

H_a: p₁ > p₂

H_a: p₁ = p₂

H_a: p₁ < p₂

Incorrect. We want to test whether aspirin reduces the heart attack rate. So, we want the alternative hypothesis to say the heart attack rate for those taking aspirin is less than the heart attack rate of those taking a placebo.

Correct. We want to test whether aspirin reduces the heart attack rate. So, we want the alternative hypothesis to say the heart attack rate for those taking aspirin is less than the heart attack rate of those taking a placebo.

Incorrect. Try again.

Question 6

425

Question 6.

For these data, what is the value of $\widehat{p}_{1}$ ?

\frac{189}{22071}

$\frac{189}{22071}$

\frac{104}{22071}

$\frac{104}{22071}$

\frac{104}{293}

$\frac{104}{293}$

\frac{104}{11037}

$\frac{104}{11037}$

Incorrect.

{\hat{p}}_{1}

$\widehat{p}_{1}$ =

\frac{c o u n t i n a s p i r i n g r o u p w h o h a d h e a r t a t t a c k}{c o u n t o f d o c t o r s i n a s p i r i n g r o u p}

$\frac{count\ in\ aspirin\ group\ who\ had\ heart\ attack}{count\ of\ doctors\ in\ aspirin\ group}$ =

\frac{104}{11037}

$\frac{104}{11037}$

Correct.

{\hat{p}}_{1}

$\widehat{p}_{1}$ =

\frac{c o u n t i n a s p i r i n g r o u p w h o h a d h e a r t a t t a c k}{c o u n t o f d o c t o r s i n a s p i r i n g r o u p}

$\frac{count\ in\ aspirin\ group\ who\ had\ heart\ attack}{count\ of\ doctors\ in\ aspirin\ group}$ =

\frac{104}{11037}

$\frac{104}{11037}$

Incorrect. Try again.

Question 7

532

Question 7.

Why is data collection appropriate for this two-sample proportion procedure?

Because the doctors were randomly allocated to the two treatments (aspirin and placebo)

Because two separate simple random samples were takenóone from each population

Because the order of applying the two treatments to each doctor was randomized in this matched pairs design

Incorrect. This study is a completely randomized experiment. The doctors were randomly assigned to the two treatment groups.

Correct. This study is a completely randomized experiment. The doctors were randomly assigned to the two treatment groups.

Incorrect. Try again.

Question 8

577

Question 8.

For these data, what is the heart attack rate for all doctors in the study?

0.009

0.013

0.017

Incorrect.

\hat{p}

$\widehat{p}$ is the heart attack rate for all doctors in the study and equals 0.013.

Correct.

\hat{p}

$\widehat{p}$ is the heart attack rate for all doctors in the study and equals 0.013.

Incorrect. Try again.

Question 9

580

Question 9.

Knowing that 99.7% of all z-scores are between -3 and +3, what is the P-value for z = -5.14?

0.2981

0.000

0.0514

Incorrect. A z-score of -5.14 is off the standard Normal table; the area of a standard Normal curve less than z = –5.14 is essentially zero.

Correct. A z-score of -5.14 is off the standard Normal table; the area of a standard Normal curve less than z = –5.14 is essentially zero.

Incorrect. Try again.

Question 10

592

Question 10.

With P-value = 0.000 and $\alpha$ = 0.05, should we reject H₀?

Yes

Incorrect. P-value = 0.000 is less than

α

$\alpha$ = 0.05, so we reject H₀.

Correct. P-value = 0.000 is less than

α

$\alpha$ = 0.05, so we reject H₀.

Incorrect. Try again.

Question 11

647

Question 11.

Can we conclude that taking aspiring causes a reduction in the heart attack rate?

Yes

Incorrect. Since this was a valid experiment and the results are significant, we can conclude causation.

Correct. Since this was a valid experiment and the results are significant, we can conclude causation.

Incorrect. Try again.

StatTutor Lesson - Significance Tests for Comparing Proportions

Question 1

Question 1.

Question 2

Question 2.

Question 3

Question 3.

Question 4

Question 4.

Question 5

Question 5.

Question 6

Question 6.

Question 7

Question 7.

Question 8

Question 8.

Question 9

Question 9.

Question 10

Question 10.

Question 11

Question 11.