(a) The proportions of the adults who took the 1946 survey and the recent survey who...

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In 1946, an organization surveyed 1100 adults and asked, Are you a total abstainer from, or do you on occasion consume, alco

(a) The proportions of the adults who took the 1946 survey and the recent survey who were total abstainers are ______ and ______, respectively. (Round to three decimal places as needed.)

(b) Has the proportion of adults who totally abstain from alcohol changed? Use the α = 0.01 level of significance. Identify the t-value, P-value, average of differentiated data, and standard deviation of differentiated data. (Round to six decimal places as needed.)

(c) Has the proportion of adults who totally abstain from alcohol changed? Use the α = 0.05 level of significance. Identify the t-value, P-value, average of differentiated data, and standard deviation of differentiated data. (Round to six decimal places as needed.)

(d) Has the proportion of adults who totally abstain from alcohol changed? Use the α = 0.10 level of significance. Identify the t-value, P-value, average of differentiated data, and standard deviation of differentiated data. (Round to six decimal places as needed.)

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Answer 1

Answer #1

(a)

The proportions of the adults who took the 1946 survey and the recent survey who were total abstainers are 363/1100 = 0.330 and 286/1100 = 0.260 respectively.

(b)

(i)

H0: Null Hypothesis: p1 =p2 ( the proportion of adults who totally abstain from alcohol has not changed )

HA:Alternative Hypothesis: p1 $\neq$ p2 ( the proportion of adults who totally abstain from alcohol has changed ) (claim)

n1 = 1100

$\hat{p}$ 1 = 363/1100 = 0.330

n2 = 1100

$\hat{p}$ 2 =286/1100= 0.260

Pooled Proportion is given by:

$\bar{p}=\frac{363+286}{1100+1100}=0.295000$

$SE=\sqrt{0.295000\times (1-0.295000)\times (1/1100+1/1100)}=\sqrt{0.000378}=0.019442$

Test Statistic is given by:

$Z=\frac{0.330-0.260}{0.019442}=3.600411$

(ii)

By Technology,

p - value = 0.000318

(iii)

average of differentiated data is given by:

$\bar{p}=\frac{363+286}{1100+1100}=0.295000$

(iv)

standard deviation of differentiated data is given by:

$SE=\sqrt{0.295000\times (1-0.295000)\times (1/1100+1/1100)}=\sqrt{0.000378}=0.019442$

(v)

Since p - value = 0.000318 is less than $\alpha$ = 0.01, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that the proportion of adults who totally abstain from alcohol has changed.

(c)

(i)

H0: Null Hypothesis: p1 =p2 ( the proportion of adults who totally abstain from alcohol has not changed )

HA:Alternative Hypothesis: p1 $\neq$ p2 ( the proportion of adults who totally abstain from alcohol has changed ) (claim)

n1 = 1100

$\hat{p}$ 1 = 363/1100 = 0.330

n2 = 1100

$\hat{p}$ 2 =286/1100= 0.260

Pooled Proportion is given by:

$\bar{p}=\frac{363+286}{1100+1100}=0.295000$

$SE=\sqrt{0.295000\times (1-0.295000)\times (1/1100+1/1100)}=\sqrt{0.000378}=0.019442$

Test Statistic is given by:

$Z=\frac{0.330-0.260}{0.019442}=3.600411$

(ii)

By Technology,

p - value = 0.000318

(iii)

average of differentiated data is given by:

$\bar{p}=\frac{363+286}{1100+1100}=0.295000$

(iv)

standard deviation of differentiated data is given by:

$SE=\sqrt{0.295000\times (1-0.295000)\times (1/1100+1/1100)}=\sqrt{0.000378}=0.019442$

(v)

Since p - value = 0.000318 is less than $\alpha$ = 0.05, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that the proportion of adults who totally abstain from alcohol has changed.

(d)

(i)

H0: Null Hypothesis: p1 =p2 ( the proportion of adults who totally abstain from alcohol has not changed )

HA:Alternative Hypothesis: p1 $\neq$ p2 ( the proportion of adults who totally abstain from alcohol has changed ) (claim)

n1 = 1100

$\hat{p}$ 1 = 363/1100 = 0.330

n2 = 1100

$\hat{p}$ 2 =286/1100= 0.260

Pooled Proportion is given by:

$\bar{p}=\frac{363+286}{1100+1100}=0.295000$

$SE=\sqrt{0.295000\times (1-0.295000)\times (1/1100+1/1100)}=\sqrt{0.000378}=0.019442$

Test Statistic is given by:

$Z=\frac{0.330-0.260}{0.019442}=3.600411$

(ii)

By Technology,

p - value = 0.000318

(iii)

average of differentiated data is given by:

$\bar{p}=\frac{363+286}{1100+1100}=0.295000$

(iv)

standard deviation of differentiated data is given by:

$SE=\sqrt{0.295000\times (1-0.295000)\times (1/1100+1/1100)}=\sqrt{0.000378}=0.019442$

(v)

Since p - value = 0.000318 is less than $\alpha$ = 0.10, the difference is significant. Reject null hypothesis.

Conclusion:
The data support the claim that the proportion of adults who totally abstain from alcohol has changed.

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