Question

Sleep deprivation, CA vs. OR. For a recent report on sleep deprivation, the Centers for Disease...

Sleep deprivation, CA vs. OR. For a recent report on sleep deprivation, the Centers for Disease Control and Prevention interviewed 11536 residents of California and 4556 residents of Oregon. In California, 923 respondents reported getting insufficient rest or sleep during each of the preceding 30 days, while 396 of the respondents from Oregon reported the same. Round each calculation to 4 decimal places.

1. Using California as population 1 and Oregon as population 2, what are the correct hypotheses for conducting a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states?

A. ?0:?1−?2=0H0:p1−p2=0, ??:?1−?2>0HA:p1−p2>0
B. ?0:?1−?2=0H0:p1−p2=0, ??:?1−?2<0HA:p1−p2<0
C. ?0:?1−?2=0H0:p1−p2=0, ??:?1−?2≠0HA:p1−p2≠0

2. Calculate the pooled estimate of the proportion for this test. ?̂ p^ =

3. Calculate the standard error. SE =

4. Calculate the test statistic for this hypothesis test.  ? z t X^2 F  =

5. Calculate the p-value for this hypothesis test. p-value =

6. Based on the p-value, we have:
A. some evidence
B. extremely strong evidence
C. strong evidence
D. little evidence
E. very strong evidence
that the null model is not a good fit for our observed data.

Homework Answers

Answer #1

Given:

California

n1=11536, X1=923

Oregon

n2=4556, X2= 396

Calculation:

For California:

For Oregon:

Hypothesis:

ANSWER: C

C. H0:p1−p2=0, HA:p1−p2 ≠ 0

2) Pooled estimate of Proportion:

Q = 1-P = 1-0.08 = 0.92

3)

Standard Error (SE):

4)

Test statistic:

5)

P-value = 0.1502 ......................Using standard Normal table

6)

P-value > , 0.1502 > 0.05, That is Fail to Reject Ho at 5% level of significance.

ANSWER: C. strong evidence

that the null model is not a good fit for our observed data.

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