6.24 Sleep deprivation, CA vs. OR, Part II. Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents.(a)  Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. (Reminder: Check conditions)(b)  It is possible the conclusion of the test in part (a) is incorrect. If this is the case, what type of error was made?

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Asked Dec 4, 2019
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6.24 Sleep deprivation, CA vs. OR, Part II. Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents.

  1. (a)  Conduct a hypothesis test to determine if these data provide strong evidence the rate of sleep deprivation is different for the two states. (Reminder: Check conditions)

  2. (b)  It is possible the conclusion of the test in part (a) is incorrect. If this is the case, what type of error was made?

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Expert Answer

Step 1

a). From the provided information, the proportion (1) of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 0.08 and the proportion (2) of Oregonians residents is 0.088.

The sample size of California residents (n1) = 11545

The sample size of Oregonians residents (n2) = 4691

The hypotheses can be constructed as:

Null hypothesis, H0:1 = 2 (There is no significance difference between both the proportions.)

Alternative hypothesis, Ha: 12 (There is a significance difference between both the proportions.)

Step 2

The test statistic can be obtained as:

P (1- 2,) P:(1- Þ, )
P:(1– p.)
п,
п,
0.08 – 0.088
0.08(1–0.08) 0.088(1–0.088)
11545
4691
=-1.65
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P (1- 2,) P:(1- Þ, ) P:(1– p.) п, п, 0.08 – 0.088 0.08(1–0.08) 0.088(1–0.088) 11545 4691 =-1.65

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Step 3

The level of significance is not provided so it can be assume as 0.05. The P value at test statistic -1.65 with significance level 0.05 from standard normal table is 0.099 which...

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