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joja [24]
2 years ago
13

Exercise 6.22 provides data on sleep deprivation rates of Californians and Oregonians. The proportion of California residents wh

o 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?
Mathematics
1 answer:
kherson [118]2 years ago
5 0

Answer:

a. The alternative hypothesis H₀: p'₁ ≠ p'₂ is accepted

b. Type I error

Step-by-step explanation:

Proportion of California residents who reported insufficient rest = 8.0%

Proportion of Oregon  residents who reported insufficient rest = 8.8%

p'₁ = 0.08 * 11545 =923.6

p'₂ = 0.088 * 4691=412.81

σ₁ = \sqrt{n*p_1*q_1}  = \sqrt{n*p_1*(1-p_1)} = \sqrt{11545*0.08*(1-0.08)} = 29.15

σ₂ = \sqrt{n*p_2*q_2}  = \sqrt{n*p_2*(1-p_2)}= \sqrt{4691*0.088*(1-0.088)} = 19.40

Samples size of California residents n₁ = 11,545

Samples size of Oregon residents n₂ = 4,691

Hypothesis can be constructed thus

Let our null hypothesis be H ₀: p'₁ = p'₂

and alternative hypothesis H ₐ: p'₁ ≠ p'₂

Then we have  

z =\frac{(p'_1 -p'_2)-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2} } }

The test statistics can be computed by

     

t₀ = \sqrt{\frac{n_1n_2(n_1+n_2-2)}{n_1+n_2} } *\frac{p_1'-p_2'}{\sqrt{(n_1-1)\sigma_1^2+(n_2-1)\sigma_2^2} } =      1104.83

c from tables is   P(T ≤ c) = 1 - α where α = 5% and c = 1.65

since t₀ ≥ c then then the hypothesis is rejected which means the alternative hypothesis is rejected

b. Type I error, rejecting a true hypothesis

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