Question

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?

Answer 1

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