Hw7_skeleton

8.2

• 1. 0.44; p=220/500=.44
• 2. sample proportion, p-hat = x/n
• 3. False; while it is possible for the sample proportion to have the same value as the population proportion, it will not always have the same value.
• 4. True; the mean of the sampling distribution of p-hat is p regardless of whether np(1-p) ≥ 10 or not.
• 5. The sampling distribution of p-hat is approximately normal when n ≤ 0.05N and np(1-p) ≥ 10.
• 6. As the sample size increases the standard deviation of p-hat decreceases. If the sample size is increased by a factor of 4 then the standard deviation of p-hat is reduced to half.
• 7. 25,000(0.05)=1,250; the sample size n=500, is less than 5% of the population size and np(1-p)=500(0.4)(0.6) = 120 ≥ 10. The distribution of p-hat is approximately normal, with mean mu sub p-hat = p = 0.4 and standard deviation ≈ 0.022.
• 8. N = 25,000 and that n=300, p=.07. 300≤1250 and 300(0.4)(0.6) ≥ 10. So the mean of the sampling distribution p-hat is approximately normal. The mean of the sampling distribution of mu sub p-hat = p = 0. The standard deviation of the sampling distribution = 0.0264.
• 1. 1,000,000(0.05) = 50,000; the sample size n=1,000, is less than 5% of the population size and np(1-p)=1,000(0.35)(0.65)=227.5 ≥ 10. The distribution of p-hat is aprroximately normal, with mean mu sub p-hat = p = 0.35 and standard deviation ≈ 0.015

  2. p-hat = x/n = 390/1,000=0.39. P(X ≥ 390) = P(p-hat ≥ 0.39). P(Z ≥ 2.65) = 1-P(Z < 2.65). 1-0.9960 = 0.0040

  3. p-hat=x/n=320/1000=0.32 P(X ≤ 320) = P(p-hat ≤ 0.32) P(Z ≤ -1.99) = 0.0233. About 2 out of 100 random samples of size n=1000 will result in 320 or fewer individuals (32% or less) with the characteristic

• 1. 1,500,000(0.05)=75,000; the sample size n=1,460 is less than 5% of the population and np(1-p)=1,460(.42)(.58)=355.656 ≥ 10. The distribution of p-hat is approximately normal with mean mu sub p-hat = p = .42 and standard deviation ≈ .013

  2. x = 657 or more. p-hat=657/1460 = 0.45 or more. (Hard to show my work for the next part on r-studio, I apologize). =P(z ≥ 2.32) = 1 - 0.9898 = 0.0102

  3. x = 584 or less p-hat = 584/1460 or less (Hard to show my work for the next part on r-studio, I apologize). = P (z ≤ -0.55) = 0.0606

• 1. Sample size, n=500, is less than 5% of the population size and np(1-p)=500(0.39)(0.61) = 118.95 ≥ 10. The distribution of p-hat is approximately normal, with mean mu sub p-hat = p = 0.39 and standard deviation ≈ 0.022

  2. (Hard to show work for beginning part) = P(Z < -0.46) =0.3228 About 32 out of100 random samples of size n=500 adults wil result in less than 38% who believe marriage is osbsolete.

  3. P(0.40 < p-hat < 0.45) = P(0.46 < Z < 2.75) =0.9970-0.6772 =0.3198 So about 32/100 random samples of size n=500 adults will have between 40% and 45% of the respondents who say that marriage is obsolete.

  4. P(X ≥ 210) = P (p-hat ≥ .42) 1-P(p-hat < 0.42) (Hard to show work for this part) =1-P (Z < 1.38) =1-0.9162 =0.0838 So about 8 out of 100 random samples of size n=500 adults will have at least 210 of the responsdents who say that marriage is obsolete.