Bob selects independent random samples from wo populations and obtains the values pi 0.700 and p2 confidence interval for p1 – P2 and gets: 0.500. He constructs the 95% 0.200 ± 1.96(0.048) = 0.200 ±0.094. Note that 0.048 is called the estimated standard error of pi – P2 (the ESE of the estimate). Tom wants to estimate the mean of the success rates: Pi + p2 (a) Calculate Tom's point estimate. (b) Given that the estimated standard er- ror of (p1 + p2)/2 is 0.024, calculate the 95% confidence interval estimate of (Pi + p2)/2. Hint: The answer has our usual form: Pt. est. +1.96 × ESE of the estimate.
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- The percentage score of the male and female students is normally distributed with unknown means and with a known s.d of 1.2, 1.3 respectively. A random sample of size 30 is drawn with a mean value of 67.8, and 70.4. Construct a 95% C.I for the difference between means of the percentage scores. Q4.T3.6As a bonus assignment a former student checked if your professor gave a statisticallysignificant difference in grades between his male and female students. She based herstudy based on grades assigned in intermediate Econ courses (Econ 303, 305 and 317)and her sample included nm = 485 male students and nf = 264 female students. Theaverage grades received were xm = 84.6 and xf = 85.8 The population standad deviation were σ m = 12.0 and σ f = 11.4 8. From the same extra-credit study as in question 7 see above, this former student found that the proportion of female students in principle courses (Econ 203, 205) was ?̅? = 0.380, while the proportion of female students in intermediate courses (Econ 303, 305, 317) was ?̅? = 0.352. The principle courses sample size was np = 782, while the intermediate courses sample size was ni = 749. Test the hypothesis that female students are less in intermediate courses using a 90% confidence level and the p-value approach.
- a=1 b=8 c=6 d=0A sample of n = 64 scores has a mean of M = 68. Assuming that the population mean is p = 60, find the z-score for this sample: If it was obtained from a population with o = 16 Z = If it was obtained from a population with o = 32 Z = If it was obtained from a population with o = 48 Z =Suppose in a local Kindergarten through 12th grade (K -12) school district, 49% of the population favor a charter school for grades K through 5. A simple random sample of 144 is surveyed. a. Find the mean and the standard deviation of X of B(144, 0.49). Round off to 4 decimal places. O = b. Now approximate X of B(144, 0.49) using the normal approximation with the random variable Y and the table. Round off to 4 decimal places. Y - N( c. Find the probability that at most 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X 75) - P(Y > a (Z > e. Find the probability that exactly 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X = 81) - P(Consider the scenario ai from the previous question. Suppose you take an SRS of 50 avocados and find that the mean is 1.1 inches, which yields a p-value of 0.0542. What conclusion would you draw at the =0.05 level? At the =0.10level?Suppose in a local Kindergarten through 12th grade (K -12) school district, 49% of the population favor a charter school for grades K through 5. A simple random sample of 144 is surveyed. a. Find the mean and the standard deviation of X of B(144, 0.49). Round off to 4 decimal places. O = b. Now approximate X of B(144, 0.49) using the normal approximation with the random variable Y and the table. Round off to 4 decimal places. Y - N( c. Find the probability that at most 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X 75) - P(Y > a (Z > e. Find the probability that exactly 81 favor a charter school using the normal approximation and the table. (Round off to z-values up to 2 decimal places.) P(X = 81) - P(A sample of n=4 scores has a mean of M=75. Find the z-score for this sample: a. If it was obtained from a population with u= 80 and o=10. b. If it was obtained from a population with u=80 and o=20.The random sample of n=75�=75 yielded a sample mean of x¯=26�¯=26.Assume that we know the population standard deviation is σ=30�=30. Step 2 of 3 : Calculate a 95% confidence interval for the population mean. Round your answers to 2 decimals places.A sample of n = 7 scores is selected from a population with an unknown mean ( µ). The sample has a mean of M = 40 and a variance of s 2 = 63. Which of the following is the correct 95% confidence interval for µ?A random sample of size 10 yielded roughly "mound-shaped" data with a sample mean of 63.5 and a sample variance of 60.8. Let (L, OU) be the interval estimate that contains the population mean with 95% probability. Find the width of the interval. That is, find 0 – 0₁. 4.52 5.49 5.58 5.70 9.04 10.99 11.16 11.40 none of the other answers give the correct widthRecommended textbooks for youA First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSONA First Course in Probability (10th Edition)ProbabilityISBN:9780134753119Author:Sheldon RossPublisher:PEARSON