Given the discrete uniform population shown to the right, find the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 9.5 but less than 10.3. Assume the means are measured to the nearest tenth. f(x) = 1 x=4, 9, 14 0, elsewhere
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- Two separate samples are selected from the same population. Each have an n=8. Both samples receive a different treatment. After the treatment, the first sample has SS=16 and the second sample has SS=24. With a mean of 3 for the first group and mean of 6 for the second group, compute the t statistic. State the conclusion of these findings.Find the corresponding z-score for a sample of n = 16 and M = 38 taken from a population with a mean of μ= 40 and σ = 8.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.
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The degrees of freedom for the numerator, d.f.N, is ____ and the degrees of freedom for the denominator, d.f.D, is _____ The critical…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(The dean of a university estimates that the mean number of classroom hours per week for full-time faculty is 11.0. As a member of the student council, you want to test this claim. A random sample of the number of classroom hours for eightfull-time faculty for one week is shown in the table below. At α=0.10, can you reject the dean's claim? Assume the population is normally distributed. 12.3 9.1 11.8 6.7 6.2 10.6 14.6 9.1 What is the P value? Reject or fail to reject Ho? At α=0.10, can you reject the dean's claim?A sample of n = 64 scores has a mean of M = 68. Assuming that the population mean is μ = 60, find the z-score for this sample: If it was obtained from a population with σ = 16 z = If it was obtained from a population with σ = 32 z = If it was obtained from a population with σ = 48Suppose 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 random sample of size n1 =15 is selected from a normal population with a mean of 76 and standard deviation of 9. A second random sample size n2 =10 is taken from another normal population with mean 71 and a standard deviation 13. Let x barı and x bar2 be the sample means. Find: a. The probability that xbar1 – xbar2 exceeds 4.5 b. The probability that 3.5 s xbar1– xbar2 s 5.5ndependent random samples are selected from two populations and are used to test the hypothesis Hn: (H, - H2) = 0 against the alternative H,: (4, - H2) #0. An analysis of 232 observations from population 1 and 311 from population 2 yielded p-value of 0.113. Complete parts a and b below. a. Interpret the results of the computer analysis. Use as0.10. A. Since this p-value exceeds the given value of a, there is insufficient evidence to indicate that the population means are different. O B. Since the given a value exceeds this p-value, there is insufficient evidence to indicate that the population means are different. OC. Since the given a value exceeds this p-value, there sufficient evidence to indicate that the population means are different. O D. Since this p-value exceeds the given value of , there is sufficient evidence to indicate that the population means are different. b. If the alternative hypothesis had been Ha: (H1 - H2) <0, how would the p-value change? Interpret the p-value…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 youMATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
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