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- A random sample of size n is chosen from a normally distributed population with a mean of µ and known standard deviation 6. a) Compute a 95% CI for µ if the sample mean is 41 and n = 25. b) Compute a 95% CI for µ if the sample mean is 41 and n is 49. c) What sample size is needed to create a 95% CI for µ with a width that is no greater than 2? d) Check your work by calculating a 95% CI for µ given x = 41 using the sample size from part c).A random sample of n = 16 scores is obtained from a population with a mean of u ment is administered to the individuals in the sample, the sample mean is found to be M = 49.2. a. Assuming that the sample standard deviation is s = 8, compute r and the estimated Cohen's d to measure the size of the treatment effect. %3D 45. After a treat- b. Assuming that the sample standard deviation is s = 20, compute r and the estimated Cohen's d to measure the size of the treatment effect. c. Comparing your answers from parts a and b, how does the variability of the scores in the sample influence theneasures of effect size?A researcher predicts that scores in treatment A will be higher than scores in treatment B. If the mean for the 10 participants in treatment A is 4 points higher than the mean for the 10 participants in treatment B and the data produced t = 2.095 What decision should be made? Choose one a. With an alpha = .05 reject the null for either a one tail or two tails b. With an alpha = .05 fail to reject the null for either a one tail or two tails c. With an alpha = .05 reject the null for one tail but fail to reject with two tails d. With an alpha = .05 reject the null for two tails but fail to reject for a one tail
- The sample mean and standard deviation from a random sample of 30 observations from a normal population were computed as x¯=31x¯=31 and s = 9. Calculate the t statistic of the test required to determine whether there is enough evidence to infer at the 3% significance level that the population mean is greater than 28. Test Statistic =17. A 20,000-piece manufacturing run of 2200 resistors has a mean value of 2200 (naturally!) and a standard deviation of 2.50. They are to be sold as a 2% tolerance product, which means that the values must be between 0.98 x220 N = 215.60 and 1.02 x 220 0 = 224.40.25. The State of California claims the population average of the amount of ice cream each Californian eats in the month of September is 6.85 pints with population standard deviation of 1.35 pints. An SRS of 500 Californians resulted in a sample average of 6.75 pints eaten per person in the month of September . At alpha=0.05, is there evidence to support the State of California's claim that Californians eat an average of 6.85 pints of ice cream in the month of September? Write a conclusion using the context of the problem.
- A random sample of n = 12 individuals is selected from a population with μ = 70, and a treatment is administered to each individual in the sample. After treatment, the sample mean is found to be M = 74.5 with SS = 297.a. How much difference is there between the mean for the treated sample and the mean for the original population? (Note: In a hypothesis test, this value forms the numerator of the t statistic.)b. How much difference is expected just by chance between the sample mean and its population mean? That is, find the standard error for M. (Note: In a hypothesis test, this value is the denominator of the t statistic.)c. Based on the sample data, does the treatment have a significant effect? Use a two-tailed test with α = .05.23. The State of California claims the population average of the amount of ice cream each Californian eats in the month of September is 6.85 pints with population standard deviation of 1.35 pints. An SRS of 500 Californians resulted in a sample average of 6.75 pints eaten per person in the month of September At alpha = 0.05 , is there evidence to support the State of California's claim that Californians eat an average of 6.85 pints of ice cream in the month of September? Find the p-value ?a random sample of n = 25 individuals is selected from a population with u = 20 and a treatment administered to each individual in the sample. After treatment the sample mean is found to be m =22.2 with SS =384. a) how much difference is there between the mean for the treated sample and the mean for the original population? (Note: in a hypothesis test this value forms the numerator of the drastic). b) if there is no treatment effect how much difference is expected between the sample mean and is population mean? That is to find the standard error for m. c) based on the sample data does a treatment have a significant effect? use a two tailed test with (x = 0.05.
- Assume that you have a sample of n1=7 , with the sample mean X1=44 , and a sample standard deviation of S1=6 , and you have an independent sample of n2=6 from another population with a sample mean of X2=32 and the sample standard deviation S2=5 . Assuming the population variances are equal, at the 0.01 level of significance, is there evidence that μ1>μ2?Suppose that X1, X2, X3, and X4 represent a simple random sample selected from a population with mean, μ = 2.82 and standard deviation, σ = 1.87. Let the statistic W = (X1 + 2X2 + 2X3 + X4)/4 be used as a point estimator for μ. Calculate the bias of W. Blank 1. Calculate the answer by read surrounding text. Round your answer to 3 decimal places. Hint: Recall than E(aX + bY) = aE(X) + bE(Y), where a and b are constants and X and Y are random variables.