7.4 Find the p-value, Part II. An independent random sample is selected from an approximately normal population with an unknown standard deviation. Find the p-value for the given sample size and test statistic. Also determine if the null hypothesis would be rejected at a = 0.01. (a) n = 26, T = 2.485 (b) n = 18, T = 0.5
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- In a test of hypothesis, the null hypothesis is that the population mean is less than or equal to 32 and the alternative hypothesis is that the population mean is greater than 32. The population has a normal distribution with variance of 12.5. The test is to be made at the 10% significance level. A sample of size 41 is selected from this population. The approximate value of ßatu = 32. 35 is a. 0.5403 O d. 0.7414 O e. 0.9000 O b. 0.2586 O c. 0.9723-What is the test statistic, t ? -The P-value is l ?On a certain hearing ability test, the mean is 300 and the standard deviation is 20. Thebetter you can hear, the higher your score. Can people who clean their ears frequentlyhear better than others? You take a sample of 31 people who clean their earsfrequently. Their sample mean test score is 308. Do a hypothesis test with α = 0.01.H 0 : People who clean their ears regularly hear at the same level that everyone elsedoes: µ = 300. c. Calculate your test statistic and draw it on your null hypothesis curve. This is justyour sample statistic ( or ) converted to a Z-score on your null hypothesis samplingdistribution. Make sure you are using the appropriate SE formula! Use forproportions and for means. d. Treat your test statistic as a Z-score and look up the area beyond it in the Z-table.That is your p-value. It may be helpful to shade this on your curve. (Remember, ifyou have a < in your alternative hypothesis, you’re looking at the area below yourtest statistic. If you have a…
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- Q10 = In a one-tailed hypothesis test with a sample size of 300, the test statistic is determined to be Z = 1.53 Assuming population standard deviation o is known , The p-value for this test is: Q11 = In a two-tailed hypothesis test with a sample size of 300, the test statistic is determined to be Z = 2.21 . Assuming population standard deviation is known, The p-value for this test isA sample of n=4 scores has ΣX=8 and ΣX2=40. What is the value of the sample variance, s2?Q1: A researcher was interested in the effect of exercise on stress levels. For the general population, the distribution of the Stress Battery Scale scores is normal with the mean of µ = 25 and the standard deviation of σ = 5. A sample of n = 100 participants was asked to exercise for three weeks and then reported their stress levels. The sample mean was ?X = 23. Conduct a hypothesis test and determine the effect of exercise. Use the two-tailed test α = .01 1.Type the null and research hypotheses in complete sentences. 2.Cut off points 3.Standard error and z-score -show your computation process 4.Your conclusion thoroughly.
- 14Researchers conducted a study to determine whether magnets are effective in treating back pain. The results are shown in the table for the treatment (with magnets) group and the sham (or placebo) group. The results are a measure of reduction in back pain. Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. Complete parts (a) and (b) below. Treatment Sham μ μ1 μ2 n 13 13 x 0.48 0.41 s 0.76 1.42 a. Use a 0.05 significance level to test the claim that those treated with magnets have a greater mean reduction in pain than those given a sham treatment. What are the null and alternative hypotheses? A. H0: μ1=μ2 H1: μ1>μ2 B. H0: μ1<μ2 H1: μ1≥μ2 C. H0: μ1=μ2 H1: μ1≠μ2 D. H0: μ1≠μ2 H1: μ1<μ2 The test statistic, t, is nothing. (Round…n=10 b = 0.50 sb = 0.02 α=0.05 find the confidence interval for β.