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- Ch 9, #10 (pg. 264) Stats Question: A random sample of n=4 individuals is selected from a population with mu = 35, and a treatment is administred to each individual in the sample. After treatment, the sample mean is found to be M = 40.1 with SS = 48. Based on the sample data, does the treatment have a significant effect? Use a two-tailed test with alpha = .05. I understand how to determine that t = 2.55. What I don't understand is how the critical value of 3.182 was determined in the explanation posted on this site.Statistics and probabilityTest the claim that the mean GPA of night students is larger than 2 at the 0.025 significance level. The null and alternative hypothesis would be: Но: и 0.5 Но: n> 2 H1: µ > 2 H1:p + 0.5 H1:µu # 2 H1:p > 0.5 H1:p < 0.5 H1: µ < 2 The test is: left-tailed right-tailed two-tailed Based on a sample of 40 people, the sample mean GPA was 2.01 with a standard deviation of 0.06 The p-value is: (to 2 decimals) Based on this we: Reject the null hypothesis OFail to reject the null hypothesis
- Remaining Time: 1 hour, 36 minutes, 24 seconds. v Question Completion Status: QUESTION 9 Let A and B be the two events of a sample S such that P(AUB) = 0.85, P(ANB) = 0.35,P(BNA) =0.30 What is P(A|B)? 11 11 5. 11 QUESTION 10 A normally distributed random variable X with variance o2 = 400 and P(x> 130) = 0.6554. Then the mean of X is O 122 O OPopulation 1 Population 2 Use the given statistics to complete parts (a) and (b). Assume that the populations are normally distributed. (a) Test whether H, > H, at the a = 0.01 level of significance for the given sample data. (b) Construct a 90% confidence interval about u, -H2. 28 16 45.3 43.2 9.7 6.7 (a) Identify the null and alternative hypotheses for this test. O A. Ho: H1 > H2 O B. H: H1 =H2 OC. Hg: H1 = H2 H;: H1 # H2 O F. Ho: H1 H2. O B. Do not reject Hg. There is not sufficient evidence at the a = 0.01 level of significance to conclude that u, > H2. O C. Do not reject H,. There is sufficient evidence at the a= 0.01 level of significance to conclude that u, > H2- O D. Reject Hn. There is not sufficient evidence at the a= 0.01 level of significance to conclude that u, > H2- (b) The 90% confidence interval about u, - H, is the range from a lower bound of to an upper bound of (Round to three decimal places as needed.)Compute the z-scores and the population probabilities for these normally distributed events: a) mx = 96, sx = 14, z = _______ P(x < 81) = _____________ b) mx = 0.42, sx = 0.11, z = _______ P(x > 0.60) = _____________ c) mx = 1250, sx = 850, z = _______ P(x > 650) = _____________ d) mx = 9.0, sx = 3.4, z = _______ P(x < 11.8) = _____________
- Test the claim that the mean GPA of night students is smaller than 2.5 at the 10% significance level. The population standard deviation for this group is known to be 0.16. The null and alternative hypotheses for this test are: Но: д 2.5 Нд:д > 2.5 НА:Д < 2.5 The test is: left-tailed right-tailed Based on a sample of 30 people, the sample mean GPA was 2.47. Hint: Before you find the p-value below, calculate the test statistic to two decimal places. The p-value is: (to 4 decimals) The significance level as a decimal is: Based on this we O fail to reject the null hypothesis, Ho O reject the null hypothesis, HoQUESTION 6 What hypothesis test should be used to test H1: X1 - x2 o O Left tailed, one-sample test of means Right tailed, one-sample test of means Left tailed, one-sample test of proportions Right tailed, one-sample test of proportions Left tailed, one-sample test of variances Right tailed, one-sample test of variances Left tailed, two-sample test of means (independent samples) Right tailed, two-sample test of means (independent samples) Left tailed, two-sample test of means (paired samples) Right tailed, two-sample test of means (paired samples) Left tailed, two-sample test of proportions Right tailed, two-sample test of proportions O Left tailed, two-sample test of variances O Right tailed, two-sample test of variances(a) State the null and alternative hypotheses.Ho= Ha= (b) Give the notation for the sample statistic.statistic symbol-statistic symbol (c) Give the value for the sample statistic. Value=# Use the randomization distribution to estimate the p-value? p-value=#
- 9.6 In a certain population the random variable Y has variance equal to 490. Two independent random samples, each of size 20, are drawn. The first sample mean is used as the predictor of the second sample mean. (a) Calculate the expectation, expected square, and variance, of the prediction error. (b) Approximate the probability that the prediction error is less than 14 in absolute value. Show Transcribed TextTest the claim that the mean GPA of night students is larger than 2 at the 0.005 significance level. The null and alternative hypothesis would be: Но:р 2 Нo:д 0.5 Но:и — 2 Нi:р> 0.5 H:n 2 Hi:p#0.5 Hі:р<0.5 Hі:р +2 The test is: two-tailed left-tailed right-tailed Based on a sample of 65 people, the sample mean GPA was 2.02 with a standard deviation of 0.08 The p-value is: (to 2 decimals) Based on this we: Fail to reject the null hypothesis O Reject the null hypothesisrandom sample of 136 kissing couples, both people in 84 of the couples tended to lean more to the right than to the left. (Use ? = 0.05.) (a) If 2/3 of all kissing couples exhibit this right-leaning behavior, what is the probability that the number in a sample of 136 who do so differs from the expected value by at least as much as what was actually observed? (Round your answer to four decimal places.) (b) Does the result of the experiment suggest that the 2/3 figure is implausible for kissing behavior?State the appropriate null and alternative hypotheses. H0: p = 2/3Ha: p > 2/3H0: p = 2/3Ha: p ≤ 2/3 H0: p = 2/3Ha: p < 2/3H0: p = 2/3Ha: p ≠ 2/3 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value =