Output from a software package follows:
One-sample Z:
Test of mu=35 vs not =35
The assumed standard deviation = 3.4
Variable=x N=25 Mean =36.2 StDev.= 1.62 SE Mean = ? Z= ? Pvalue= ?
(a) Fill in the SE Mean, Z, and Pvalue. Conduct the test using the pvalue.
(b) Conduct the test using the critical value approach. Draw the rejection region and mark your test statistic Z* on the graph. Is your conclusion consistent with the one from (a)?
(c) Construct a 95% two sided z confidence interval on the mean.
(d) If you use the CI from (c) to make a decision on the test of mu=35 vs not =35, what is your conclusion?
(e) What would the P-value be if the alternative hypothesis is mu > 35?
Output from a software package follows: One-sample Z: Test of mu=35 vs not =35 The assumed...
Q5. The following output from MINITAB presents the results of a hypothesis test. Test of mu = 4.7 vs not = 4.7 The assumed standard deviation = 2.0 N Mean SE Mean 35 5.401 0.3381 95% CI (4.738, 6.064) Z 2.074 0.038 a. What are the null and altemate hypothesis? b. What is the value of the test statistic? c. What is the P-value? d. Do you reject He at the a=0.05 level? Why? e. Do you reject H at...
Really just need the LAST part but you can
answer all if u want (D)
Output from a software package follows: One-Sample 2: Test of mu = 14.5 vs > 14.5 The assumed standard deviation = 1.1 Variable N Mean StDey SE Mean 16 15.016 1.015 2 (a) [1 point) Is this a one-sided or a two-sided test? Z 2 P ? (b) (3 points) Fill in the SE Mean, Z, and P. (c) [3 points] Use the normal table...
Quiz 6: The Minitab output for the packing time example is as follows: Two-Sample T-Test and CI: New machine, Old machine Two-sample T for New machine va 0ld machine StDev 42.140 0.683 0.750 SE Mean 0.22 0.24 N Mean New machine old machine 10 43.230 10 Difference u (New machine) H (0ld machine) Estimate for difference: -1.090 T-Test of difference 0(vs ): T-Value -3.40 P-Value 0.003 DE= 18 Both use Pooled StDev 0. 7174 Test the variances of this test.
Two-sample T for height gender N Mean StDev SE Mean male 19 5.73 0.51 0.117 female 24 5.25 0.47 0.096 Difference = mu (male) - mu (female) Estimate for difference: 0.48 90% lower bound for difference: 0.283 T-Test of difference = 0 (vs >): T-Value = 3.17 P-Value = 0.002 DF = 18 Fill in the blanks: Ho: Ha: Type of test: t- test – SRS, Normal ?=?=.05 t-value = P-value: look up on your t-table Decision: Conclusion:
1 A sample of 35 observations is selected from a normal population. The sample mean is 16, and the population sta Conduct the following test of hypothesis using the 005 significance level. ndard deviation is 4 H1: μ > 15 a. Is this a one- or two-tailed test? One-tailed test Two-tailed test b. What is the decision rule? Reject He when z> 1645 Reject He when z s1.645 Prey 1012 İİİ Next > Type here to search Quiz 103 6...
A sample of 35 observations is selected from a normal population. The sample mean is 29, and the population standard deviation is 2. Conduct the following test of hypothesis using the 0.02 significance level. H0: ? ? 28 H1: ? >28 1. a. Is this a one- or two-tailed test? Two-tailed test One-tailed test 2. b. What is the decision rule? Reject H0 when z ? 2.054 Reject H0 when z > 2.054 3. c. What is the value of...
Gun Murders - Texas vs New York - Significance Test In 2011, New York had much stricter gun laws than Texas. For that year, the proportion of gun murders in Texas was greater than in New York. Here we test whether or not the proportion was significantly greater in Texas. The table below gives relevant information. Here, the p̂'s are population proportions but you should treat them as sample proportions. The standard error (SE) is given to save calculation time...
Absentee rates - Friday vs Wednesday: We want to test whether or not more students are absent on Friday afternoon classes than on Wednesday afternoon classes. In a random sample of 302 students with Friday afternoon classes, 48 missed the class. In a different random sample of 297 students with Wednesday afternoon classes, 32 missed the class. The table below summarizes this information. The standard error (SE) is given to save calculation time if you are not using software. Data...
The MINITAB printout shows a test for the difference in two population means. Two-Sample T-Test and CI: Sample 1, Sample 2 Two-sample T for Sample 1 vs Sample 2 N Mean StDev SE Mean Sample 1 6 28.00 4.00 1.6 Sample 2 9 27.86 4.67 1.6 Difference = mu (Sample 1) - mu (Sample 2) Estimate for difference: 0.14 95% CI for difference: (-4.9, 5.2) T-Test of difference = 0 (vs not =): T-Value = 0.06 P-Value = 0.95...
A sample of 35 observations is selected from a normal population. The sample mean is 16, and the population standard deviation is 4. Conduct the following test of hypothesis using the 0.05 significance level. Họ: H = 15 Hi: 4 > 15 a. Is this a one-or two-tailed test? One-tailed test Two-tailed test b. What is the decision rule? Reject Ho when z> 1.645 Reject Ho when zs 1.645 c. What is the value of the test statistic? (Round your...