The following MINITAB output presents the results of a hypothesis test for the differenceHx - Hy between two population means:Two-sample T for X vs YNMeanStDevSE Mean1353.942.650.231804.432.380.18Difference = mu (X) – mu (Y)Estimate for difference: -0.48444295% upper bound for difference: -0.007380T-Test of difference = 0 (vs <): T-Value = -1.68 P-Value = 0.047 DF = 270Is this a one-tailed or two-tailed test?a.b.What is the null hypothesis?Can H, be rejected at the 5% level? How can you tell?C.d.The output presents a Student's t test. Compute the P-value using a z test. Are the tworesults similar?Use the output and an appropriate table to compute a 99% confidence interval for Hy-Hy based on the z statistic.e.

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Answered: The following MINITAB output presents…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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ou wish to test the following claim (Ha) at a significance level of α=0.005. Ho:μ1=μ2 Ha:μ1≠μ2You believe both populations are normally distributed, but you do not know the standard deviations for either, and you have no reason to believe the variances of the two populations are equal. You obtain a sample of size n1=18 with a mean of M1=81.6 and a standard deviation of SD1=8.2 from the first population. You obtain a sample of size n2=23 with a mean of M2=88.9 and a standard deviation of SD2=5.8 from the second population. What is the test statistic for this sample? (Report answer accurate to two decimal places.) test statistic =
In 1990, the average duration of long-distance telephone calls originating in one town was 9.4 minutes. A long-distance telephone company wants to perform a significance test to determine whether the average duration of long-distance phone calls has changed from the 1990 mean of 9.4 minutes. The hypotheses are: H0: μ = 9.4 minutes HA: μ ≠ 9.4 minutesSuppose that the results of the sample lead to non-rejection of the null hypothesis. Classify that conclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean duration of long-distance phone calls has changed from the 1990 mean of 9.4 minutes. Choose the best asnwer below: A. Neither B. Type I error C. Type II error D.No error
Based on the photo given, what are the interpretation of the coefficient on public debt and human capital? Please. Asap.
After conducting your one-sample z-test, you find an obtained z-statistic of 1.08 and a critical z-statistic of 1.64. What do you conclude? Group of answer choices Fail to reject the null hypothesis; the sample mean and population mean are significantly different. Reject the null hypothesis; the sample mean and population mean are significantly different. Fail to reject the null hypothesis; the sample mean and population mean are not significantly different. The results of the test are inconclusive.
Consider the following hypothesis test. a. What is your conclusion if = Calculate the value of the test statistic (to 2 decimals). 21, s = 8.2, n₂ = 26, and s=4.0? Use a = The p-value is between 0.05 and 0.10 What is your conclusion? Cannot conclude that the two population variances are different b. Repeat the test using the critical value approach. What is the rejection rule for this hypothesis test? Round the critical values to 2 decimal places. Reject Ho if F greater than or equal to ✓ What is your conclusion? Hide Feedback Partially Correct Use Table 4 of Appendix B. Cannot conclude that the two population variances are different Ho: ₁ = 0 H₂: 002/ 0.05 and the p-value approach.
Two groups of individuals were compared with respect to a high carbohydrate low-fat diet (LF) and a high-fat low carbohydrate diet (LC). Mood was assessed using a total mood disturbance score where a lower score is associated with a less negative mood. Assuming that the random samples come from independent normal distributions with a common but unknown variance, construct a 95% confidence interval for the difference between the two population means, LC less LF. Group LC LF 28+18.56 28±8.96 28±22.37 28+22.30 n 12 13 mean 47.3 19.3 standard deviation 28.3 25.8
My answer is as follows: T critical = 2.70 T value= 7.775 P< a (0.01) Therefore we reject the null. Is this correct?
5. We don't want predictor variables that are too highly correlated with each other or they are redundant (say ≥±.70) or multicollinear (2±.90), which means the variables are measuring pretty much the same thing so why have them all in the equation. What was the finding for redundancy between SPI Scientist and SPI Practitioner? a. No issue with redundancy since the shared variance between the two variables was only 4% (r2 = .04). b. There was an issue with redundancy between the two variables shared 62% of the variance in the relationship.
A researcher randomly assigns college freshmen to either of two experimental conditions. Because both groups consist of college freshmen, someone claims that it is appropriate to use a t test for the two related samples. Comments?
8.2 Use technology to find the P-value for the hypothesis test described below. The claim is that for a smartphone carrier's data speeds at airports, the mean is μ=17.00 Mbps. The sample size is n=21 and the test statistic is t=−1.798
Using a sample that has n=50 pair of data, we get a correlation coefficient r=-0.269 for the following null and alternative hypotheses determine and compute the test statistic value: HO: p=0 H1: pe 0 O a t=7.51 O b. z=7.51 O o.z=-0. 16 Odt=-1.08
A new method of postoperative treatment was evaluated for patients undergoing a certain surgical procedure. Under the old method, the mean length of hospital stay was 6.3 days. The sample mean for the new method was 6.1 days. A hypothesis test was performed in which the null hypothesis stated that the mean length of stay for the new method was greater than or equal to the mean length for the old method, and the alternate hypothesis stated that the mean stay for the new method was lower. The P-value was 0.002. True or false: a) Because the P-value is very small, we can conclude that the new method provides an important reduction in the mean length of hospital stay. b) Because the P-value is very small, we can conclude that the mean length of hospital stay is less for patients treated by the new method than for patients treated by the old method.

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