Given test statistic t = 2.83 and p - value = 0.003
For a = 0.05
Decision rule : If p-value < 0.05 we reject the null hypothesis Ho otherwise we fail to reject the null hypothesis.
Our p-value = 0.003 < 0.05 we reject the null hypothesis
Conclusion : There is a difference in effectiveness between the two methods.
Question 3 of 8 When the two-sample t-test is carried out the researchers find that the...
These two groups are two samples representing the population of
workers in the economy. We want to know if the workers who take the
training (treatment sample) have higher earnings than the group
that do not take the training (control sample). If we find that the
trained workers have higher earnings it would indicate that the
training is effective.[1]In terms of statistics, we will do a
hypothesis test on the difference between the mean earnings in the
treatment population and...
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...
I think this question should use t-Test: paired two sample for means, but my friend think it should use t-test: t-test: Two-sample assuming equal variances. So, I need someone to help me explain which method is correct and answer the following question. Thanks!! A professor in the School of Business wants to investigate the prices of new textbooks in the campus bookstore and the Internet. The professor randomly chooses the required texts for 12 business school courses and compares the...
I. , 0.6 OF 12 pts 9.4.7-T Question Help 0 Researchers conducted an experiment to test the effects of alcohol. Errors were recorded in a test of visual and motor skills for a treatment group of 22 people who drank ethanol and another group of 22 people given a placebo. The errors for the treatment group have a standard deviation of 240, and the errors for the placebo group have a standard deviation of 0.83. Use a 0.05 significance level...
1) Find the test statistic, t, to test the hypothesis that . Two samples are randomly selected and come from populations that are normal. The sample statistics are given below. Do not pool the variances. 25 30 1.5 1.9 A. 3.287 B. 4.361 C. 1.986 D. 2.892 2) The following data represent the muzzle velocity (in feet per second) of rounds fired from a 155-mm gun. For each round, two measurements of the velocity were recorded using two different measuring...
Assume that the assumptions and conditions for inference with a two-sample t-test are met. Test the indicated claim about the means of the two populations. State your conclusion.A researcher wishes to determine whether people with high blood pressure can reduce their blood pressure by following a particular diet. Use the sample data below to test the claim that the treatment population mean μ1 is smaller than the control population mean μ2. Test the claim using a significance level of 0.01. Treatment...
Question 3 1 pts The following APA summary was reported for a one-sample t-test (two-tailed, alpha =.05). t(20) = 1.5, p > .05 Which of the following is an appropriate conclusion? The difference is statistically significant. The critical t is 1.5. 20 individuals participated in the study. The difference is not statistically significant.
8. AM -vs- PM sections of Stats - Significance test (Raw Data, Software Required): There are two sections of statistics, one in the afternoon (PM) with 30 students and one in the morning (AM) with 22 students. Each section takes the identical test. The PM section, on average, scored higher than the AM section. The scores from each section are given in the table below. Test the claim that the PM section did significantly better than the AM section, i.e.,...
is that enough for you
Two-Sample T-Test and CI: 2011, 2012 Two-sample T for 2011 vs 2012 2011 2012 N 75 93 Mean 6.466 6.604 St Dev 0.352 0.398 SE Mean 0.041 0.041 Difference=mu (2011) -mu (2012) Estimate for difference: -0.1376 95% CI for difference: (-0.2535, -0.0217) T-Test of difference=0 (vs not =) : T-Value = -2.34 P-Value=0.020 DF = 166 Both use pooled StDev=0.3782 Manutan International S.A. is a specialist mail-order business providing industrial and office equipment and supplies...
MAKE SURE YOU FULLY EXPLAIN PART (c). The mentioning of
"incorrect" use of two sample t test in (c) means (a) would be some
other t test.
40. Lactation promotes a temporary loss of bone mass to provide adequate amounts of calcium for milk produc tion. The Lactation to Postweaning in Adolescent Mothers with Low Calcium Intakes" (Amer. J. of Clinical Nutr. 2004: 1322-1326) gave the following data on total body bone mineral content (TBBMC) (g) for a sample both...