1. The null hypothesis: There are no promising variables that seem to have a relationship with the amount of the prepaid cards The alternative hypothesis: There are variables that have a relationship to the amount stored on the prepaid card. To find out of if we are to accept or reject the null hypothesis we are first to determine the critical value‚ this was done by using the denominators (20)‚ and then aligning with the number of degrees of freedom (4) at the 0.05 significance level. The
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means are not equal. Conclusion‚ at the .05 significance level‚ 5.24 F exceeds the F-critical value of 3.159. There is sufficient evidence to reject the Null hypothesis in favor of the alternative. The P-value is .00081399‚ the ANOVA P-value is significantly less than alpha‚ and therefore there is strong evidence to reject the null hypothesis. Using the inferences about individual treatment means is appropriate‚ when we look at the health of the patients. It is reasonable to believe that all the
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required for reactions to progress in the cells. Null hypothesis states that there will be no difference between the result of two separate variables A and B. The null hypothesis states in regards to the enzyme experiment that lactase will not bind preferentially‚ or more specifically‚ to maltose or lactose. Before a null hypothesis can be rejected we must notice a large difference between glucose produced from maltose versus lactose. Alternate Hypothesis states the opposite of null in that there will be
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months that has positive investment returns on AT&T are equal to months of negative returns. Therefore he sets a hypothesis; the null hypothesis H0 :ppositive = pnegative = 0.5 versus the alternative hypothesis Ha: not H0. The number of months that has the positive returns on AT&T is 32‚ and the number of months that has the negative returns is 22. In order to test the hypothesis‚ he should do chi-square test. First of all he calculates the expected value that has the positive and negative returns
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results were similar to my hypothesis. My hypothesis was that the densest object would be the rubber stopper. In the lab results the densest object was the penny and the rubber stopper was second. My hypothesis was switched. My hypothesis was correct when I stated that the macaroni was the third densest object. My hypothesis was also correct for the dice‚ but the bead and the dice share the same density. So they are both the fourth densest item. I was wrong with my hypothesis when I placed the popsicle
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can be applied to a relatively small number of cases. It was specifically designed to evaluate statistical differences for samples of 30 or less. 2.0 DEFINITION OF T-TEST A t-Test is any statistical hypothesis test in which the test statistics follows a Student’s t distribution if the null hypothesis is supported. 3.0 A BRIEF EXPLANATION One of the most commonly used statistical procedures is the t-test. There are actually three variations of the t-test that we will consider. These are the single-sample
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the logic of a hypothesis test? “If our sample result is very unlikely under the assumption of the null hypothesis‚ then the null hypothesis assumption is probably false. Thus‚ we reject the null hypothesis and infer the alternative hypothesis.” • What is the logic of using a CI to do a HT? We are 95% confident the proportion is in this interval… if the sample mean or proportion is in the confidence interval the null hypothesis could be inferred. Else‚ the alternative hypothesis would. •
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write on the space below‚ for markers only. Page 2-7 8 9-10 11 12-13 14-15 Question 1-17 18 19abc 20ab 21abc 22abcd Total Max 51 9 9 9 9 13 100 Mark 1 Part 1. Multiple Choice (3 marks each‚ no part marks. 51 marks in total) 1. In testing H 0 : p 0.7 against H a : p 0.7 ‚ the p-value is 0.0231. Which one of the following statements is true? (a) Reject H 0 if the significance level is less than 0.01. (b) Reject H 0 if the significance level is less than 0.05. (c) A 99% confidence interval
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significantly different and the null hypothesis should be rejected. 2. State the null hypothesis for the Baird and Sands (2004) study that focuses on the effect of the GI with PMR treatment on patients’ mobility level. Should the null hypothesis be rejected for the difference between the two groups in change in mobility scores over 12 weeks? Provide a rationale for your answer. * Treatment group mean = control group mean * With the p-value being < the alpha‚ the null hypothesis would be rejected indicating
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Introduction The purpose of this report is to show if the new golf ball coating design does induced resist cuts and provide a more durable ball. Hypothesis µ1 is the mean driving distance of currently produced golf balls µ2 is the mean driving distance of newly designed golf balls Null hypothesis (H0): µ1- µ2 ≤ 0 Alternate hypothesis (Ha): µ1- µ2 > 0 Analysis The sample mean for current golf balls is 270.275‚ while the new golf balls have a sample mean of 267.50. So on average the current
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