Use Minitab to calculate the appropriate hypothesis test for the given problems, paste the corresponding Minitab output from the session window for each problem. State the null and alternative hypotheses, state α, state the equation of the test statistic, check assumptions,…
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To determine whether the wording of the cover letter influences the response rate, three different cover letters were used in a survey of students at a Midwestern university. Suppose that each of the three cover letters accompanied questionnaires sent to an equal number of randomly selected students. Returned questionnaires were then classified according to the type of cover letter (I, II, or III). Use the accompanying data to test the hypothesis that the true proportions of all returned questionnaires accompanied by cover letters I, II, and III are the same. Use a 0.05 significance level.
As shown in Minitab output, all expected frequencies are greater than 1 and no expected frequency is less than 5. Furthermore, as stated in the problem, the sample is a simple random sample. Therefore, all the assumptions of the test are met.
At the 5% significance level, the data provide sufficient evidence to conclude that the hypothesis that the true proportions of all returned questionnaires accompanied by cover letters I, II, and III are the same.
2. A survey was conducted in the San Francisco Bay area in which each participating individual was classified according to the type of vehicle used most often and city of residence. A subset of the resulting data is given in the accompanying table. Do the data provide convincing evidence of an association between city of residence and vehicle type? Use a significance level of 0.05. You may assume that it is reasonable to regard the sample as a random sample of Bay area residents.
As shown in Minitab output, all expected frequencies are greater than 1 and no expected frequency is less than 5. Furthermore, as stated in the problem, it is reasonable to regard the sample as a random sample of Bay area residents. Therefore, all the assumptions of the test are met.
Standardized residuals greater than 2 in absolute value provide evidence
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(Minitab Assignment Example | Topics and Well Written Essays - 250 Words - 1)
“Minitab Assignment Example | Topics and Well Written Essays - 250 Words - 1”, n.d. https://studentshare.org/statistics/1682611-minitab.
From the graph, it there seems a linear relationship between the two variables as shown below. 2. Equation of the "Best Fit" Line for Relationship between CREDIT BALANCE and SIZE In next step, regression model was determined between Credit Balance and Size using the Regression Analysis in Minitab.
Is there evidence to suggest that mean has changed?
iv) Perform a hypothesis test to see whether the mean number of male cattle is equal to the mean number of female cattle. To three decimal places, what is the p value from the test? Is there evidence to reject the null
Then the model fit can be checked using the F-test. It is also important that the right model is selected in the beginning.
c) The variability of this data would be looked into by first calculating the standard deviation for each level of absorbance with
The p-value is given as 0.000 (a value less than α=0.05); we thus reject the null hypothesis and conclude that there is significant difference in the proportion of those who gave the right answers and those
Based on the results provided by the author, it is evident that more money was spent on newspaper advertisements than was spent on radio advertisements. Since he rejected the null hypothesis, he concludes that indeed the expenditure made on radios was significantly different from that made on newspapers.
Do you think the sample would have been randomly selected from this target population? Explain your answer. Yes. Every person in the city has an equal chance of being selected.As a result the random sampling will not be biased (Sweeney, et al., 2011, pp. 142-145)
At 95% CI [16.6, 19 ] the sample mean lies within the parameters. Thus we reject null hypothesis that the average jail sentence of traffic offenders is 16.7.
The sample N = 500 is a large size thus we have to detect the gross non-normality,