In a hypothesis test, if the p-value is less than the chosen significance level (), what decision should be made regarding the null hypothesis? a) Reject the null hypothesis b) Fail to reject the null hypothesis c) Accept the null hypothesis d) Rerun the experiment.
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- Provide an example of the null hypothesis and a corresponding example of a alternative Hypothesis. Use example related to a business or a school.Which best describes the null hypothesis associated with an Analysis of Variance (ANOVA)? Group of answer choices a. Ho: Variance 1 = Variance 2 = Variance 3 b. Ho: Standard Deviation 1 = Standard Deviation 2 = Standard Deviation 3 c. Ho: Proportion 1 = Proportion 2 = Proportion 3 d. Ho: Median 1 = Median 2 = Median 3 e. Ho: Mean 1 = Mean 2 = Mean 3A publisher reports that 67% of their readers own a personal computer. A marketing executive wants to test the claim that the percentage is actually over the reported percentage. A random sample of 160 found that 71% of the readers owned a personal computer. Is there sufficient evidence at the 0.01 level to support the executive's claim? Step 1 of 6: State the null and alternative hypotheses. Step 2 of 6: Find the value of the test statistic. Round your answer to two decimal places. Step 3 of 6: Specify if the test is one-tailed or two-tailed. Step 4 of 6: Determine the decision rule for rejecting the null hypothesis, H0H0. Step 5 of 6: Make the decision to reject or fail to reject the null hypothesis. Step 6 of 6: State the conclusion of the hypothesis test.
- A sample of twenty automobiles was taken, and the miles per gallon (MPG), horsepower, and total weight were recorded. Develop a linear regression model to predict MPG using horsepower as the only indepen- dent variable. Develop another model with weight as the independent variable. Which of these two models is better? Explain. MPG 44 44 40 37 37 34 35 32 30 28 26 26 25 22 20 21 18 18 16 16 4 HORSEPOWER 67 50 62 69 66 63 90 99 63 91 94 88 124 97 114 102 114 142 153 139 WEIGHT 1,844 1,998 1,752 1,980 1,797 2,199 2,404 2,611 3,236 2,606 2,580 2,507 2,922 2,434 3,248 2,812 3,382 3,197 4,380 4,036An advertising executive claims that there is a difference in the mean household income for credit cardholders of Visa Gold and of MasterCard Gold. A random survey of 15 Visa Gold cardholders resulted in a mean household income of $68,420 with a standard deviation of $9100. A random survey of 6 MasterCard Gold cardholders resulted in a mean household income of $59,200 with a standard deviation of $9200. Is there enough evidence to support the executive's claim? Let μ1 be the true mean household income for Visa Gold cardholders and μ2 be the true mean household income for MasterCard Gold cardholders. Use a significance level of α=0.05 for the test. Assume that the population variances are not equal and that the two populations are normally distributed. Step 1 of 4: State the null and alternative hypotheses for the test. Step 2 of 4: Compute the value of the t test statistic. Round your answer to three decimal places. Step 3 of 4: Determine the decision rule for rejecting the…Evaluate the following statement: There is strong evidence of a systematic difference in gym attendance across groups prior to the experiment. Group of answer choices Yes, because it is very unlikely that chance alone would produce differences this large. No, it is not unusual for chance alone to produce differences this large.
- a test of a hypothesis to determine if the two groups differ with respect to the score on one variable is called options are test of difference test of similarity test of mean test of standard deviationGiven two independent random samples with the following results: n1=441 x1=185 n2=560 x2=351 Can it be concluded that there is a difference between the two population proportions? Use a significance level of α=0.05 for the test. Step 1 of 6: State the null and alternative hypotheses for the test. Step 2 of 6: Find the values of the two sample proportions, p^1 and p^2. Round your answers to three decimal places. Step 3 of 6: Compute the weighted estimate of p,p‾. Round your answer to three decimal places. Step 4 of 6: Compute the value of the test statistic. Round your answer to two decimal places. Step 5 of 6: Determine the decision rule for rejecting the null hypothesis H0. Round the numerical portion of your answer to two decimal places. Reject H0 if ...... ...... ...... Step 6 of 6: Make the decision for the hypothesis test.In light of the fact that research always contains the possibility of error, we have to be careful not to present research results as causal relationships when they are only correlations. And yet, sometimes correlation is used. For example, in car insurance, male drivers are correlated with more accidents, so insurance companies charge them more. It would be impossible to actually test this for causation. Given the above, how do we balance this possibility of error with the need to have statistical results to inform our business decisions?
- According to Stuart (2010), which of the following statements about observational study designs for causal inference are typically true? Choose all that apply. Including propensity score estimates as a predictor in a regression model on outcomes can help resolve imbalance between treatment and control groups for covariates used in the propensity score model Standard diagnostic tools for binary prediction or classification models (e.g. logistic regression or classification trees) are similarly used in evaluating propensity score estimates When conditional ignorability holds given the observed covariates, then the treatment assignment will also be ignorable conditioned on the propensity scores Overfitting in propensity score estimation can achieve more efficient estimates of treatment effects than using propensity score estimates that are closer to the true propensities Unlike full matching, subclassification, and weighting methods, nearest neighbor matching does not necessarily use all…Experimental Study: In the experimental study, the researcher would first identify a sample population of individuals who are willing to participate in the study. The participants would be randomly assigned to two groups: intervention vs. control. The intervention group would consume a specific amount of fried food each day, while the control group would not consume any fried food. Both groups would undergo testing to measure cardiovascular health. In this study, the independent variable is fried food consumption, which is measured by recording the amount and type of fried food consumed by the intervention group. Cardiovascular health would be the dependent variable. The research would be conducted as a randomized controlled trial, a type of experimental study. The study could last several weeks or months, and participants would be closely monitored for changes in their cardiovascular health. Following the intervention, both groups would be subjected to post-intervention testing to…. What are the advantages and disadvantages of Regression Model, Econometric Model, Driving Indicator Models?
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