Suppose that we reject a null hypothesis at the 0.1 level of significance. Then for which of the following α−values do we also reject the null hypothesis? A. 0.2 B. 0.02 C. 0.01 D. 0.002 E. 0.001
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Suppose that we reject a null hypothesis at the 0.1 level of significance. Then for which of the following α−values do we also reject the null hypothesis?
A. 0.2
B. 0.02
C. 0.01
D. 0.002
E. 0.001
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- Consider the following two formulations of the bivariate PRF, where ui and εi are both mean-0 stochastic disturbances (i.e random errors): yi = β0 + β1xi + u yi = α0 + α1(xi − x¯) + ϵ a) Write the OLS estimators of β1 and α1. Are the two estimators the same? b) What is the advantage, if any, of the second model over the first?Using the information provided in the SPSS output. You are asked to make a decision about whether to reject the null hypothesis that the population value of gamma equals 0 (i.e., that there is no ordinary association in the population between frequency of prayer and health). If alpha = 0.05 and you use a non-directional alternative hypothesis, which of the following is true?) A. Since the p-value is larger than alpha, you do not reject the null hypothesis B. Since the p-value is less than alpha, you do not reject the null hypothesis C. Since the p-value is larger than alpha, you reject the null hypothesis. D. Since the p-value is less than alpha, you reject the null hypothesisIf the value of Cronbach’s alpha is 0.07, it means ___________; a. Research instrument is not reliable b. Research instrument is internally consistent c. Data is reliable d. Data is internally consistent
- Assume we wanted to show that the true proportion of virginica iris seeds that fail to germinate is less than 20%. Let's say we tested at an alpha of 0.05, and found a p-vale of 0.082. Which of the following would be an appropriate conclusion and interpretation? A) With an alpha of 0.05, and a p-value of 0.082, we fail to reject the null and state we have insufficient evidence to support that the true proportion of virginica iris seeds that fail to germinate is less than 20%. B) With an alpha of 0.05, and a p-value of 0.0082, we reject the null and state we have sufficient evidence to support that the true proportion of virginica iris seeds that fail to germinate is less than 20%. C) With an alpha of 0.05, and a p-value of 0.082, we reject the null and state we have sufficient evidence to support that the true proportion of virginica iris seeds that fail to germinate is less than 20%. D) With an alpha of 0.05, and a p-value of 0.082, we fail to reject the null and state…. In general, a ____ t-statistic and a _____ p-value (relative to a chosen critical value and alpha) indicate that our estimated statistic is ____ to occur by chance if the null is true. a. Large; large; likely b. Small; large; likely c. Small; small; unlikely d. Large; large; likelyData from the 2014 General Social Survey (GSS) show that out of n=1,606 adult respondents, the proportion who reported having voted for Barack Obama in the 2008 presidential election was 0.61 Q: If I calculate a z test static of 4.84, and my alpha = 0.05, what decision should I make about the hypothesis? (Calculate the p-value for a 2-sided test to make your decision.) Options: A: fail to reject the null hypothesis that the proportion of American adults who voted for Obama in 2008 was 0.55 B: reject the null hypothesis that the proportion of American adults who voted for Obama in 2008 was 0.55 C: Reject the alternative hypothesis that the proportion of American Adults who voted for Obama in 2008 was not equal to 0.55 D: Fail to reject the alternative hypothesis that the proportion of American adults who voted for Obama in 2008 was not equal to 0.55
- If the critical t is ±1.796 and the obtained t is -2.09, what decision would you make regarding the null hypothesis?Dr. Chapman conducted an experiment in which participants watched paint dry for 30 minutes twice, once being paid $1 and once being paid $30. When comparing the samples, he calculated t = 3.57. He assumed α = .01 with df = 5, so the tcv = ±4.032. Because the calculated value was: A. greater than the critical value, Dr. Chapman can reject the null hypothesis. B. greater than the critical value, Dr. Chapman failed to reject the null hypothesis. C. less than the critical value, Dr. Chapman failed to reject the null hypothesis. D. less than the critical value, Dr. Chapman can reject the null hypothesis.A researcher predicts that scores in treatment A will be higher than scores in treatment B. If the mean for the 10 participants in treatment A is 4 points higher than the mean for the 10 participants in treatment B and the data produced t = 2.095 What decision should be made? Choose one a. With an alpha = .05 reject the null for either a one tail or two tails b. With an alpha = .05 fail to reject the null for either a one tail or two tails c. With an alpha = .05 reject the null for one tail but fail to reject with two tails d. With an alpha = .05 reject the null for two tails but fail to reject for a one tail
- If after performing a t-test for comparison of means (alpha= .05) we obtain p=0.0256, what is our conclusion? a. Fail to reject H0 b. Reject H0 c. Reject H1 d. Unable to determine from the information providedIf X1 and X2 constitute a random sample of size n = 2from an exponential population, find the efficiency of 2Y1relative to X, where Y1 is the first order statistic and 2Y1and X are both unbiased estimators of the parameterAssume that on average undergraduate students drink 7.5 cups of coffee per week (with a population standard deviation of 4.7). A class of 19 psychology students drinks 5 cups on average per week. Which of the following statistical tests should you run given this information? a. One sample z-test b. One sample t-test c. One sample p-test d. None of the above Given the statistical test you are running for this example, what are the (two-tailed) critical values (alpha = .05)? a. -2.09 & 2.09 b. -1.96 & 1.96 c. Can't calculate from the information provided d. None of the above