Rejecting the null hypothesis when the research hypothesis is false is? a. Effect size b. Power c. A type
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- Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .005 significance level. The null and alternative hypothesis would be: Ho: µM = µF Ho: HM = µuF Ho:PM = PF Ho:PM = PF Ho:HM = µf Ho:pM = PF H1: HM PF H1:PM µF H1:PM # PF The test is: left-tailed two-tailed right-tailed Based on a sample of 60 men, 30% owned cats Based on a sample of 40 women, 55% owned cats The test statistic is: (to 2 decimals) The critical value is: (to 2 decimals) Based on this we: O Reject the null hypothesis O Fail to reject the null hypothesis > Next QuestionA community college institutional researcher wants to analyze a data set with age (in years) predicting years spent in school. What inferential statistical procedure is appropriate for this study? Chi-square test for independence (2-way contingency table) Regression O One-sample z-test for mu (because sigma is known) O One-sample t-test Chi-square goodness-of-fit test Two-group t-test OZ-test for two proportions O One-way ANOVA O Correlation OZ-test for one proportion O Matched-pairs t-test Suemit Question MacBook Air esc F6 tab caps lock shift F1 Q A alt @ 2 Z F2 W S 20 # 3 F3 X E H $ LA 4 D F4 R с % 5 FO F5 T Y & 7 F7 G V B H * U ESuppose that you want to perform a hypothesis test based on independent random samples to compare the means of two populations. You know that the two distributions of the variable under consideration have the same shape and may be normal. You take the two samples and find that the data for one of the samples contain outliers. Which procedure would you use? Explain your answer.
- Identify the type I error and the type II error that correspond to the given hypothesis. The percentage of adults who have a job is less than 88%. dentify the type I error. Choose the correct answer below. A. Fail to reject the null hypothesis that the percentage of adults who have a job is less than 88% when the percentage is actually equal to 88%. B. Reject the null hypothesis that the percentage of adults who have am jobm is equal to 88% when that percentage is actually equal to 88%. C. Fail to reject the null hypothesis that the percentage of adults who have a job is equal to 88% when that percentage is actually less than 88%. D. Reject the null hypothesis that the percentage of adults who have a job is less than 88% when that percentage is actually less than 88%. Identify the type II error. Choose the correct answer below. A. Reject the null hypothesis that the percentage of adults who have a job is less than 88% when that percentage is actually less…Test the claim that the proportion of men who own cats is significantly different than the proportion of women who own cats at the 0.2 significance level. The null and alternative hypothesis would be: Ho: M = μF Ho: PM = PF H₁:µm > µF H₁:PM > pF Ho: PM = PF H₁:PMA researcher is interested in investigating whether living situation and pet ownership are dependent. The table below shows the results of a survey. Frequencies of Living Situation and Pet Single Family Couple Dog 87 81 95 Cat 118 70 100 Various 34 46 57 None 60 57 46 What can be concluded at the αα = 0.10 significance level? What is the correct statistical test to use? Independence Goodness-of-Fit Paired t-test Homogeneity What are the null and alternative hypotheses?H0:H0: The distribution of pets is not the same for each living situation. Pet ownership and living situation are independent. Pet ownership and living situation are dependent. The distribution of pets is the same for each living situation. H1:H1: Pet ownership and living situation are independent. The distribution of pets is not the same for each living situation. Pet ownership and living situation are dependent. The distribution of pets is the same for each living situation. The…Raw data in the table below come from an independent-measures experiment comparing three different treatment conditions. Treatment 1 0 1 0 3 X=1 Treatment 2 1 4 1 2 X=2 Should you reject or fail to reject the null hypothesis? O Reject O Fail to reject Treatment 3 3 3 4 5 X=3.75 To evaluate the effect of the independent variable, researchers conducted the independent-measures one-way ANOVA. They computed the sum of squares between groups to be 15.5 and the sum of squares within groups 14.75.Test the claim that the mean GPA of night students is larger than 2 at the 0.10 significance level. The null and alternative hypothesis would be: Họ :p = 0.5 Họ:µ = 2 Ho:u 0.5 Ho:p 2 H1 :p + 0.5 H1:µ # 2 H1:µ > 2 H1:p 0.5 H1:µ < 2 The test is: left-tailed right-tailed two-tailed Based on a sample of 25 people, the sample mean GPA was 2.02 with a standard deviation of 0.05 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: Reject the null hypothesis Fail to reject the null hypothesisIn a study of the effectiveness of an insecticide against a certain insect a large area of land was sprayed. Later the area was examined for live insects by randomly selecting squares and counting the number of live insects per square. Past experience has shown the average number of live insects per square after spraying to be 0.5. If the number of live insects per square follows a Poisson distribution, what is the probability that a selected square will contain: (b) Exactly one live insect? (a) No live insects? (c) Exactly four live insects? (d)One or more live insectsYou'd like to test if a sample of prisoners is more likely to have PTSD (diagnosed, not diagnosed) than a sample of non-incarcerated individuals. O One Sample Z test One Sample t test O Two Sample Z test O Independent two Sample t test Paired t test O One-way ANOVA Chi Square Test Pearson's Correlation Coefficient O Simple Linear RegressionThe hypothesis that suggests that an experimental treatment has no effect on a dependent variable would be known as the __________ hypothesis. reliability alternative statistical null validityTest the claim that the mean GPA of night students is larger than 2.6 at the 0.01 significance level. The null and alternative hypothesis would be: Ho:p > 0.65 Ho:p 2.6 H1:p 0.65 H1:p+ 0.65 H1:µ > 2.6 H1:µ + 2.6 H:µ < 2.6 %3D The test is: two-tailed right-tailed left-tailed Based on a sample of 65 people, the sample mean GPA was 2.63 with a standard deviation of 0.07 The test statistic is: (to 2 decimals) The p-value is: (to 2 decimals) Based on this we: O Fail to reject the null hypothesis O Reject the null hypothesisSEE MORE QUESTIONS
