The test statistic for a left-tailed test is Z = -2.65. Determine the P-value. 0.0080 0.0040 0.9960 0.9920
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- stat 1bQuiz navigatio Assume that a chi-square test based on three degrees of freedom yielded a test statistic = 6.5. Based on the above information, the p-value of the test |1 2 3 4 Select one: 8 9 10 1 15 16 17 18 less than 0.01 Ob. Finish attempt .. between 0.025 and 0.05 Time left 0:55:40 between 0.10 and 0.90 Od. between 0.05 and 0.10 Next page fin Chapter 8 a.docx Chapter 8 assig.docx Chapter 4.docx Assignment 4 A..docx MAY 30 PVitamin C in Fruits and Vegetables The amounts of vitamin C (in milligrams) for 100g (3.57ounces) of various randomly selected fruits and vegetables are listed. Is there sufficient evidence to conclude that the standard deviation differs from 12mg? Use =α0.01. 7.9 16.3 12.8 13.0 32.2 28.1 34.4 46.4 53.0 (c)Compute the test value. Round the answer to at least three decimal places. =χ2
- Find the p-value for the z-test. (Round your answer to four decimal places.) a left-tailed test with observed z = -1.83 In USE SALT p-value = Determine the significance of the results. O The results are highly significant. The results are statistically significant. The results are only tending toward statistical significance. The results are not statistically significant.Summer high temperatures are distributed normally with a mean of 92.3 and a standard deviation of 4. NOTE: Round your z-score to 2 decimal places before calculating a probability.What is the summer high temperature that is the 39th percentile of this distribution? 93.4 91.2 90.7 93.9 None of the above What is the probability that a randomly selected summer day has a high temperature of 94? 0.9292 0.0708 0 0.6664 0.3336 What is the probability that a randomly selected summer day has a high temperature greater than 94? 0.0708 0.3336 0.9292 0 0.6664 What is the probability that a randomly selected group of 12 summer days have an average high temperature greater than 94? 0 0.0708 0.6664 0.3336 0.9292Determine the value of the test statistic. F STAT = _____ (Simplify your answer.) Determine the p-value. p-value = (Round to three decimal places as needed.) At the 0.05 level of significance, is there an effect due to factor A? State the result of this test. b. At the 0.05 level of significance, is there an effect due to factor B? Determine the hypotheses.D. H0: There is no effect due to factor B. Determine the value of the test statistic. F STAT = ________ (Simplify your answer.) Determine the p-value. p-value = ________ (Round to three decimal places asneeded.) At the 0.05 level of significance, is there an effect due to factor B? State the result of this test. c. At the 0.05 level of significance, is there an interaction effect? Determine the hypotheses. Determine the value of the test statistic. F STAT = _____ (Simplify your answer.) Determine the p-value. p-value = _____ (Round to three decimal places as needed.) At the 0.05 level of significance, is there…
- GA Golf Scores At a recent PGA tournament (the Honda Classic at Palm Beach Gardens, Florida) the following scores were posted for nine randomly selected golfers for two consecutive days. At α=0.20, is there evidence of a difference in mean scores for the two days? Assume the variables are normally or approximately normally distributed and use μD for the mean on Thursday minus the mean on Friday. Compute the test value. Round the answer to at least three decimal places. Golfer 1 2 3 4 5 6 7 8 9 Thursday 65 64 66 70 68 69 70 68 68 Friday 68 70 69 70 71 72 69 70 75Z score options: -0.13, 0.13, 0.5517, 0.4483 Probability that X is less than 270 options: -0.13, 0.13, 0.5517, 0.44833.03, find the p-value to 4 decimal You are performing a two-tailed test with test statistic z = places
- 4736 You'd like to test the null hypothesis that the means of the two samples (column A and column B) are the same. The alternative hypothesis is that they are not the same. You have no reason to believe that the standard deviations of the two samples are equal. Test at the alpha = 0.10 level. After using Excel, what do you conclude? Are the means the same? Group of answer choices You reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same. You reject the null hypothesis. Therefore, you conclude that the means of the two populations are the same. You cannot reject the null hypothesis. Therefore, you conclude that the means of the two populations are different. X1 X2 97.54 109.27 101.15 97.42 94.22 97.79 98.49 98.93 94.51 108.67 99.72 107.85 96.52 98.51 99.78 102.50 99.64 108.56 103.69…Test the claim that the mean GPA of night students is smaller than 3 at the 0.10 significance level.Based on a sample of 30 people, the sample mean GPA was 2.99 with a standard deviation of 0.03The test statistic is: (to 2 decimals)The p-value is: (to 2 decimals)
