7) Consider the computer output shown below: One-Sample T: Y Test of mu = 91 vs. not = 91 Variable N Mean Std. Dev. SE Mean 95% CI т 25 92.5805 0.4675 (91.6160, ? ) 3.38 0.002 (d) What is the P-value if the alternative hypothesis is H,: u> 91?
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- Q2) The frequency table below shows the Compressive strength of concrete samples results: Compressive strength (MPa) 25 - 34 35 - 44 45 - 54 55 – 64 65 - 74 No. of samples 10 9. 12 8 6. 1- Calculate the Mathematical Mean, Median and Mode. 2- Calculate Variance, Standard Deviation and Mean Absolute Deviation.66. :- . ; The ascorbic acid concentration of five different brands of orange juice was measured. Six replicate samples of each brand were analyzed. The following partial ANOVA table was obtained. Variation Source S df MS F Between juices Within juices 8.45 0.913 Total a. Fill in the missing entries in the table. b. State the null and alternative hypothesis. c. Is there a difference in the ascorbic acid content of the five juices at the 95% confidence level?
- Q2) Which distribution is used to determine whether a correlation coefficient is statistically significant? 1. P-distribution 2. t-test distribution 3. One-sample t-test 4. Chi-squared testsisin, 2 Calculate Quartile Deviation and its Coefficient from the following data : Weight (in pounds) 140 150 160 120 122 124 126 130 No, of Students 1 1 3. 10 SolutianFrom the data please state the intervals and what this tells us about the silver content between coins 1 and 2
- In 2000, the mean math SAT score for students at one school was 472. Five years later, in 2005, a teacher performed a hypothesis test to determine whether the average math SAT score of students at the school had changed from the 2000 mean of 472. The hypotheses were as follows where u is the mean math SAT score, in 2005, for students at the school. Ho: H = 472 Hạ: u # 472 A type Il error would occur if, in fact, p = 472, but the results of the sampling do not lead to rejection of that fact u # 472, and the results of the sampling lead to that conclusion µ # 472, but the results of the sampling fail to lead to that conclusion p = 472, but the results of the sampling lead to the conclusion that p # 4724. Tensile strength tests were carried out on two difference grades of wire rod, resulting in the accompanying data. Tensile strength is normally distributed for both grades of wire rod. Assume the population standard deviations are unequal. df= 22.51 Sample Mean (kg/mm²) 123.6 Grade Sample Size Sample SD AISI 1078 AISI 1064 14 2.0 12 107.6 1.3 c) Estimate with the mean difference in strength between the 1078 grade and the 1064 grade with 90% confidence. d) Does the data provide compelling evidence for concluding that the true average strength of the 1078 grade is higher than 123 kg/mm?? Test the appropriate hypotheses using the p-value approach at the .01 level.A researcher was doing a study for a new Coronavirus Drug and conducted a hypothesis test like this one: H0: μ = 18 Ha: μ ≠ 18 a. What type of test is this? left-tailed right-tailed two-tailed b. She obtained a test statistic of 0.75, with a p-value of 0.4648. So she concluded that her true mean must be equal to 18. Is this interpretation correct? Yes No Why or why not?
- Please answer the incorrect ones only.1.A chi-square test for independence is being used to evaluate the relationship between two variables, one of which is classified into 2 categories and the second of which is classified into 4 categories. What is the df value for the chi-square statistic? a. df = 2 b. df = 6 c. df = 7 d. df = 3A clinical study was conducted on a medicine and the data obtained as follows: calculate Sensitivity Diseased •True Positive: 1000 •True negative: 50 Non diseased •True positive: 100 •True negative: 800