Find the value of a, b, ..., f in the one-way ANOVA table below. SS df MS F Source 361.5 b 60.25 Treatment a 21 d. Error 821.7 e Total
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- Use the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P82 Question content area bottom Part 1 The z-score that corresponds to P82 is enter your response here . (Round to two decimal places as needed.) z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664…Please help answer all parts of my homework questionchoose the best answer, thank you much
- Answer correctly Please provide typed answer. ONLY TYPED ANSWERSUse the standard normal table to find the z-score that corresponds to the given percentile. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. P20 z 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 −3.4 0.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 −3.3 0.0003 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0005 0.0005 0.0005 −3.2 0.0005 0.0005 0.0005 0.0006 0.0006 0.0006 0.0006 0.0006 0.0007 0.0007 −3.1 0.0007 0.0007 0.0008 0.0008 0.0008 0.0008 0.0009 0.0009 0.0009 0.0010 −3.0 0.0010 0.0010 0.0011 0.0011 0.0011 0.0012 0.0012 0.0013 0.0013 0.0013 −2.9 0.0014 0.0014 0.0015 0.0015 0.0016 0.0016 0.0017 0.0018…Katie is studying aggression among adolescent girls. She believes that there is a relationship between the level of interaction a girl has with her mother and the level of aggression. She has identified 5 girls who fall into each of 4 interaction level and has measured their aggression scores. Her data are given below. Complete table below by filling in boxes with correct value. Show calculation. No Low Moderate High interaction interaction interaction interaction A, A A1 A2 A3 A4 4 16 4 16 4 16 25 36 4 16 4 16 36 25 3 4 16 36 4 16 25 3. 5 25 25 4 16 3 9 Σ Α82 n2=5; n3=5; n4=5; N= O; X = D; X;=4.80; X 3=4.00; X 4=3.40 Σ Α-26 Σ Α-138 Σ Α Σ Α-118 Σ A-20 Σ Α-17 ΣΑ ; ΣΑΟ397; ni- tot
- The height of women ages 20-29 is normally distributed, with a mean of 63.6 inches. Assume o = 2.5 inches. Are you more lrely to randomly select 1 woman with a height less than 64.1 inches or are you more likely to select a sample of 20 women with a mean height less than 64.1 inches? Explain. E Click the icon to view page 1 of the standard normal table. E Click the icon to view page 2 of the standard normal table. What is the probability of randomly selecting 1 woman with a height less than 64.1 inches? (Round to four decimal places as needed.) Enter your answer in the answer box and then click Check Answer. Check Answer Clear All parts 2 FemainingOlease show all your workData were collected on the heights of singers' and are summarized and displayed below. Does average height differ by voice? mean sd alto 64.88571 2.794653 35 bass 70.71795 2.361408 39 soprano 64.25000 1.872737 36 tenor 69.15000 3.216323 20 bass tenor alto soprano (a) Are the conditions for using the F-distribution satisfied? Yes No Height (inches) 60 65 70 75
- Consider the following ANOVA table below. Some parts are blank intentionally. The number of observations is 249. Source Model (Regression) Residual (Error) Total The MSE is: SS 1,332 DF MS 921 Round your answer to two decimal places.A company manufactures printers and fax machines at plants located in Atlanta, Dallas, and Seattle. To measure how much employees at these plants know about quality management, a random sample of 6 employees was selected from each plant and the employees selected were given a quality awareness examination. The examination scores for these 18 employees are shown in the following table. The sample means, sample variances, and sample standard deviations for each group are also provided. Managers want to use these data to test the hypothesis that the mean examination score is the same for all three plants. Sample mean Sample variance Sample standard deviation Plant 1 Atlanta 84 76 81 76 70 81 78 25.2 5.02 Plant 2 Plant 3 Dallas Seattle 72 74 72 74 70 88 75 42.8 6.54 60 64 62 68 76 66 66 32.0 5.66What is the mean of the standard normal distribution? 0 01 OZ