The accompanying data set lists full IQ scores for arandom sample of subjects with medium lead levels intheir blood and another random sample of subjects withhigh lead levels in their blood. Use a 0.05 significancelevel to test the claim that IQ scores of subjects withmedium lead levels vary more than IQ scores of subjectswith high lead levels."U ~12H₁:0² <0²C. Ho: o² = 0²H₁:0² > 0²Identify the test statistic.The test statistic is(Round to two decimal places as needed.)MediumHigh72 90 92 71 86 79 83 114100 93 91 98 91 46 85 78U12H₁: 0² = 0²D. Ho: o² = 0²H₁:0² #0²82 93 85 75 85 80 101 8980 94 88 104 88 88 83

F test are used to understand the variance of two samples are equal or not. It is the type of…

Answered: The accompanying data set lists full IQ…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x=101.22, y=102.45, r=0.916, P-value=0.000, and y=−8.44+1.1x, where x represents the IQ score of the older child. Find the best predicted value of y given that the older child has an IQ of 99? Use a significance level of 0.05.
Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x = 101.13, y = 103.5, r=0.867, P-value = 0.000, and y = 6.25 +0.96x, where x represents the IQ score of the younger child. Find the best predicted value of y given that the younger child has an IQ of 101? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of ŷ is. (Round to two decimal places as needed.)
You recently had a cholesterol panel completed and see that the results for your High Density Lipoprotien (HDL) level comes back with z = -1.7 among people of your stature. Your doctor is going to review the results with you but based on what you know about z-scores you can infer: a.The average person of your stature is 1.7 deviations below you. b.Your HDL level is slightly below average. c.The score is negative so this will be good news. d.Your HDL level is extremely low for a person of your stature.
Cadmium, a heavy metal, is toxic to animals. Mushrooms, however, are able to absorb and accumulate cadmium at high concentrations. Some governments have a safety limit for cadmium in dry vegetables at 0.56 part per million (ppm). A research team measured the cadmium levels in a random sample of edible mushrooms, where the hypothesis test is to decide whether the mean cadmium level in the sample is greater than the governments' safety limit. The null and alternative hypotheses are H0: μ=0.56 ppm, Ha: μ>0.56 ppm. Complete parts (a) through (e) below. d. Now suppose that the results of carrying out the hypothesis test lead to nonrejection of the null hypothesis. Classify that conclusion by error type or as a correct decision if in fact the mean cadmium level in that type of mushrooms equals the safety limit of 0.56 ppm. Choose the correct answer below. A. type I error B. correct decision because a true null hypothesis is not rejected C. correct decision because…
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x= 100.1, y = 98.3, r=0.961, P-value = 0.000, and y = - 18.02 + 1.16x, where x represents the IQ score of the wife. Find the best predicted value of y given that the wife has an IQ of 100? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is (Round to two decimal places as needed.)
Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x = 102.13, y = 101.35, r= 0.927, P-value = 0.000, and y = 2.05 + 0.97x, where x represents the IQ score of the older child. Find the best predicted value of y given that the older child has an IQ of 103? Use a significance level of 0.05. Click the icon to view the critical values of the Pearson correlation coefficient r. Critical Values of the Pearson Correlation Coefficient r The best predicted value of y is (Round to two decimal places as needed.) Critical Values of the Pearson Correlation Coefficient r a = 0.05 a = 0.01 NOTE: To test Ho: p=0 Jagainst H,: p#0, reject H, lif the absolute value of r is greater than the critical value in the table. 4. 0.950 0.990 0.878 0.959 6 0.811 0.917 0.754 0.875 8. 0.707 0.834 0.666 0.632 0.602 9 0.798 10 0.765 11 0.735 12 0.576 0.708 13 0.553 0.684 14 0.532 0.661 15 0.514 0.641 16 0.497 0.623 17 0.482 0.606 0.468 0.456 18 0.590 19 0.575…
A new medication is being developed to treat severe nausea and vomiting among chemotherapy patients. The information below was generated from a small sample size (n = 11) and shows the number of times a chemotherapy patient vomited both before and after treatment due to the side effects of chemotherapy. Set up and interpret the results of a sign test at the alpha equals .05 level to determine if there is a difference in symptoms before and after treatment. Chemotherapy Patient Before Treatment After Treatment Difference Sign 1 9 3 2 12 5 3 14 7 4 9 4 5 8 2 6 11 3 7 15 7 8 12 8 9 7 8 10 9 5 11 8 7 A) We fail to reject H0, which states the median difference in the number of vomiting episodes in a day before and after treatment are equal because 1 is less or equal to critical value 1 for a sign test…
A random sample of vehicle speeds were collected. Use the data to test the claim that the mean vehicle speed is more than 40 mph. Use a 0.05 level of significance. Sample data: x¯=42 mphs=7 mphn=34 Identify the tail of the test. Find the P-value Will the null hypothesis be rejected? Is the initial claim supported?
A data set includes weights of garbage discarded in one week from 62 different households. The paired weights of paper and glass were used to obtain the results shown to the right. Is there sufficient evidence to support the claim that there is a linear correlation between weights of discarded paper and glass? Use a significance level of α = 0.05. Click here to view a table of critical values for the correlation coefficient. Determine the null and alternative hypotheses. Ho: P H₁: p (Type # V 11 > cimals. Do not round.) ... Correlation matrix: Variables Paper Glass 10.1768 Paper Glass 0.1768 1
A data set includes weights of garbage discarded in one week from 62 different households. The paired weights of paper and glass were used to obtain the results shown to the right. Is there sufficient evidence to support the claim that there is a linear correlation between weights of discarded paper and glass? Use a significance level of α = 0.05. Click here to view a table of critical values for the correlation coefficient. Determine the null and alternative hypotheses. Ho: P H₁: P (Type integers or decimals. Do not round.) Identify the test statistic, r. r= (Round to three decimal places as needed.) Identify the critical value(s). (Round to three decimal places as needed.) A. There are two critical values at r = ± B. There is one critical value at r = State the conclusion. Because the absolute value of the test statistic is the positive critical value, there there is a linear correlation between the weights of discarded paper and glass for a significance level of α = 0.05.…
A study was conducted of 90 adult male patients following a new treatment for congestive heart failure. One of the variables measured on the patients was the increase in exercise capacity (in minutes) over a 4-week treatment The previous treatment regime had produced an average increase of μ = 2 minutes. The researchers wanted to evaluate whether the new treatment had increased the value of μ in comparison to the previous treatment. The sample data yielded ?̅ = 2.17 and ? = 1.05: Using α = 05, what conclusions can you draw about the research hypothesis?

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