Question 3From the data we find thatSxx = 0.11273, Syy = 11,807,324,805, Sxy = 34,422.28. n=29,test the null hypothesis that p = 0.9 against the alternative that p > 0.9. Use a 0.05 level of significance. Use 2 decimal places for the answers.Critical Region: z>Test statistic: z =P-value: p=Decision: There is certainly some evidence that the correlation coefficient does not exceed 0.9. True or False?Type T or F for the answer.

Calculate the Sample Correlation

Answered: From the data we find that Sxx =…

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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To build a 86% CI for the mean of a population, you take a random sample of 31 values and find s = 18. What is the Margin of Error, E? Round the Critical Value and your answer to three decimals. Submit Question tor
Suppose IQ scores were obtained for 20 randomly selected sets of siblings. The 20 pairs of measurements yield x = 99.37, y = 98.55, r=0.859, P-value = 0.000, and y = 10.94 +0.88x, 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 105? Use a significance level of 0.05. 10CritPearson.pdf O A. 98.55 OB. 0.88 O C. 99.37 O D. 103.34
The p-value for a t-statistic was .008. What is the best way to report the level of significance? p<.008 p<.05 p<.01 p<.001
For each of the following hypotheses in Questions 6-8, (i) provide the p-value (or its bounds) and (ii) draw a basic conclusion (REJECT or DO NOT REJECT only) with respect to Ho with 99% confidence. Handwritten notes are not necessary for these questions.
A doctor tests a claim that an average adult above 50 consumes no more than 3-1/2 pills per day. Standard Dev is 0.7. A sample of 45 adults avg 3.72 per day. Test the claim. -Standard Dev of sample mean -Z-score for sample mean -P-Value
The corrosive effects of various soils on coated and uncoated steel pipe was tested by using a dependent sampling plan. The data collected are summarized below, where d is the amount of corrosion on the coated portion subtracted from the amount of corrosion on the uncoated portion. Does this random sample provide sufficient reason to conclude that the coating is beneficial? Use ? = 0.01 and assume normality. n = 36, Σd = 227, Σd2 = 6244(a) Find t. (Give your answer correct to two decimal places.)(ii) Find the p-value. (Give your answer correct to four decimal places.)(b) State the appropriate conclusion. Reject the null hypothesis, there is significant evidence that the coating is beneficial.Reject the null hypothesis, there is not significant evidence that the coating is beneficial. Fail to reject the null hypothesis, there is significant evidence that the coating is beneficial.Fail to reject the null hypothesis, there is not significant evidence that the coating is beneficial.
n 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 25 30 35 40 45 50 60 70 80 90 100 n x=0.05 0.950 0.878 0.811 0.754 0.707 0.666 0.632 0.602 0.576 0.553 0.532 0.514 0.497 0.482 0.468 0.456 0.444 0.396 0.361 0.335 0.312 0.294 0.279 0.254 0.236 0.220 0.207 0.196 α = 0.05 α = 0.01 0.990 0.959 0.917 0.875 0.834 0.798 0.765 0.735 0.708 0.684 0.661 0.641 0.623 0.606 0.590 0.575 0.561 0.505 0.463 0.430 0.402 0.378 0.361 0.330 0.305 0.286 0.269 0.256 α = 0.01 NOTE: To test Ho: p=0 against H₁: p0, reject Ho if the absolute value of ris greater than the critical value in the table.
Please answer the incorrect ones only.
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Suppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 101.02, y = 100.75, r = 0.852, P-value = 0.000, and y = 5.93 + 0.94x, where x represents the IQ score of the husband. Find the best predicted value of y given that the husband has an lQ of 100? Use a significance level of 0.05. %3D Click the icon to view the critical values of the Pearson correlation coefficient r. The best predicted value of y is Critical values of the pearson correlation coefficient r (Round to two decimal places as needed.) Critical Values of the Pearson Correlation Coefficient r X = 0.05 a = 0.01 INOTE: To test Ho: p=0 n Jagainst H,: p+0, reject Ho if the absolute value of r is greater than the critical value in the table. 4 0.950 0.990 0.878 0.959 0.811 0.917 7 0.754 0.875 0.707 0.834 0.666 0.798 10 0.632 0.765 11 0.602 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 18 0.468 0.590 19 0.456 0.575 20 0.444…

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