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- Suppose IQ scores were obtained for 20 randomly selected sets of twins. The 20 pairs of measurements yield x=98.34, y=100.45, r=0.940, P-value=0.000, and y=−3.86+1.06x, where x represents the IQ score of the twin born second. Find the best-predicted value of y given that the twin born second has an IQ of 103? Use a significance level of 0.05.In a certain school district, it was observed that 24% of the students were classified as only children (only siblings). However, in the special program for talented and gifted children, 80 out of 286 are only children. The school administrators want to know if the proportion of only the children in the special program is significantly different from the proportion for the school district. Test at the a = 0.02 level of significance. c. Hypotheses for this situation Using the normal approximation for the binomial distribution (without the continuity correction), was is the test statistic for the sample based on the sample proportion? z = To 3 decimal places P-value = To 4 decimal placesSuppose IQ scores were obtained for 20 randomly selected sets of couples. The 20 pairs of measurements yield x = 101.16, y = 102.3, r= 0.810, P-value = 0.000, and y = 23.09+0.78x, where x represents the IQ score of the wife. Find the best %3D %3D %3D predicted value of y given that the wife has an IQ of 109? 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.) Critical values of the pearson correlation coefficient r NOTE: To test Ho: p = 0 against H,: p#0, reject Ho |if the absolute value of r is greater than the critical value in the table. a = 0.05 a = 0.01 4 0.950 0.990 0.878 0.959 0.811 0.917 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 0.561 25 0.396 0.505 30 0.361 0.463 35 0.335 0.430 40…
- In a random sample of males, it was found that 25 write with their left hands and 217 do not. In a random sample of females, it was found that 58 write with their left hands and 450 do not. Use a .05 significance level to test the claim that the rate of left-handedness among males is less than that among females. Identify the test statistic.z=?Suppose that there are two borrowing strategies from a commercial bank as short-term and long-term. We have two small samples (n1=8 and n2=8) and sampled populations are normal. Standard deviation for the first sample is 200 and for the second one is 150. The researcher wants to determine whether the variation in the customers preferring short-term borrowing differs from the variation in the customers preferring long-term borrowing. (Use 0.10 significance level)a. Find the t values that form the boundaries of the critical region for a two-tailed test with ? = .05 for a sample size of n = 12. b. Repeat the above for a one-tailed test, with ? = .05.
- In 1990, the average duration of long-distance telephone calls originating in one town was 9.4 minutes. A long-distance telephone company wants to perform a significance test to determine whether the average duration of long-distance phone calls has changed from the 1990 mean of 9.4 minutes. The hypotheses are: H0: μ = 9.4 minutes HA: μ ≠ 9.4 minutesSuppose that the results of the sample lead to non-rejection of the null hypothesis. Classify that conclusion as a Type I error, a Type II error, or a correct decision, if in fact the mean duration of long-distance phone calls has changed from the 1990 mean of 9.4 minutes. Choose the best asnwer below: A. Neither B. Type I error C. Type II error D.No errorThe 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.Recently, a simple random sample of 572 adult males showed that 124 of them smoke. It has been established that in 2008. 20.4% of adults smoke. Use a 0.05 significance level to test the claim that the rate of smoking by adult males is now the same as in 2008. What is the value of the critical z tabular value?
- 5. We don't want predictor variables that are too highly correlated with each other or they are redundant (say ≥±.70) or multicollinear (2±.90), which means the variables are measuring pretty much the same thing so why have them all in the equation. What was the finding for redundancy between SPI Scientist and SPI Practitioner? a. No issue with redundancy since the shared variance between the two variables was only 4% (r2 = .04). b. There was an issue with redundancy between the two variables shared 62% of the variance in the relationship.You have data drawn from a normal distribution with a known variance of 16. You set up the following NHST: • Ho: data follows a N(2, 4²) • HÃ: data follows a N(µ, 4²) where µ ‡ 2. Test statistic: standardized sample mean z. Significance level set to a = .05. You then collected n = 16 data points with sample mean 1.5. (a) Find the rejection region. Draw a graph indicating the null distribution and the rejection region. (b) Find the z-value and add it to your picture in part (a). (c) Find the p-value for this data and decide whether or not to reject Ho in favor of HA.Sgdhfjyjygfgdvdvrhhti5oyoggadadsbgngbdfegrhfgrhefefegthgjglhlyhsdCfbvngkyhadafdhfjgkghadqsagdjgkhdvsvdbgnfvsvdbdbdbdbfbdvdgdgeggdgc fsvdbfhth sbdvdgrgrhg13. Six random determinations of sulfur content in steel at a particular point in a process gave the values 3.07, 3.11, 3.14, 3.24, 3.16, and 3.08. Assume the values are normally distributed. A previous study based on a sample of 21 random observations gave an estimate of variance of 1.51 × 10–3. Use the 10% level of significance. a. Compute for the new sample variance. b. Using the variance ratio, is the variance significantly higher now? Justify your answer.grhhjhmhbdcscfngbsvngbevfngnfjgjyoisfaxacsvdbfhfbdvsvsvdvdvdbdbźvdbfngnmklljfhsfadsfsvdbfbfngngkyoosdadwdsvdbfnfnfjgnjfjdvevdbbfngntjfhrhrhdgdgdvdhdhdbgdvgdfbfhfhtj