The interval from -2.07 to 1.87 captures an area of 0.95 under the z curve. This implies that another large-sample 95% confidence interval for u has lower limit x - 2.07- 0 you recommend using this 95% interval over the 95% interval x + 1.96- discussed in the text? Explain. (Hint: Look at the width of each interval.) √n * √ The 95% interval discussed in the text has width 3.94 The new 95% interval has width 3.92 the text, so the new interval would not be recommended ✔ . 0 Vn *√n The new interval is wider than 0 and upper limit x + 1.87-. Would Vn ✔the interval discussed in
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- 2.4N(6.2, 2.5)Using the empirical rule, approx what percentage of people spent less than 6.3 hours?Approximately what percentage of people spent more than 8.6 hours?Assume that the entire sample has 8 positive observations and 4 negatives observations. Variable X1: at the left branch has 9 positive observations and 3 negatives observations;at the right branch has 8 positive observations and 8 negative observations. What is the information gain or reduction in uncertainty of X1 using the Gini index? (round to two decimal spaces)
- Cardiologists use the short-range scaling exponent α1, which measures the randomness of heart rate patterns, as a tool to assess risk of heart attack. The article “Applying Fractal Analysis to Short Sets of Heart Rate Variability Data” compared values of α1 computed from long series of measurements (approximately 40,000 heartbeats) with those estimated from the first 300 beats to determine how well the long-term measurement (y) could be predicted the short-term one (x). Following are the data (obtained by digitizing a graph). Short Long 0.54 0.55 1.02 0.79 1.4 0.81 0.88 0.9 1.68 1.05 1.16 1.05 0.82 1.05 0.93 1.07 1.26 1.1 1.18 1.19 0.81 1.19 0.81 1.2 1.28 1.23 1.18 1.23 0.71 1.24 Note: This problem has a reduced data set for ease of performing the calculations required. This differs from the data set given for this problem in the text. Find a 95% confidence interval for the mean long-term measurement for those with short-term measurements…Cardiologists use the short-range scaling exponent α1, which measures the randomness of heart rate patterns, as a tool to assess risk of heart attack. The article “Applying Fractal Analysis to Short Sets of Heart Rate Variability Data” compared values of α1 computed from long series of measurements (approximately 40,000 heartbeats) with those estimated from the first 300 beats to determine how well the long-term measurement (y) could be predicted the short-term one (x). Following are the data (obtained by digitizing a graph). Short Long 0.54 0.55 1.02 0.79 1.4 0.81 0.88 0.9 1.68 1.05 1.16 1.05 0.82 1.05 0.93 1.07 1.26 1.1 1.18 1.19 0.81 1.19 0.81 1.2 1.28 1.23 1.18 1.23 0.71 1.24 Note: This problem has a reduced data set for ease of performing the calculations required. This differs from the data set given for this problem in the text. A. Compute the least-squares line for predicting the long-term measurement from the short-term measurement.…Suppose f(x) = 1/20 from 5 to 25. What is the 25th percentile of X? 10 1/20 None of the other choices is correct .25
- Find a recent journal article(peer reviewed scientific publication) in a field of interest to you which contains references to p-values. I will help you find an article if you want, tell me what you’re interested in. How large is the p-value? Do you think the results of this study are strong? How might they be stronger? Is there anything the authors should have included?For the situation described below, state the null and alternative hypotheses to be tested. (Enter != for ≠ as needed.) A new variety of pearl millet is expected to provide an increased yield over the variety presently in use, which has a mean yield of about 55 bushels per acre.Stoaches are fictional creatures, sometimes mistaken for mome raths. Suppose a two-tailed t-test is conducted to test for a difference between the population means of adult female stoach weights and adult male stoach weights. The female sample has 13 weights, and the male sample has 7 weights. The t-statistic for the test is -0.20542. What is the p-value? (Give your answer to 4 decimal places.) p-value:
- A researcher wants to see if there is a difference in APGAR scores of newborns depending on whether the mother smoked during pregnancy or not. She collects data and analyzes the results using SPSS. What is the correct way to report the results in APA style? Question 23 options: t(58) = 3.288, p>.05 t(58) = 3.288, p= .002 t(26.5) = 3.278, p<.05 t(26.5) =3.278, p=.003A chi‑square statistic (?2) from a 4×3 two-way (contingency) table is 15.15661. Determine the upper and lower limits for the range of ?-values with this chi-square table. Please use the number of decimal places shown in the table. If the value of ?2 is too large to be in the table, use 0 for the lower limit. If the value of ?2 is too small to be in the table, use 1 for the upper limit. lower limit: upper limit:A study is conducted to determine factors in solving missing persons cases. The outcome is Y; which is equal to 1 if the i-th missing person case is unsolved and 0 if it is solved. Let covariate X1; be an indicator equal to 1 if the i-th case is female and 0 for male. Let X2i be a categorical covariate of age group for the i-th case. Categories are less than 14 years old (base level/group), between 14 and 19 years old, and older than 19 years. Set u = P(Y = 1) and the following model is fit: logit(u) = Bo + BịI(Female) + B2I(14 to 19 yrs old) + B3I(> 19 yrs old) The output from the model is below: Estimate S.E. p-value Bo -4.565 0.128 0.380 0.087 B2 Вз -0.198 0.042 0.163 1.128 0.133 a. Using the logistic regression output, calculate the estimated odds ratio of a case being unsolved (again Y=1 is unsolved and Y=0 is solved) comparing females to males (female is the numerator odds) of the same age. Interpret this odds ratio in context of the problem. iii b. Create a 95% confidence…