Construct both a 98% and a 80% confidence interval for β1. β^1=44, se=6.1, SSxx=44, n=23 98% : ≤β1≤ 80% : ≤β1≤
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Construct both a 98% and a 80% confidence interval for β1.
β^1=44, se=6.1, SSxx=44, n=23
98% : ≤β1≤
80% : ≤β1≤
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- A study was conducted that measured the total brain volume (TBV) (in mm³) of patients that had schizophrenia and patients that are considered normal. The first table contains the TBV of the non- schizophrenic patients and the second table contains the TBV of schizophrenia patients. Compute a 99% confidence interval for the difference in TBV of non-schizophrenic patients and patients with Schizophrenia. Prior studies say nothing about the equality of variances. Round to the nearest whole number. Total Brain Volume (in mm³) of Non-Schizophrenic Patients 1371703 1409670 1528399 1541583 1652404 1668272 1606774 1497601 1525728 1487617 1556951 1484397 1455624 1430001 1297891 1365193 1362972 1353428 1371403 1663617 1415307 1756698 1340146 1673225 1456134 1427036 1144683 1542673 1549862 1472556 1483211 1528697 1381758 1265598 Total Brain Volume (in mm³) of Schizophrenia Patients 1348795 1093447 1575546 1183986 1363837 1137626 1342077 1339149 1317181 1570744 1409139 1242976 1511787 1171081…The campus paper reported that the average hours of sleep for students at a neighboring university is 6.5 hours. This has made you curious about the amount of sleep students at your university get. You randomly selected four students on your campus and their amounts of sleep are 5, 3, 6 and 2 and has a sample mean of 4.0 hours. What is the 95% confidence interval for μ? I ended up with an answer of 95% CI [-1,5] Is this correct? do I need to do a hypothesis t-test? It's only asking for the CI so that's what I did.A study was conducted that measured the total brain volume (TBV) (in mm³) of patients that had schizophrenia and patients that are considered normal. The first table contains the TBV of the non- schizophrenic patients and the second table contains the TBV of schizophrenia patients. Compute a 95% confidence interval for the difference in TBV of non-schizophrenic patients and patients with Schizophrenia. Prior studies say nothing about the equality of variances. Round to the nearest whole number. Total Brain Volume (in mm³) of Non-Schizophrenic Patients 1493670 1505354 1642063 1269967 1520437 1581952 1550295 1549683 1338983 1422347 1492283 1195153 1396509 1478392 1485412 1618109 1459455 1717233 1357073 1576932 1417992 1548420 1484842 1595374 1679448 1569374 1370347 1389635 1396599 1533894 1565016 1340012 Total Brain Volume (in mm³) of Schizophrenia Patients 1274453 1212643 1523072 1543675 1466250 1299458 1283251 1592091 1730346 1074105 1362385 1435833 1603966 1523104 1248181 1424469…
- H: H=3 versus H: u# 3 Since the 95% confidence interval (Choose one) Hr then H (Choose one) be rejected at the 0.05 level. contains does not contain Will/will not111) Give a 99.5% confidence interval, for μ1−μ2μ1-μ2 given the following information. n1=30n1=30, ¯x1=2.58x¯1=2.58, s1=0.316s1=0.316n2=40n2=40, ¯x2=2.68x¯2=2.68, s2=0.988s2=0.988 blank ± blank Rounded to 2 decimal places.We have a gantry crane that handles one TEU container in 4 minutes. A new gantry crane developer claims that their machine handles one TEU container in 3 minutes. If this is statistically significant, we will purchase this new machine. We have selected 10 random container handling operation of this machine and tested their difference from the 4 minutes by ONE SAMPLE T TEST in the 5% confidence level. We have found T STAT as -3.323. What are the H0 and Ha for this question? Should we purchase this new machine or not? Why?
- A 95% confidence interval of the mean is (10, 15). The result of testing Ho:H=12 against H:u#12, at 0.05 level of significance, will be Do not reject the null hypothesis because 12 is between 10 and 15 There is insufficient information to make any conclusions about the hypothesis test Reject the null hypothesis because 12 is more than 10 Do not reject the the null hypothesis because 12 is less than 15A magazine provided results from a poll of 2000 adults who were asked to identify their favorite pie. Among the 2000 respondents, 13% chose chocolate pie, and the margin of error was given as ±3 percentage points. What values do p(^), q(^), n, E, and p represent? If the confidence level is 95%, what is the value of α?A study was conducted that measured the total brain volume (TBV) (in mm³) of patients that had schizophrenia and patients that are considered normal. The first table contains the TBV of the non- schizophrenic patients and the second table contains the TBV of schizophrenia patients. Compute a 99% confidence interval for the difference in TBV of non-schizophrenic patients and patients with Schizophrenia. Prior studies say nothing about the equality of variances. Round to the nearest whole number. Total Brain Volume (in mm³) of Non-Schizophrenic Patients 1487997 1611227 1527495 1409812 1297379 1400278 1724615 1249097 1309037 1546157 1352710 1620645 1446958 1540957 1535216 1303042 1363737 1318121 1460479 1602502 1353583 1292976 1558222 1474289 1541397 1373993 1398046 1282528 1516312 1567897 1532468 1381812 Total Brain Volume (in mm³) of Schizophrenia Patients 1653156 1037545 1063003 1287704 1262253 1455977 1438274 1280706 1451005 1283081 1214368 1536994 1082431 1372061 1096716 1699683…
- Suppose you have a model Y = B₁ + B₂X₂ + uż that satisfies all CNLRM assumption with 8 observations, and the o2 = 36. What is the 95% confidence interval of o²? (17.17, 98.19) (18.63, 132.46) (17.75, 145.56) (14.95, 174.62) Do not have enough informationA random sample of size 30 from a normal population yields = 39 and s = 4.9. The lower bound of a 95 percent confidence interval is (Round off upto 2 decimal places).Genetic studies have shown that 4% of Americans with European descent also have African ancestry. A random sample of DNA from 500 Americans has been chosen, calculate a 95% confidence interval for the proportion of Americans that with European descent who also have African ancestry. Genetic studies have shown that 4% of Americans with European descent also have African ancestry. A random sample of DNA from 500 Americans has been chosen, calculate a 95% confidence interval for the proportion of Americans that with European descent who also have African ancestry. (1.92%, 1.581%) (-0.23%, 1.826%) (2.282%, 5.718%) (1.742%, 6.257%)