CT 60, 894 NY 1,029,888 MO 245,425 MS 92,831 CO 251,595 ME 60,894 PA 573,815 NE...
|
CT 60, 894 |
NY 1,029,888 |
MO 245,425 |
MS 92,831 |
CO 251,595 |
|
ME 60,894 |
PA 573,815 |
NE 80,432 |
NC 370,883 |
ID 64,514 |
|
MA 380, 277 |
IL 574,429 |
ND 36,863 |
SC 176,365 |
MT 45,359 |
|
NH 70,699 |
IN 252,150 |
SD 38,950 |
TN 246,025 |
UT 119, 240 |
|
RI 47.731 |
MI 407,475 |
AL 165, 716 |
VA 384,479 |
WY 25,707 |
|
VT 31,284 |
OH 477, 536 |
AR 102,755 |
WV 65,138 |
AK 37,053 |
|
DE 42.655 |
WI 246,818 |
FL 876,422 |
AZ 257,505 |
CA 1,858,421 |
|
DC 44,361 |
IA 122,914 |
GA 383,834 |
NM 75,977 |
HI 69,262 |
|
MD 282, 921 |
KS 11,339 |
KY 157,066 |
OK 136,037 |
NV 125,801 |
|
NJ 469,735 |
MN 266,161 |
LA 168,245 |
TX 1,120,665 |
OR 179,015 |
|
WA 350,959 |
What is the confidence interval of 97%? What is the margin of error? Please show your work.
Homework Answers
using excel we have
| Confidence Interval Estimate for the Mean | ||
| formula used | ||
| Data | ||
| Sample Standard Deviation | 331713.9889 | STDEV.S(A2:A53) |
| Sample Mean | 268055.3846 | AVERAGE(A2:A53) |
| Sample Size | 52 | COUNT(A2:A53) |
| Confidence Level | 97% | |
| Intermediate Calculations | ||
| Standard Error of the Mean | 46000.45369 | C4/SQRT(C6) |
| Degrees of Freedom | 51 | C6-1 |
| t Value | 2.2325 | T.INV.2T(1-C7, C11) |
| Margin of Error | 102696.1653 | C12*C10 |
| Confidence Interval | ||
| Interval Lower Limit | 165359.22 | C5-C13 |
| Interval Upper Limit | 370751.55 | C5+C13 |
the confidence interval of 97% is (165359.22,370751.55)
the margin of error is 102696.1653
Add Answer to:
CT 60, 894
NY 1,029,888
MO 245,425
MS 92,831
CO 251,595
ME 60,894
PA 573,815
NE...
and MT NE NV ND NH NJ NM NY NC OH OK OR PA RI SC...
and MT NE NV ND NH NJ NM NY NC OH OK OR PA RI SC SD TN TX UT VA VT WA WV WI WY EM 95813624 7262 1430278 3 22896 03171 12111 1165380 2111211 1.0160 72221 8 tat-AL AK AZ CA CO CN DE DC FL GA HI ID IL IN IA KA KY LA ME MD MA MI MN MS MO ONEC EDA-NSO git M w w M w w E E E E E W W...
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AK 119.1 24.8 5.6 603.3 14.1 4638 69.31
AL 93.3 19.4 4.4 840.9 7.8 2892 69.05
AR 94.1 18.5 4.8 569.6 6.7 2791 70.66
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College Graduation and Other State Characteristics (n = 50, k = 8) omit State ColGrad% Dropout EdSpend Metro% Age LPRFem Neast Seast West Midwest AL 19.8 19.1 1221 89.2 37.4 55.8 0 1 0 0 AK 28.7 8.3 2187 74.7 33.9 65.6 0 0 1 0 AZ 27.9 14.3 1137 96.7 34.5 57.4 0 0 1 0 AR 17.5 18.6 1158 78.8 37.0 54.9 0 1 0 0 CA 30.4 19.7 1657 99.3 34.4 57.6 0 0 1 0 CO...
and MT NE NV ND NH NJ NM NY NC OH OK OR PA RI SC SD TN TX UT VA VT WA WV WI WY EM 95813624 7262 1430278 3 22896 03171 12111 1165380 2111211 1.0160 72221 8 tat-AL AK AZ CA CO CN DE DC FL GA HI ID IL IN IA KA KY LA ME MD MA MI MN MS MO ONEC EDA-NSO git M w w M w w E E E E E W W...
can someone help with step by step answers to these problems
Name: aux pobles Show all relevant work for full credit. Ch.3 08/26/2019 Homework #3 - Math 227 Use the following data to answer the questions. 1.) Consider the following murder rates (per 100.000 people) for each state: AL 13.3 AK 12.9 AZ 941 AR 9.1 FL 11. 0 AZ 9.4 GA 14.4 AR 9.1 ICA 11.7 CO 73 CA 11.7 CT 4.2 TDE 6.7 CO 3 MA 3.7 MI...
CPS 276 DOC02-- Assignment 02 Cost of Living Calculator Your
client maintains job search website. As a service on their website,
they need an application that compares costs of living between
different metropolitan areas. Client will pay $780 for a complete
application. Background Information: A person living in an area
with a high cost of living will need higher income to maintain the
same standard of living as someone living in an area with a low
cost of living. For...
We consider a multiple linear regression model with LIFE (y) as
the response variable, and MALE (x1), BIRTH (x2), DIVO (x3), BEDS
(x4), EDUC (x5), and INCO (x6), as predictors.
"STATE" "MALE" "BIRTH" "DIVO" "BEDS" "EDUC" "INCO" "LIFE"
AK 119.1 24.8 5.6 603.3 14.1 4638 69.31
AL 93.3 19.4 4.4 840.9 7.8 2892 69.05
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