MyLab Statistics with Pearson eText -- Standalone Access Card -- for Biostatistics for the Biological and Health Sciences (My Stat Lab)
2nd Edition
ISBN: 9780134748870
Author: Marc M. Triola, Mario F. Triola, Jason Roy
Publisher: PEARSON
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Chapter 4.5, Problem 1SLCT
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Calculate the 90% confidence interval for the population mean difference using the data in the attached image. I need to see where I went wrong.
Chapter 4 Solutions
MyLab Statistics with Pearson eText -- Standalone Access Card -- for Biostatistics for the Biological and Health Sciences (My Stat Lab)
Ch. 4.1 - Prob. 1SLCTCh. 4.1 - Prob. 2SLCTCh. 4.1 - Prob. 3SLCTCh. 4.1 - Prob. 4SLCTCh. 4.1 - Prob. 5SLCTCh. 4.1 - Prob. 6SLCTCh. 4.1 - Prob. 7SLCTCh. 4.1 - Prob. 8SLCTCh. 4.1 - Prob. 9SLCTCh. 4.1 - Prob. 10SLCT
Ch. 4.1 - Prob. 11SLCTCh. 4.1 - Prob. 12SLCTCh. 4.1 - Prob. 13SLCTCh. 4.1 - Prob. 14SLCTCh. 4.1 - Prob. 15SLCTCh. 4.1 - Prob. 16SLCTCh. 4.1 - Prob. 17SLCTCh. 4.1 - Prob. 18SLCTCh. 4.1 - Prob. 19SLCTCh. 4.1 - Prob. 20SLCTCh. 4.1 - Prob. 21SLCTCh. 4.1 - Prob. 22SLCTCh. 4.1 - Prob. 23SLCTCh. 4.1 - Prob. 24SLCTCh. 4.1 - Prob. 25SLCTCh. 4.1 - Prob. 26SLCTCh. 4.1 - Prob. 27SLCTCh. 4.1 - Prob. 28SLCTCh. 4.1 - Prob. 29SLCTCh. 4.1 - Prob. 30SLCTCh. 4.1 - Prob. 31SLCTCh. 4.1 - Prob. 32SLCTCh. 4.1 - Prob. 33SLCTCh. 4.1 - Prob. 34SLCTCh. 4.1 - Prob. 35SLCTCh. 4.1 - Prob. 36SLCTCh. 4.1 - Prob. 37SLCTCh. 4.1 - Prob. 38SLCTCh. 4.1 - Prob. 39SLCTCh. 4.1 - Prob. 40SLCTCh. 4.2 - Prob. 1SLCTCh. 4.2 - Prob. 2SLCTCh. 4.2 - Prob. 3SLCTCh. 4.2 - Prob. 4SLCTCh. 4.2 - Prob. 5SLCTCh. 4.2 - Prob. 6SLCTCh. 4.2 - Prob. 7SLCTCh. 4.2 - Prob. 8SLCTCh. 4.2 - Prob. 9SLCTCh. 4.2 - Prob. 10SLCTCh. 4.2 - Prob. 11SLCTCh. 4.2 - Prob. 12SLCTCh. 4.2 - Prob. 13SLCTCh. 4.2 - Prob. 14SLCTCh. 4.2 - Prob. 15SLCTCh. 4.2 - Prob. 16SLCTCh. 4.2 - Prob. 17SLCTCh. 4.2 - Prob. 18SLCTCh. 4.2 - Prob. 19SLCTCh. 4.2 - Prob. 20SLCTCh. 4.2 - Prob. 21SLCTCh. 4.2 - Prob. 22SLCTCh. 4.2 - Prob. 23SLCTCh. 4.2 - Prob. 24SLCTCh. 4.2 - Prob. 25SLCTCh. 4.2 - Prob. 26SLCTCh. 4.2 - Prob. 27SLCTCh. 4.2 - Prob. 28SLCTCh. 4.2 - Prob. 29SLCTCh. 4.2 - Prob. 30SLCTCh. 4.2 - Prob. 31SLCTCh. 4.2 - Prob. 32SLCTCh. 4.2 - Prob. 33SLCTCh. 4.2 - Prob. 34SLCTCh. 4.3 - Prob. 1SLCTCh. 4.3 - Prob. 2SLCTCh. 4.3 - Prob. 3SLCTCh. 4.3 - Prob. 4SLCTCh. 4.3 - Prob. 5SLCTCh. 4.3 - Prob. 6SLCTCh. 4.3 - Prob. 7SLCTCh. 4.3 - Prob. 8SLCTCh. 4.3 - Prob. 9SLCTCh. 4.3 - Prob. 10SLCTCh. 4.3 - Prob. 11SLCTCh. 4.3 - Prob. 12SLCTCh. 4.3 - Prob. 13SLCTCh. 4.3 - Prob. 14SLCTCh. 4.3 - Prob. 15SLCTCh. 4.3 - Prob. 16SLCTCh. 4.3 - Prob. 17SLCTCh. 4.3 - Prob. 18SLCTCh. 4.3 - Prob. 19SLCTCh. 4.3 - Prob. 20SLCTCh. 4.3 - Prob. 21SLCTCh. 4.3 - Prob. 22SLCTCh. 4.3 - Prob. 23SLCTCh. 4.3 - Prob. 24SLCTCh. 4.3 - Prob. 25SLCTCh. 4.4 - Prob. 1SLCTCh. 4.4 - Prob. 2SLCTCh. 4.4 - Prob. 3SLCTCh. 4.4 - Prob. 4SLCTCh. 4.4 - Prob. 5SLCTCh. 4.4 - Prob. 6SLCTCh. 4.4 - Prob. 7SLCTCh. 4.4 - Prob. 8SLCTCh. 4.4 - Prob. 9SLCTCh. 4.4 - Prob. 10SLCTCh. 4.4 - Prob. 11SLCTCh. 4.4 - Prob. 12SLCTCh. 4.4 - Prob. 13SLCTCh. 4.4 - Prob. 14SLCTCh. 4.4 - Prob. 15SLCTCh. 4.4 - Prob. 16SLCTCh. 4.4 - Prob. 17SLCTCh. 4.4 - Prob. 18SLCTCh. 4.5 - Prob. 1SLCTCh. 4.5 - Prob. 2SLCTCh. 4.5 - Prob. 3SLCTCh. 4.5 - Prob. 4SLCTCh. 4.5 - Prob. 5SLCTCh. 4.5 - Prob. 6SLCTCh. 4.5 - Prob. 7SLCTCh. 4.5 - Prob. 8SLCTCh. 4.5 - Prob. 9SLCTCh. 4.5 - Prob. 10SLCTCh. 4.5 - Prob. 11SLCTCh. 4.5 - Prob. 12SLCTCh. 4.5 - Prob. 13SLCTCh. 4.5 - Prob. 14SLCTCh. 4.5 - Prob. 15SLCTCh. 4.5 - Prob. 16SLCTCh. 4.5 - Prob. 17SLCTCh. 4.5 - Prob. 18SLCTCh. 4.5 - Prob. 19SLCTCh. 4.5 - Prob. 20SLCTCh. 4.5 - Prob. 21SLCTCh. 4.6 - Prob. 1SLCTCh. 4.6 - Prob. 2SLCTCh. 4.6 - Prob. 3SLCTCh. 4.6 - Prob. 4SLCTCh. 4.6 - Prob. 5SLCTCh. 4.6 - Prob. 6SLCTCh. 4.6 - Prob. 7SLCTCh. 4.6 - Prob. 8SLCTCh. 4.6 - Prob. 9SLCTCh. 4.6 - Prob. 10SLCTCh. 4.6 - Prob. 11SLCTCh. 4.6 - Prob. 12SLCTCh. 4.6 - Prob. 13SLCTCh. 4.6 - Prob. 14SLCTCh. 4.6 - Prob. 15SLCTCh. 4.6 - Prob. 16SLCTCh. 4.6 - Prob. 17SLCTCh. 4.6 - Prob. 18SLCTCh. 4.6 - Prob. 19SLCTCh. 4.6 - Prob. 20SLCTCh. 4.6 - Prob. 21SLCTCh. 4.6 - Prob. 22SLCTCh. 4.6 - Prob. 23SLCTCh. 4.6 - Prob. 24SLCTCh. 4.6 - Prob. 25SLCTCh. 4.6 - Prob. 26SLCTCh. 4.6 - Prob. 27SLCTCh. 4.6 - Prob. 28SLCTCh. 4.6 - Prob. 29SLCTCh. 4.6 - Prob. 30SLCTCh. 4.6 - Prob. 31SLCTCh. 4.6 - Prob. 32SLCTCh. 4 - Prob. 1CQQCh. 4 - Prob. 2CQQCh. 4 - Prob. 3CQQCh. 4 - Prob. 4CQQCh. 4 - Prob. 5CQQCh. 4 - Prob. 6CQQCh. 4 - Prob. 7CQQCh. 4 - Prob. 8CQQCh. 4 - Prob. 9CQQCh. 4 - Prob. 10CQQCh. 4 - Prob. 1RECh. 4 - Prob. 2RECh. 4 - Prob. 3RECh. 4 - Prob. 4RECh. 4 - Prob. 5RECh. 4 - Prob. 6RECh. 4 - Prob. 7RECh. 4 - Prob. 8RECh. 4 - Prob. 9RECh. 4 - Prob. 10RECh. 4 - Prob. 11RECh. 4 - Prob. 12RECh. 4 - Prob. 13RECh. 4 - Prob. 14RECh. 4 - Prob. 1CRECh. 4 - Prob. 2CRECh. 4 - Prob. 3CRECh. 4 - Prob. 4CRECh. 4 - Prob. 5CRE
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- Microsoft Excel snapshot for random sampling: Also note the formula used for the last column 02 x✓ fx =INDEX(5852:58551, RANK(C2, $C$2:$C$51)) A B 1 No. States 2 1 ALABAMA Rand No. 0.925957526 3 2 ALASKA 0.372999976 4 3 ARIZONA 0.941323044 5 4 ARKANSAS 0.071266381 Random Sample CALIFORNIA NORTH CAROLINA ARKANSAS WASHINGTON G7 Microsoft Excel snapshot for systematic sampling: xfx INDEX(SD52:50551, F7) A B E F G 1 No. States Rand No. Random Sample population 50 2 1 ALABAMA 0.5296685 NEW HAMPSHIRE sample 10 3 2 ALASKA 0.4493186 OKLAHOMA k 5 4 3 ARIZONA 0.707914 KANSAS 5 4 ARKANSAS 0.4831379 NORTH DAKOTA 6 5 CALIFORNIA 0.7277162 INDIANA Random Sample Sample Name 7 6 COLORADO 0.5865002 MISSISSIPPI 8 7:ONNECTICU 0.7640596 ILLINOIS 9 8 DELAWARE 0.5783029 MISSOURI 525 10 15 INDIANA MARYLAND COLORADOarrow_forwardSuppose the Internal Revenue Service reported that the mean tax refund for the year 2022 was $3401. Assume the standard deviation is $82.5 and that the amounts refunded follow a normal probability distribution. Solve the following three parts? (For the answer to question 14, 15, and 16, start with making a bell curve. Identify on the bell curve where is mean, X, and area(s) to be determined. 1.What percent of the refunds are more than $3,500? 2. What percent of the refunds are more than $3500 but less than $3579? 3. What percent of the refunds are more than $3325 but less than $3579?arrow_forwardA normal distribution has a mean of 50 and a standard deviation of 4. Solve the following three parts? 1. Compute the probability of a value between 44.0 and 55.0. (The question requires finding probability value between 44 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 44, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the answer of the second part.) 2. Compute the probability of a value greater than 55.0. Use the same formula, x=55 and subtract the answer from 1. 3. Compute the probability of a value between 52.0 and 55.0. (The question requires finding probability value between 52 and 55. Solve it in 3 steps. In the first step, use the above formula and x = 52, calculate probability value. In the second step repeat the first step with the only difference that x=55. In the third step, subtract the answer of the first part from the…arrow_forward
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