Listed below are amounts of strontium-90 (in millibecquerels or mBq per gram of calcium) in a simple random sample of baby teeth obtai and state B. Use a 0.05 significance level to test the claim that amounts of Strontium-90 from state A residents vary more than amounts f that both samples are independent simple random samples from populations having normal distributions. E Click the icon to view the sample data. Identify the test statistic. F = |(Round to two decimal places as needed.) Use technology to identify the P-value. The P-value is . (Round to three decimal places as needed.) What is the conclusion for this hypothesis test? O A. Fail to reject Ho. There is sufficient evidence to support the claim that amounts of Strontium-90 from state A residents vary more residents. Clrontium 90 from state A residents yary more than ar

Listed below are amounts of strontium-90 (in millibecquerels or mBq per gram of calcium) in a simple random sample of baby teeth obtai and state B. Use a 0.05 significance level to test the claim that amounts of Strontium-90 from state A residents vary more than amounts f that both samples are independent simple random samples from populations having normal distributions. E Click the icon to view the sample data. Identify the test statistic. F = |(Round to two decimal places as needed.) Use technology to identify the P-value. The P-value is . (Round to three decimal places as needed.) What is the conclusion for this hypothesis test? O A. Fail to reject Ho. There is sufficient evidence to support the claim that amounts of Strontium-90 from state A residents vary more residents. Clrontium 90 from state A residents yary more than ar

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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You recently had a cholesterol panel completed and see that the results for your High Density Lipoprotien (HDL) level comes back with z = -1.7 among people of your stature. Your doctor is going to review the results with you but based on what you know about z-scores you can infer: a.The average person of your stature is 1.7 deviations below you. b.Your HDL level is slightly below average. c.The score is negative so this will be good news. d.Your HDL level is extremely low for a person of your stature.
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 4.8 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 242samples is 5.2 ppm with a variance of 1.001.00. Does the data support the claim at the 0.05 level? Assume the population distribution is approximately normal. Find the value of the test statistic. Round your answer to three decimal places. Specify if the test is one-tailed or two-tailed. Determine the decision rule for rejecting the null hypothesis. Round your answer to three decimal places. Make the decision to reject or fail to reject the null hypothesis.
The accompanying data table contains chest deceleration measurements (in g, whereg is the force of gravity) from samples of small, midsize, and large cars. Shown are the technology results for analysis of variance of this data table. Assume that a researcher plans to use a 0.05 significance level to test the claim that the different size categories have the same mean chest deceleration in the standard crash test. Complete parts (a) and (b) below. Click the icon to view the table of chest deceleration measurements A Click the icon to view the table of analysis of variance results. ..... a. What characteristic of the data specifically indicates that one-way analysis of variance should be used? O A. There are three samples of measurements. O B. The population means are approximately normal. O C. The measurements are categorized according to the one characteristic of size. O D. Nothing specifically indicates that one-way analysis of variance should be used. b. If the objective is to test…
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Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.97.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 2424 samples is 8.18.1 ppm with a variance of 0.250.25. Does the data support the claim at the 0.10.1 level? Assume the population distribution is approximately normal. Step 3 of 5: Specify if the test is one-tailed or two-tailed.
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Unfortunately, arsenic occurs naturally in some ground water. A mean arsenic level of u = 8 parts per billion (ppb) is considered safe for agricultural use. A well in Los Banos is used to water cotton crops. This well is tested on a regular basis for arsenic. A random sample of 37 tests gave a sample mean of = 7.3ppb arsenic. It is known that o = 1.9 ppb for this type of data. Does this information indicate that the mean level of arsenic in this well is less than 8 ppb? Use the classical approach. Use a = 0.01 What is the Decision (step 5) for this problem? There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 7.3 ppb. O There is sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. There is not sufficient evidence at the 0.01 level of significance to show the mean arsenic level in the Los Banos well is less than 8.0 ppb. O There is…
You run a PPM correlation on sprint speed and vertical jump height in NFL running backs and find an r value of 0.89. Your sample has a mean sprint speed of 11 m/s ± 0.3 m/s and a mean jump height of 0.49 m ± 0.3 m. What is the effect size estimate? 2.96 0.89 -35.033 35.033 0.267
Our environment is very sensitive to the amount of ozone in the upper atmosphere. The level of ozone normally found is 7.9 parts/million (ppm). A researcher believes that the current ozone level is at an excess level. The mean of 24 samples is 8.1 ppm with a variance of 0.25. Does the data support the claim at the 0.1 level? Assume the population distribution is approximately normal. Step 5 of 5: Make the decision to reject or fail to reject the null hypothesis.