Check whether the data provide sufficient evidence to conclude that the mean number of seeds detected differs for the two methods.

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Blood cocaine concentration (mg/L) was determinedboth for a sample of individuals who had died fromcocaine-induced excited delirium (ED) and for a sampleof those who had died from a cocaine overdose withoutexcited delirium; survival time for people in bothgroups was at most 6 hours. The accompanying datawas read from a comparative boxplot in the article“Fatal Excited Delirium Following Cocaine Use” (J.of Forensic Sciences, 1997: 25–31). ED 0 0 0 0 .1 .1 .1 .1 .2 .2 .3 .3.3 .4 .5 .7 .8 1.0 1.5 2.7 2.83.5 4.0 8.9 9.2 11.7 21.0Non-ED 0 0 0 0 0 .1 .1 .1 .1 .2 .2 .2.3 .3 .3 .4 .5 .5 .6 .8 .9 1.01.2 1.4 1.5 1.7 2.0 3.2 3.5 4.14.3 4.8 5.0 5.6 5.9 6.0 6.4 7.98.3 8.7 9.1 9.6 9.9 11.0 11.512.2 12.7 14.0 16.6 17.8 a. Determine the medians, fourths, and fourth spreadsfor the two samples.b. Are there any outliers in either sample? Any extremeoutliers?c. Construct a comparative boxplot, and use it as abasis for comparing and contrasting the ED andnon-ED samples.

Answer 1

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Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 11, Problem 71CR

To determine

Check whether the data provide sufficient evidence to conclude that the mean number of seeds detected differs for the two methods.

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Blood cocaine concentration (mg/L) was determinedboth for a sample of individuals who had died fromcocaine-induced excited delirium (ED) and for a sampleof those who had died from a cocaine overdose withoutexcited delirium; survival time for people in bothgroups was at most 6 hours. The accompanying datawas read from a comparative boxplot in the article“Fatal Excited Delirium Following Cocaine Use” (J.of Forensic Sciences, 1997: 25–31). ED 0 0 0 0 .1 .1 .1 .1 .2 .2 .3 .3.3 .4 .5 .7 .8 1.0 1.5 2.7 2.83.5 4.0 8.9 9.2 11.7 21.0Non-ED 0 0 0 0 0 .1 .1 .1 .1 .2 .2 .2.3 .3 .3 .4 .5 .5 .6 .8 .9 1.01.2 1.4 1.5 1.7 2.0 3.2 3.5 4.14.3 4.8 5.0 5.6 5.9 6.0 6.4 7.98.3 8.7 9.1 9.6 9.9 11.0 11.512.2 12.7 14.0 16.6 17.8 a. Determine the medians, fourths, and fourth spreadsfor the two samples.b. Are there any outliers in either sample? Any extremeoutliers?c. Construct a comparative boxplot, and use it as abasis for comparing and contrasting the ED andnon-ED samples.

Suppose the National Transportation Safety Board (NTSB) wants to examine the safety of compact cars, midsize cars, and full-size cars. It collects a sample of three for each of the treatments (cars types). Using the hypothetical data provided below, test whether the mean pressure applied to the driver's head during a crash test is equal for each types of car. Use a = 0.05, ssw=6300 and SST=6350. 1) The null and the alternative hypotheses are : a) Ho: H1 = H2 = Hg. H: At most one of uj:j = 1,2,3 is different. H : At least one of u;:j = 1,2,3 is different. H: At least one of H;:j = 1,2,3 is different. H: 41 = 42 = Hz. vs b) Ho: 41 = H2 = Hg. Vs %3D Vs d) Ho: At least one of u;:j = 1,2,3 is different. %3D VS 2) The F test statistics= a) 0.0228 b) 0.0218 c) 0.0208 d) 0.0238 3) The critical value = a) 4.3433 b) 5.5433 c) 5.1433 d) 5.3433 4) The decision of the test is: a) Do not reject Ho. There is insufficient evidence that at least one of the population means is different. b) Reject Hg.…

Five samples of a ferrous-type substance were used to determine if there is a difference between a laboratory chemical analysis and an X-ray fluorescence analysis of the iron content. Each sample was split into two subsamples and the two types of analysis were applied, with the accompanying results. Assuming that the populations are normal, test at the 0.02 level of significance whether the two methods of analysis give, on the average, the same result. Click here to view the sample analyses. Click here to view page 1 of the table of critical values of the t-distribution. Click here to view page 2 of the table of critical values of the t-distribution. Let sample 1 be the X-ray fluorescence results and let sample 2 be the laboratory chemical results. State the null and alternative hypotheses. Ho: HD H1: HD + Identify the critical region. Select the correct choice below and fill in the answer box(es) to complete your choice. (Round to three decimal places as needed.) O A. t> В. t O C. t<

Answer 2

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Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 11, Problem 71CR

To determine

Check whether the data provide sufficient evidence to conclude that the mean number of seeds detected differs for the two methods.

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Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 11, Problem 71CR

To determine

Check whether the data provide sufficient evidence to conclude that the mean number of seeds detected differs for the two methods.

Expert Solution & Answer

Trending nowThis is a popular solution!

See solution

Check out sample textbook solution

Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 11, Problem 71CR

To determine