2. A new insecticidelaboratory test, theimmediately after contact.Is this sufficient evidence to support the advertised claim? Use ax=0.05.(b) Test statistic for the problem above is%00 %0O %020-=0.9181.021is advertised to kill more than 95% roaches upon contact. In ainsecticide was applied to 400 roaches and 384 died- 0.918- 1.021

Hypothesized population proportion Sample size Number of successes Significance level

Answered: 2. A new insecticide laboratory test,…

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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The U.S. Department of Transportation, National Highway Traffic Safety Administration, reported that 77% of all fatally injured automobile drivers were intoxicated. A random sample of 52 records of automobile driver fatalities in a certain county showed that 33 involved an intoxicated driver. Do these data indicate that the population proportion of driver fatalities related to alcohol is less than 77% in Kit Carson County? Use ? = 0.05. (a) level of significance: .05 | H0: p = 0.77; H1: p < 0.77 (b) The standard normal, since np > 5 and nq > 5. What is the value of the sample test statistic? (Round your answer to two decimal places.) Find the P-value of the test statistic. (Round your answer to four decimal places.)
10. Claim: P₁ = P₂; α = 0.05. Sample statistics: x1 = 265, n₁ - X2 204, 2 - 301, 247 'D
Random samples of 324 are taken from a large population and studied. It was found that x = 7.41. If 22.96% of all sample means were greater than 288.6834. What is μ? μ =
A population has a mean of µ = 20. A sample is selected from this population and a treatment is administered to each individual in the sample. The scores from this sample are summarized as follows: n = 16 ,X = 28 ,s2 = 64. Do the sample data support the conclusion that the treatment has a significant effect? Test with α = .05.
A sample of n = 20 scores is transformed in to z-scores. The mean for the 20 z-scores is ____? If a researcher reports t(18) = 1.30, p > .05 for a two-sample independent-measures experiment, the results were significant. T or F?
1.3.2 Consider the output given below that was obtained using the One Proportion applet. Use information from the output to find the standardized statistic for a sample propor- tion value of 0.45. Probability of success (n): 0.30 Sample size (n): 25 Number of samples: 1000 O Animate Draw Samples Total = 1000 180 Mean = 0.301 SD = 0.091 120 60 0.08 0.16 0.24 0.32 0.40 0.48 0.56 0.64 Proportion of success
16. A one-way ANOVA test is conducted at the 10% level of significance and finds the following results: n = 34, K = 4, x̅ = 16.62, MST = 30.303, and MSE = 12.356. What is the tail area of the ANOVA test? A. 0.083 D. 0.097 B. 0.178 C. 0.071
3. On the average 35 % of the people afflicted with a certain disease die from it. Under a new treatment only 22% out of 200 patients die from it. Using a 0.05 level of significance, ca we say that the new treatment is effective?
What is the answer of the 7.14 question?
4. A new medication is being developed to treat severe nausea and vomiting among chemotherapy patients. The information below was generated from a small sample size (n = 11) and shows the number of times a chemotherapy patient vomited both before and after treatment due to the side effects of chemotherapy. Set up and interpret the results of a sign test at the alpha equals .05 level to determine if there is a difference in symptoms before and after treatment. Chemotherapy Patient Before Treatment After Treatment Difference Sign 1 9 3 6 + 2 12 5 8 + 3 14 7 7 + 4 9 4 5 + 5 8 2 6 + 6 11 3 8 + 7 15 7 8 + 8 12 8 4 + 9 7 8 -1 - 10 9 5 4 + 11 8 7 1 + We reject H0 in favor of H1, which states the median difference in the number of vomiting episodes in a day before and after treatment are not equal because 1 is less or equal to the critical…
10. A random sample of n =4 individuals is selected from a population with u = 35, and a treatment is administered to each individual in the sample. After treatment, the sample mean is found to be M = 40.1 with SS = 48. %3| a. How much difference is there between the mean for the treated sample and the mean for the original population? (Note: In a hypothesis test, this value forms the numerator of the t statistic.) b. If there is no treatment effect, how much difference is expected between the sample mean and its population mean? That is, find the standard error for M. (Note: In a hypothesis test, this value is the denominator of the t statistic.) c. Based on the sample data, does the treatment have a significant effect? Use a two-tailed test with a = .05.

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