6. A random sample is selected from a normal popula-tion with a mean of u = 30 and a standard deviationof o = 8. After a treatment is administered to theindividuals in the sample, the sample mean is found%3Dto be M = 33.%3Da. If the sample consists of n = 16 scores, is thesample mean sufficient to conclude that the treat-ment has a significant effect? Use a two-tailed testunwith a = .05. b. If the sample consists of n = 64 scores, is thesample mean sufficient to conclude that the treat-ment has a significant effect? Use a two-tailed testwith a = .05.c. Comparing your answers for parts a and b, ex-plain how the size of the sample influences theoutcome of a hvpothesis test.111

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Answered: A random sample is selected from a…

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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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…
Currently patrons at the library speak at an average of 61 decibels. Will this average increase after the installation of a new computer plug in station? After the plug in station was built, the librarian randomly recorded 57 people speaking at the library. Their average decibel level was 63.8 and their standard deviation was 9. What can be concluded at the the αα = 0.10 level of significance?
We collected a sample of twenty-five observations. We assumed that the observations are from a population whose values follow a normal distribution with a mean of u and a standard deviation of five. We performed a test with these hypotheses Ho: μ = 0H₁ = 0 and H₁ : μ0H₁0. The test statistic is the sample mean . The 5% rejection region is 1.959964 > 1.959964. What is the power of this test when μ = -2μ = -2? Please state the statistical power up to the fourth decimal place.
It takes an average of 10.8 minutes for blood to begin clotting after an injury. An EMT wants to see if the average will increase if the patient is immediately told the truth about the injury. The EMT randomly selected 46 injured patients to immediately tell the truth about the injury and noticed that they averaged 11.7 minutes for their blood to begin clotting after their injury. Their standard deviation was 2.98 minutes. What can be concluded at the the αα = 0.05 level of significance? The null and alternative hypotheses would be: Please provide number H0:H0: mean =Correct __?__ H1:H1: mean > Correct_ __?__ The test statistic t = __?__ (please show your answer to 3 decimal places.) The p-value = __?__ (Please show your answer to 4 decimal places.)
The mean age of a random sample of 70 adults is 55yrs with SD=7yrs •Find a 95%CI, 90% & a 99%
process has been developed that can transform ordinary iron into a kind of super iron called metallic glass. Metallic glass is three to four times stronger than the toughest steel alloys. To estimate the mean temperature at which a particular type of metallic glass becomes brittle, 20 pieces of this material were randomly sampled from a recent production run. Each piece was subjected to higher and higher temperatures until it became brittle. The temperature at which brittleness first appeared was recorded for each piece in the sample. The result suggests that the average temperature which brittleness first appeared is about 472 degrees Fahrenheit with a standard deviation of 28 degrees Fahrenheit. Calculate the length of the 95% confidence interval for the mean temperature at which a particular type of metallic glass becomes brittle. Note: Round your answer to 4 decimal places.
A trial study suggest that the standard deviation for the half life for the drug chlordiazepoxide is 8 hours. With 99% confidence, find the sample size needed to determine the mean half life for the drug with a margin of error of 3 hours.
A student conducts an experiment on high temperature steel. The student takes an initial set of 20 measurements and calculates a sample average of 660 oF and a sample deviation of 15.3 oF. How confident is the student that the mean temperature of the steel is above 600 oF?
A series of measurements in the lab led to an experimental result of 32.9 mL, with a calculated standard deviation of 0.3 mL. What is the standard way to report this result? Select one: 32.6-33.2 mL 32.9 ± 0.3 mL 32.9 ± 0.30 mL = 32.9 mL
The average local cell phone call length was reported to be 2.27 minutes. A random sample of 10phone calls showed an average of 2.98 minutes in length with a standard deviation of 0.98 minute. At alpha= 0.01can it be concluded that the average differs from the population average?
The braking ability was compared for two car models. Random samples of two cars were selected. The first random sample of size 64 cars yield a mean of 36 and a standard deviation of 8. The second sample of size 64 yield a sample mean of 33 and a standard deviation of 8. Do the data provide sufficient evidence to indicate a difference between the mean stopping distances for the two models? Use Alpha= 0.01. Ho: µ1 – µ2 = 0 vs. Ha: µ1 – µ2 + 0 .p-value = 0.0017. Reject Ho Но: Д, — м2 — 0 vs. Ha:M1 — M2 + 0. p-value — 0.034. Ассеpt Ho O Ho:µ1 – Hz Но: И — 2 — 0 vs. Ha:M, — нz + 0. p-value —D 0.0017. Ассept Ho Ho: H1 – U2 = 0 vs. Ha: µ1 – µ2 + 0. p-value = 0.034. Reject Ho
A research team measured tidal volume in 15 experimental animals , the mean and standard deviation were 45 and 5 cc, respectively. Do these data provide sufficient evidence t indicate that the population mean is greater than 40 cc? Let α = 0.05 *with steps please.

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A random sample is selected from a normal popula- tion with a mean of u = 30 and a standard deviation of o = 8. After a treatment is administered to the %3D individuals in the sample, the sample mean is found to be M = 33. %3D