(d) Suppose it had previously been believed that when air void increased by 1 percent, the associated true average change in dielectric constant would be at least -0.055. Does the sample data contradict this belief? Carry out a test of appropriate hypotheses using asignificance level of 0.01.State the appropriate null and alternative hypotheses.Ho: P1-0.055H:P <-0.055Ho: P₁ = -0.055H:A-0.055OH: P₁= -0.055H:A-0.055ⒸHO: ₁-0.055H: P, = -0.055Compute the test statistic value and find the P-value. (Round your test statistic to two decimal places and your P-value to three decimal places.)t =P-value=State the conclusion in the problem context.O Reject Ho. There is insufficient evidence to contradict the prior belief.O Reject Ho. There is sufficient evidence to contradict the prior belief.O Fail to reject Ho. There is insufficient evidence to contradict the prior belief.O Fail to reject H. There is sufficient evidence to contradict the prior belief. A statistical program is recommended.Electromagnetic technologies offer effective nondestructive sensing techniques for determining characteristics of pavement. The propagation of electromagnetic waves through the material depends on its dielectric properties. The following data, kindly provided by theauthors of the article "Dielectric Modeling of Asphalt Mixtures and Relationship with Density," was used to relate y dielectric constant to x = air void (%) for 18 samples having 5% asphalt content.y 4.55 4.49 4.50 4.47 4.47 4.45 4.40 4.34 4.43 4.43 4.42 4.40 4.33 4.44 4.40 4.26 4.32 4.34x 4.35 4.79 5.57 5.20 5.07 5.79 5.36 6.40 5.66 5.90 6.49 5.70 6.49 6.37 6.51 7.88 6.74 7.08The following R output is from a simple linear regression of y on x.Std. ErrorEstimate4.8586910.059768-0.0746760.009923(Intercept)AirVoidResidual standard error: 0.03551 on 16R-squared: 0.7797,F-statistic: 56.63 on 1 and 16 DF,Analysis of Variance TableResponse: Dielectric.Df1AirvoidResiduals 16Sum Sq0.0714220.20178Mean Sq0.0714220.001261t value.81.283-7.526Pr (>|t|)<2e-161.21e-06DF Multiple.Adjusted R-squared: 0.766p-value: 1.214e-06F value56.635Pr (>F)1.214e-06

The estimate of air void is -0.074676 and corresponding standard error is 0.009923.

Answered: (d) Suppose it had previously been…

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 wish to test the following claim at a significance level of α=0.001α=0.001. H0:p=0.68H0:p=0.68 H1:p≠0.68H1:p≠0.68You obtain a sample of size n=393n=393 in which there are 283 successful observations.What is the critical value for this test? (Report answer accurate to 2 decimal places.) ±±What is the test statistic for this sample? (Report answer accurate to 3 decimal places.) The test statistic is in the critical region not in the critical region This test statistic leads to a decision to reject the null accept the null fail to reject the null As such, the final conclusion is that there is sufficient evidence to conclude that the population proportion is not equal to 0.68. there is not sufficient evidence to conclude that the population proportion is not equal to 0.68. there is sufficient evidence to conclude that the population proportion is equal to 0.68. there is not sufficient evidence to conclude that the population proportion is equal to 0.68.
Currently patrons at the library speak at an average of 64 decibels. Will this average change after the installation of a new computer plug in station? After the plug in station was built, the librarian randomly recorded 69 people speaking at the library. Their average decibel level was 61.7 and their standard deviation was 6. What can be concluded at the α = 0.05 level of significance? H0:H0: ? μ p Select an answer > ≠ < = H1:H1: ? p μ Select an answer = > < ≠ The test statistic ? t z = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.)
You wish to test the following claim (HaHa) at a significance level of α=0.002α=0.002. Ho:μ1=μ2 Ha:μ1≠μ2You believe both populations are normally distributed and that the population standard deviations are known. You obtain a sample of size n1=42 with a mean of ¯x1=69.5 and a population standard deviation of σ1=16.2 from the first population. You obtain a sample of size n2=55 with a mean of ¯x2=75.4 and a population standard deviation of σ2=17.9 from the second population.What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = What is the p-value for this sample? (Report answer accurate to four decimal places.) p-value = The p-value is... less than (or equal to) αα greater than αα This test statistic leads to a decision to... reject the null accept the null fail to reject the null As such, the final conclusion is that... There is sufficient evidence to warrant rejection of the claim that the first population…
Answer part d
The claim is that for 10 AM body temperatures of males, the mean is less than 98.6°F. The sample size is n=39 and the test statistic is t=−4.041. Use technology to find the P-value. Based on the result, what is the final conclusion? Use a significance level of 0.01. Question content area bottom Part 1 State the null and alternative hypotheses. H0: μ ▼ greater than> less than< not equals≠ equals= enter your response here H1: μ ▼ greater than> less than< less than or equals≤ not equals≠ equals= greater than or equals≥ enter your response here (Type integers or decimals. Do not round. Do not include the degree symbol in your answer.) Part 2 The test statistic is enter your response here. (Round to two decimal places as needed.) Part 3 The P-value is enter your response here. (Round to three decimal places as needed.) Part 4 Based on the P-value, there ▼ is is not sufficient evidence at a significance level of 0.01 to…
AutoSave Off Ch13 60v13_10e (1) - File Home Insert Draw Page Layout Formulas Data Review View à From Text/CSV là Recent Sources T Queries & Connections là From Web LA Existing Connections Properties Get Refresh Data A From Table/Range All - A Edit Links Get & Transform Data Queries & Connections v X v fx H19 C D F ANOVA table 2 Source SS df MS 3 Regression 4 Residual 5 Total 1870.5782 1 1870.5782 41.39 1265.4934 28 45.1962 3136.0716 29 6. 7 8 Regression output 9 Variables 10 Intercept Coefficients Std. Error t(df=28) 13.76015 3.106 2.914 11 Distance-X 3.77085 7.977 6.427 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26
We want to use a z-test to determine if the sample mean is greater than 20 at the alpha = 0.05. What is the critical value(s)?
If you collect data from a random sample of 10 students and we would like test for statistical significance at the 0.05 level, what critical r value must be obtained? O .602 O.632 0.708 O .576
In a two-sided test: (H₁: μ ‡ claimed value or Ho:p ‡ claimed value) If the statement of the problem does not indicate which alternate hypothesis to use then run a two-sided test. 2. Identify the significance level (c). If the problem does not give one then take a = 0.5 3. Find the critical value. You need to determine if you are using a Z dsitribution (proportions or mean where is known) or the t-distribution (mean where you approximate with s). For a left sided test: CV = invNorm(a) or invT(α, n-1) where n is the sample size (n-1 = degrees of freedom). For a right sided test: CV = invNorm(1-x) or invT(1-a, n-1) where n is the sample size (n-1 = degrees of freedom). For a two-sided test: CV's (there are TWO) = + invnorm() or invT(n-1)where n is the sample size (n-1 = degrees of freedom). 4. Find the test statistic For means: Z = or t = where is the assumed parameter in the null hypothesis. Vi For proportions: Z=-, where p is the assumed parameter in the null hypothesis. DI V 5. Find…

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