Use technology to find the P-value for the hypothesis test described below.The claim is that for 12 AM body temperatures, the mean is µ > 98.6°F. The sample size is n = 4 and the test statistic ist=2.042.P-value =(Round to three decimal places as needed.)

Sample size(n)=4 Test statistic(t)=2.042Claim :

Answered: Use technology to find the P-value for…

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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A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean, µ, of the population from which the sample was drawn. Use the critical- value approach. Sample mean = 7.5 S = 1.8 n = 18 a = 0.01 %3D Họ: µ = 10 H1: µ <10 1. The test statistics is (Round to three decimal places) 2. The critical value(s) is/are (If there are two critical values separate each with a comma.) 3. The decision is to the null hypothesis. (Enter R if you reject and enter Fif you fail to reject)
Calculate the test statistic and p-value for each sample. Use Appendix C-2 to calculate the p-value. (Negative values should be indicated by a minus sign. Round your test statistic to 3 decimal places and p-value to 4 decimal places.) Test Statistic p-value (a) H0: ππ ≤ .65 versus H1: ππ > .65, α = .05, x = 58, n = 83 (b) H0: ππ = .30 versus H1: ππ ≠ .30, α = .05, x = 18, n = 41 (c) H0: ππ ≥ .10 versus H1: ππ < .10, α = .01, x = 5, n = 101
Also find the p value and the final conclusion
Determine if the statements below are true or false. For each false statement, suggest an alternative wording to make it a true statement. As the degrees of freedom increases, the mean of the chi-square distribution increases. If you found chi2 = 10 with df = 5 you would fail to reject H0 at the 5% significance level When finding the p-value of a chi-square test, we always shade the tail areas in both tails.4. As the degrees of freedom increases, the variability of the chi-square distribution decreases.
The manufacturer claims that your new car gets 32 mpg on the highway. You suspect that the mpg is more for your car. The 69 trips on the highway that you took averaged 35.7 mpg and the standard deviation for these 69 trips was 12.4 mpg. What can be concluded at the αα = 0.05 level of significance? For this study, we should use Select an answer z-test for a population proportion t-test for a population mean The null and alternative hypotheses would be: H0:H0: ? p μ Select an answer ≠ = < > H1:H1: ? μ p 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.) The p-value is ? > ≤ αα Based on this, we should Select an answer reject fail to reject accept the null hypothesis. Thus, the final conclusion is that ... The data suggest that the population mean is not significantly more than 32 at αα = 0.05, so there is statistically insignificant…
How do you find the p-value for this?
A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean, μ, of the population from which the sample was drawn. Use the critical-value approach. Sample mean = 43.9 s = 5.3 n = 15 α = 0.05 H0: µ = 32.6 H1: µ ≠ 32.6 The test statistics is (Round to three decimal places)NOTE: QUESTIONS 9-11 ARE RELATED)
Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is u < 98.6°F. The sample size is n = 8 and the test statistic is t= - 1.664. P-value = (Round to three decimal places as needed.)
A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean, p, of the population from which the sample was drawn. Use the critical-value approach. x= 24.4, s = 9.2, n = 25, Ho: µ= 26, H,: µ < 26, a = 0.05 O A. Test statistic: t= - 0.87. Critical value: t= - 1.708. Do not reject H,. There is not sufficient evidence to conclude that the mean is less than 26. O B. Test statistic: t= - 4.35. Critical value: t= - 1.711. Reject Hp. There is sufficient evidence to conclude that the mean is less than 26. OC. Test statistic: t= - 4.35. Critical value: t= - 1.708. Reject Hn. There is sufficient evidence to conclude that the mean is less than 26. O D. Test statistic: t= - 0.87. Critical value: t= - 1.711. Do not reject Hn. There is not sufficient evidence to conclude that the mean is less than 26.
A sample mean, sample size, and population standard deviation are provided below. Use the one-mean z-test to perform the required hypothesis test at the 1% significance level. x=34, n=20, σ=7, H0: μ=37, Ha: μ≠37 The test statistic is z=___________ (Round to two decimal places as needed.) Identify the critical value(s). Select the correct choice below and fill in the answer box within your choice. (Round to two decimal places as needed.) A. The critical values are ±zα/2=±_________ B. The critical value is zα=____________ C. The critical value is −zα=____________ ________________ (Do not reject, Reject) the null hypothesis. The data ____________ ( provide, do not provide) sufficient evidence to conclude that the mean is ______________ (not equal to 37, less than 37, equal to 37, greater than 37)
Test the hypothesis using the P-value approach. Be sure to verify the requirements of the test. Ho: p = 0.2 versus H,: p> 0.2 n= 125; x= 40; a = 0.05 Click here to view page 1 of the table. Click here to view page 2 of the table. Calculate the test statistic, zg. Zo = (Round to two decimal places as needed.)
Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is µ< 98.6°F. The sample size is n = 7 and the test statistic is t= P-value = (Round to three decimal places as needed.) hs

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Use technology to find the P-value for the hypothesis test described below. The claim is that for 12 AM body temperatures, the mean is μ> 98.6°F. The sample size is n = 4 and the test statistic is t=2.042. P-value= (Round to three decimal places as needed.)