It is a z-test. The other picture is the guide. Pls help me answer this. Thank you 1. A researcher claims that the average score of SHS population using a developed problem-solvingtest is equal to 75 and standard deviation of 15. The researcher decides to test 50 randomlyselected SHS students and found out in this sample = 80. Use the 99% confidence level to testthe claim.Answer the following:а. Но:На:b. Level of significancea=c. Type of test.Choices: one-mean, two-tailed t-testtwo-mean, one-tailed t-testone-mean, one-tailed z-testtwo-mean, two-tailed z-testd.Tabulated value and computed valuee. Decisionf. Interpretation tus tfePecision:Example # 1A Barangay Captain from a certain barangay, in Valenzuela City olaims that the average monthly income offamilies with five members from his vicinity is P12.000. But when the City Statisties Office (CSO) conducted asurvey to 10 families with fiye members, to his borangay randomly they found out that they only have anaverage monthly income of 10,800 with a standard deviation of 1.500. with this information the Cso assertthat the claim is not true. Using 0.05 level of significance test the claim of the Brgy. Captain.The average montly weomo of fundos withVHo H = P12, 000 fve members from hs woinily w P12,0004. Area of RejectionThe average monithly inoome of families withHa H# P12, 000 he members from his vionty is not P12,0002. Level of significance0.05-1.961.96Decision Rule:3. Test Statistic:h =Reject H, if the computed valueg is> +1.96 on < -1.96100I test; one mean/s; two tailed testOtherwise do not reject H,Example # 1A Barangay Coptain from a certain barangay in Valenzuela City claims that thefamilies with five members from his vicinity is P12.000. But when the City Statistsurvey to 10b families with five members, to his barangay randomly they foundaverage monthly income of 10.800 with a standard deviation of 1,50. With thisthat the claim is not true. Using 0.05 level of significance test the claim of the I5. Compute for the-value:7. Interpretation:10800 – 12000-8-Rejectionof the ni1500V100the average monthly incmembers from his viciniCalcu: (10800 - 12000) + (1500 +y100)tuo titePecision:Since, the computedbase on the sample of 100 fsignificance. Therefore, the-value -8isthe tabular value (_-1.96_).(rejectless thannot correctTherefore,the null hypothesis(Ho) at,0.05 level of significance.

Sample size is n = 50 Sample mean = 80 Population SD = 15 Confidence level = 99% = 0.99

Answer the following: а. Но: На: b. Level of significance a= c. Type of test. Choices: one-mean, two-tailed t-test two-mean, one-tailed t-test one-mean, one-tailed z-test two-mean, two-tailed z-test d.Tabulated value and computed value e. Decision f. Interpretation

Answer the following: а. Но: На: b. Level of significance a= c. Type of test. Choices: one-mean, two-tailed t-test two-mean, one-tailed t-test one-mean, one-tailed z-test two-mean, two-tailed z-test d.Tabulated value and computed value e. Decision f. Interpretation

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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