Question

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with a population size of 50 i want to test two claims on two differenf variables. the 1st claim is to test if the mean of my sample will be less than or equal to 300. the second calim is the see if the mean is equal to 25. then test them and see if we reject the null or fail to reject.

mean1= 279.99

mean2= 20.1

sd1= 321.88

sd2= 22.08

Answer 1

Answer: (i) 18t claim Crven, mean = 279.99 Standard deveation = 321.88 sample size LA) = 50. The mean of my sample will be or equal to 300. Null and altunative hypotheses: Houssoo Ha: M>300. Testistatistic (2): 2= X-M vn . 279.99-200 321.88 - - 20.01 45,5207061457 = -0.4195801756 N-0.44 [2=-0.44

P-value: C Pralue for the test statistic is the standard P-value=P(2= -0.44) From the nomal table = 0.32997 ~ 0, 3300 IP-value = 0.33007 (&) 18 0.10 Let's fake significance level C for 90s. confidence level) conclusion: It pralue less than the level of significance, than we will rehat the null hypotheses Chol otherwise, we do not refet the rull hypothes at CHO) Hur P-value (0.1800) > & 20.10). Theusson, we do not releet the nell hypothers (Ho) Hence, Accepted Ho. for the 1st celain will se less than the mean of my sample aequal to 300.

dih and charm Given, mean (a) = 20.11 Standard dewatson (o)- 22. of samplesrein) = 50. Mean or equal to 25 Wull and altanative hypotheles: Hoi M = 25 Ha: ufar. Test statistic (2): 2= Tall - 20.1-25 22. of = -1.5692134184 - -4.9 3. 1225435457 ~ -1.57 12=-1,571

P-value; P-value for the test stads stiefs P-value = P(2= -1.57) v From the standard nolmal table. 2 0.05821 ~ 0.0582 [P-value = 0,0582 Conclusion: Here, Prvalue as len than the level of significance Ca=0.10). fon 90%. confidence P-value (0.0582) << 10.10) so, we do not accept the null hypothess. Redeuted null hypothesis, Hency The the and clam, sufficent evidence to relect the null hypothesos. The mean is not equal to 25. (Reset the null hypothesis Ho.