Question

Consider the following hypotheses: Ho: 42 HA: 42 Find the p-value for this test based on the following sample information. (You may find it useful to reference the appropriate table: Z table or t table) a. 40; S = 10.4; n = 30 p-value 2 0.10 p value < 0.01 0.02 s p-value <0.05 0.05 s p-value <0.10 0.01 s p-value < 0.02 b. I = 44; s = 10.4; n = 30 p-value 2 0.10 0.05 s p value <0.10 0.02 s p-value < 0.05 0.01 s p-value < 0.02 p-value < 0.01

Answer 1

ho u = 42 Mai li tu2 this two tailed test. a test find out statisties (t- Test ) As we trest = X-u slin Pralue will be for two tailed test Pvalue = P(t It rest! Pvalue P(teattest ) +B ( ty trest ) distribution due to symmetrical ap(te-treat) Palue using Now Values. S 10.4 n = 30 a x = 40 ५० - ५ X-1 - Trest= 10.4 slun 4153 Trest 1.05 ul Pralue 2 pl t <-1.05) df= n-1 29 at df= 29 using t - Table Prabre a O. 3024 IPvalue 0.10 option otion a So

* = ༥༥ Sa 10.4 na 30 Jou t rest ५५-५१ sign 10.41530 trest a 1.05 Praluez sp( tc- trest) QPC tel.05) af- n-7 = 29 again wsing t- table Puailiuvez o. 3oqu 501 Pralue 20.10 option a x= 39 So 9.7 na 27 trest X - el 39-48 = slin 9.71527 t Test -1.61 Pralue= 2P (t< 1.61) nela 26 using t- Table Pvalue= o.1195 Puolue 118 0.10 ? option a

a ña 38 n=35 S = 9.7 39-१२ trest X-1 1 slun ۹ 4.71535 tresta -2.44 Pvalues Q Plth-2.44) dfa n-1 = 34 at df=34 using t- table Pralue= 0.02005 and 0.005 0.02 Honce bale between is 0.02 < P-value 0.05