Question

10.) to test Ho: u=35 versus H1: u unequal to 36, a simple random sample size of n=35 is obtained. complete parts a through f

To test H 36 versus Hy736, a simple random sample of size 35 is obtained. Complete parts (a) through in below. Click the loon to view the table of critical values (a) Does the population have to be normally distributed to test this hypothesis by using diatribution methode? Why? O A No--there are no constraints in order to perform a hypothesis test. OB. Nosince the sample sire is at least 30, the underlying population does not need to be normally distributed OC. Yes the population must be normally distributed in al cases in order to perform a hypothesis test OD. Yosince the sample size is ut not least 50, the underlying population does not need to be normally darbuled. (W40.2 and 9.9. compute the test statistie (Round to two decimal places as needed) (c) Draw a distribution with the area that represents the value shaded. Choose the correct gw below. ОА OB oc A A A (d) Approximate and interpret the P-value Click to select your answers

Answer 1

W = 36 Hoi H. M & 36 35 ( sample size) NO since the size us a 30 sample the doe not need clytributed atleast population normally underlying to be į = 46.2 do 9.9. © to І - и Slen 40.2-36 9.9/135 4.2 2.51 1.67340542 O -3 2 b is Option test The corred. two sided I

@ p-value is = 2P ( + 74.51) 2 (0,0085 35 -|- = 0.0170 34 at degree of freedom So p-i value is 0.017 A value B Belowe the p.value 1 gaeater than 2-0.012 is the regeit reseacha the hull The dreas cher null will not hypothes will accept the hypothes. fa and internal 99% confidence M-/= 3.5-1 35-1 = 34 tan of freedom degree 2.728 Thu 997. confidence interval wie itan S Sn 40.2 + (2.728) (9.9) X + 735 )

46.2 + 4.56505 Thi 4o.2-4.56505 L M < 40.2 + 4.56505 44.76505 35.63495 M Lowes bound Uppes bound 35.635 44.765 il Because between accept the value 36 lie the confidence internal we null hyf Dthesis