Questions that i will attach as files later

1cd3e9e2-3e2a-4884-8060-46b75110095b-fgckddpf-dutjuewy.jpeg 9221468d-49a0-43f5-934b-314685de0299-cpmjoo5e-yjy5hc5a.jpeg 159d2105-a0d3-461c-b74f-c732b296e54d-243ddrcx-hegvgoyf.jpeg 2e7424db-c198-4910-bee3-7c68711e195b-suyx1ve2-jj43clym.jpeg

Answered Same DayApr 20, 2021

Answer To: Questions that i will attach as files later

Pooja answered on Apr 20 2021

164 Votes

31)
Ho: u =500
h1: u < 500 (left tailed)
mean=     475.25
sd=     48.600
u=     500
n=     16.00
test statistic, t = (mean-u)/(sd/sqrt(n))
t = (475.25-500)/(48.6/sqrt(16))
t = -2.037
p-value
P(TP(T<-2.037)
T.DIST(-2.037,16-1,TRUE)
0.0298
32)
P = 13.00%
n = 81
E(Phat) = p =     13.00%
V(Phat) = pq/n = 0.13*(1-0.13)/81 = 0.001396296
SD(Phat) = sqrt(V(phat)) = 0.0374
Phat follows Normal with mean 13% and sd=0.0374
I know that, z = (X-mean)/(sd)
z = (0.2-0.13)/0.0374) = 1.8733

hence,
P(X>0.2)
=1 - P(X<0.2)
=1 - P(Z<1.8733)
=1 - NORMSDIST(1.8733)
=0.0305
33)

    substance abuse

    test
    yes
    no
    sum
    positive
     =0.98*5
     =1%*95
     ...

SOLUTION.PDF

Questions that i will attach as files later

Answer To: Questions that i will attach as files later

Answer To This Question Is Available To Download

Related Questions & Answers

Submit New Assignment

Sun	Mon	Tue	Wed	Thu	Fri	Sat
30	31	1	2	3	4	5
6	7	8	9	10	11	12
13	14	15	16	17	18	19
20	21	22	23	24	25	26
27	28	29	30	1	2	3