o.05cp-valuelo · homework11 solution: stat400-spring 20ha khanh nguyen ① x = weight of a "...
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Homework 11 Solution : STAT 400 - Spring 20-
Ha Khanh Nguyen
① X = weight of a"
22 - ounce"
Charlie Mulligan 's porterhouse steak .X - N ( µ , 62)
Note : I = 21.4,s'-
- O.64
,n = 6
(a) to : µ= 22
I - Ho 21.4 - 22Teststatisic: T = - =
g= -1.8370
( 6 is unknown) HT
(b) Ho : µ= 22 vs. Hi : µ 522 ( left - sided test )
Tnt* 0.1 df=5 Rejection : Reject to tf TL - to . I , df=5
⇐Ts-i.-
Decision Since 1- ⇐ - 1.837C - 1.476,- taHo at 2=0.1
.
(c) p- value = I ( Tf - 1.837 ) ⇒ O.05cp-valuelo.IT
Tutdf -- 5 Decision Since p- value > x -
- 0.05,
p- value,µ fonjectH at a -- 0.05 .
OR use R,
- 1.837
ptc - 1.837, df - 5)[ I ] 0.06281569 ← p
- value
①
(d) Ho : 6 ¥0.5
Teststatisic : x? =ln÷ = 16-47%642 =
( e) to : G f 0.5 us . Hi 70.5
Idf . 5 Rejection : Reject It ① if X'
> XIdf - 5↳ ⇒xes¥"
Decision since XE 12.8 > 1.610,XLt at a = O
. I.
(f) Suppose 6=0.5. The decision in part (e) is incorrect Type I error,reject to when to is true ( false positives ) was made .
② n = 120. To = 47
120
(a) Ho : p f 0.3 vs . It, :p > 0.3- 47p - Po Tao
- O.3
Teststatistic : Z = -_- =
#= 2.19130
✓ Po(l-p# ✓ 0.3 (0.7)n
( b)p-value = I ( Z 32.1913)
Z - N (O, I )= I - I ( Z C 2.1913)✓¥§
,
p- value
= I - 0.9857 = 0.01430=
2.1913 Decision Since p- value = 0.014342=-0.05,
Ho at a -- 0.05
.
②
(c) Suppose p= 0.25. The decision in (b) wasi Type 1 error
,
rejecting to when Ho is true , was made .
(d) Assuming to is true.
That is,
X = # of students who support universal credit / no- credit,X n Binom ( n = 120
, p= Or32, assuming to is true
in Rp-value = I C X S, 47 ) = P - I ( X f 4G ) I
= I - pbinom ( q = 46 , size-- 120
, prob -- O. 3)
= 0.01997150
③