This document discusses several statistical concepts: 1) It introduces sampling distributions of the mean, proportion, and sample variance. 2) It provides an example of calculating a 95% confidence interval for a sample mean using a t-distribution. 3) It demonstrates computing the Pearson correlation coefficient r between two variables to measure the strength and direction of their linear relationship.

1. SC 101 2 1 1

2. ~~LQZU (Sampling Distribution of the Mean) ~l~ll$~~~~~~~;lnl~~~~~i?adl;1~~~
8~42~ (Sampling Distribution of Proportion) a~~~ni:,rrsnIrs:QZlDln15~~~~i?Sl~l3
?Ei~Fl'2?kJLLni‘d~~~ (Sampling Distribution of Sample Variance)
the Mean)
2 2
LA,” Ox = c,
J
cx,- ug2 + (X2- P’,) + . . . + CT,- l.‘;)
k
2 1 2 SC 101

10. 2 2 0 SC 101

15. n"?ilWl1 6.3 &ldnl?Ml Confidence intervalB04 u lMJ%?rt-distribution
n?wd94Y sample mean (%) = 20 sample standard deviation fs) =
LLR 2 sample size (n) = 25
x+t S
OL/2,n-l I
EOf1.711 2
90% 1.711 Jn
425 19.49;d20.51
= 20~0.5133
xat s
2Ok2.0641.5
95% 2.064 4,n-l T J-E- 19.38;320.62
J" = 20+0.6192
1.52Oir2.7$?7---
&?i 19.16h20.84
= 20f0.8391

20. 2 3 0 SC 101

34. X2 z (n-l )SL
c 4
= (9HW2
0.81
= (9lC1.44) = 12.36
0.81 0.81
= 16
x: = 16.0 16.919
2 4 4 SC 101

45. wwh cx = 26, =cy 17, cx2= 158, cy2 = 71, cxy = 105
im:: n = 5 oJh~wos~~
5(105) - (26)(17)
r =
LJWW) - w21 15(71) - 07)21
= 525 - 442
d (790 - 676) (355 - 289)
= &- =
0.96
4.
cd2, n-2
M% tc > t
cd
2'
n-2
SC 101 2 5 5

46. - 2 . 1 8 9 2 . 1 8 9
o.so f/XT
4-c.5012
(.50)(4.69)
A x -
.
2 5 6 SC 101

48. 2 5 8 SC 101

50. d
2 1 . can’lPla~i~~5,“~M~~~~~~~~~ln~~~a,la~a~a,”lSin f%f
n. X Y ?I. X
34 21 3.9
30 22 4.6
40 25 6.0
34 28 2.8
39 15 3.7
35 24 3.4
42 24 4.2
45 22 4.0__
43 17- -
2 =38, ? =22 x =4,
s =
x
5,s = 4
Y
S = 1,
x
Y
46
4 6
52
50
48
4 0
42
44-
iJ = 46
S = 4a
Y
2 6 0 SC 101