friday oct 8
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Friday Oct 8. Confidence interval for How to obtain Z /2 Interpretation of confidence interval What n 30, can replace by s Read 5.2 Next time: read 5.3. Similarly P(-1.96 < Z < 1.96) = 0.95 Thus P( Ў -1.96 Ў < < Ў +1.96 Ў ) = 0.95 - PowerPoint PPT PresentationTRANSCRIPT
Friday Oct 8
Confidence interval for How to obtain Z/2
Interpretation of confidence interval What n 30, can replace by s Read 5.2 Next time: read 5.3
In generalP ( 1.65y < y < 1.65y) = 0.9
Rearranging the terms inside the probability:P(y 1.65y < < y 1.65y) = 0.9
Even though we don’t know what is, we know thatit falls within (y 1.65y, y 1.65y) 90% of thetime, where y = n
Similarly P(-1.96 < Z < 1.96) = 0.95 Thus P(Ў-1.96Ў < < Ў +1.96Ў ) = 0.95
For a 99% confidence interval:– P(-2.58 < Z < 2.58) =0.99. Z0.005=2.58
Thus P(Ў -2.58Ў < < Ў +2.58Ў) = 0.99
3
2
Z
.005.005
3
2
Z
.95.025 .025
Formula for 100(1-)% confidence interval for
– when is given is [Ў Z/2* Ў ] where Ў = /n
Z/2 is a value of Z having a tail area of /2 to its right
3
2 1-/2 /2
Z/2
3
2
1
0
Histogram of C1, with Normal Curve
f(y)
y
/2 /2
-/2 /2
A few frequently used Z value 90% C. I. = .1 Z/2 = Z.05 = 1.645
95% C. I. =.05 Z/2 = Z.025 = 1.96
99% C. I. =.01 Z/2 = Z.005 = 2.575
97% C. I. =.03 Z/2 = Z.015 = 2.17
Example
E.g. 5.10 Recent data from a national survey of 1350 women indicated that the average woman goes to a hair salon once every 5 weeks and spends on the average $26.49. With a standard deviation (s) of $12.00. Use these data to construct a 99% C.I. for (average amount spent by women in America on hair salon).
n = 1350, Ў = 26.40, = s = 12, = 0.01 99% C.I. : Ў Z0.005.121350 = 26.40 2.58 . 0.3266 (25.56, 27.24) 99% C.I. for
– Verbal interpretation of the 99% C.I. : We are 99% confident that the average amount a woman in America spends in a hair salon is between $25.56 and $27.24.
e.g. 5.15 900 high-school graduates sampled. Objected of interest: distance between the high school attended and present address. Ў = 430 miles, s = 262 miles. Find
a. 95% CI for “average distance between a person’s present address and high school” = 1 - 0.95 = 0.05 Ў ± Z/2*n, Ў =430, Z/2 = Z0.025 =1.96. n = 900 430 ± 1.96*262900 (412.88, 447.12) width = 34.24
b. 99% Confidence interval 430 ± 2.58* 262900 (407.468; 452.532) width = 45.064