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

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• Similarly P(-1.96 lt Z lt 1.96) 0.95
• Thus P(?-1.96?? lt ? lt ? 1.96?? )
  0.95
• For a 99 confidence interval
• P(-2.58 lt Z lt 2.58) 0.99. Z0.0052.58
• Thus P(? -2.58?? lt ? lt ? 2.58??) 0.99

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• Formula for 100(1-?) confidence interval for ?
• when ? is given is
• ? ? Z?/2 ?? where ?? ?/?n
• Z?/2 is a value of Z having a tail area of ?/2 to
  its right

Histogram of C1, with Normal Curve
f(y)
3
a/2
a/2
2
m
y
?Za/2 s
m-Za/2 s
1
0
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• 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

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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, ? 26.40, ? s 12, ? 0.01 99
  C.I. ? ? Z0.005.12??1350 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.

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• e.g. 5.15 900 high-school graduates sampled.
  Objected of interest distance between the high
  school attended and present address. ? 430
  miles, s 262 miles. Find
• a. 95 CI for average distance between a
  persons present address and high school ? 1 -
  0.95 0.05 ? Z?/2???n, ? 430, Z?/2
  Z0.025 1.96. n 900 430 1.96262??900
  (412.88, 447.12) width 34.24
• b. 99 Confidence interval 430 2.58
  262??900 (407.468 452.532) width 45.064