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Statistical Analysis: Confidence Intervals and Hypothesis Testing - Prof. Raoul Lepage, Study notes of Data Analysis & Statistical Methods
Solutions to various statistical problems, including calculating confidence intervals for population means and rates, estimating standard errors, and determining percentiles of the standard normal distribution. The problems involve sample sizes, means, standard deviations, and hypothesis testing.
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STT 200 Review 2-2-
- A random sample of 400 hospital admis- sions from a week's total of 5400 finds 88 were emergency contacts. Give a 98% con- fidence interval for p = rate of emergency contacts among admissions. p
`
=
88 400
=
22 100
= 0.
DF ¶ 1.96 2.326 p
`
± z p H 1 - _p_ L n N - n N - 1 Conf 95% 98%
- A random sample of 36 elk selected from the Jackon, Wy. Elk Refuge in win- ter are scored for x = lead exposure finding sample mean x = 27.
- A random sample of 36 elk selected from the Jackon, Wy. Elk Refuge in win- ter are scored for x = lead exposure finding sample mean x = 27. sample standard deviation s = 11. It is believed that x scores in this winter herd are normal distributed. Give the 80% confidence interval for population mean lead exposure m. DF 35 1.306 x ± t s n
H 1 L
¶ Conf 80%
- What does estimated margin of error of x actually estimate? population sd s sd of the list of all possible x 1.96 s 1.96 sd of the list of all possible x
- What does estimated margin of error of x actually estimate? population sd s sd of the list of all possible x 1.96 s 1.96 sd of the list of all possible x
- We have obtained estimated standard errors for rates of cracking of concrete 0.037 for p
`
mixes with latex 0.042 for p
`
mixes without latex Give the estimated margin of error for p
`
latex
- p
`
no latex .
- We have obtained estimated standard errors for rates of cracking of concrete 0.037 for p
`
mixes with latex 0.042 for p
`
mixes without latex Give the estimated margin of error for p
`
latex
- p
`
no latex . 1.96 0. 2
- We have obtained estimated standard errors for sample means of concrete hard- ness 0.037 for x mixes with latex 0.042 for x mixes without latex Give the estimated margin of error for x latex- x no latex.
We have obtained estimated standard errors for sample means of concrete hard- ness 0.037 for x mixes with latex 0.042 for x mixes without latex Give the estimated margin of error for x latex- x no latex. 1.96 0. 2 - 0. 2
Estimate the mean and sd by eye.
Estimate the mean and sd by eye. 80 100 120 140
Amount of genetic material in a given plot is normal distributed with m = 9 s = 3 Determine the standard score z of a plot with score x = 10.5.
- Amount of genetic material in a given plot is normal distributed with m = 9 s = 3 Determine the standard score z of a plot with score x = 10.5. Determine the amount x of genetic mate- rial of a plot with standard score z = 2.5.
- What is the exact chance that a 95% confidence interval for m will in fact cover m if the population is normal distributed and the t-CI is used?
- What is the exact chance that a 95% confidence interval for m will in fact cover m if the population is normal distributed and the t-CI is used?
- Use the z-table to determine P(Z < 2.43). z 0. 2.4 0.
- Use the z-table to determine P(Z < 2.43). z 0. 2.4 0.
- Determine the 86th percentile of Z. z 0. 1.0 0.
- Determine the 86th percentile of Z. z 0. 1.0 0. IQ is normal distributed and has mean 100 and sd 15. Determine the 86th percentile of IQ. IQ = 100 + z 15
- Determine the 86th percentile of Z. Calculate the sample standard deviation s for the list x = {0, 0, 4, 8}. avg = 12/4 = 3
s
x
=
H 0 - 3 L 2 +H 0 - 3 L 2 +H 4 - 3 L 2 +H 8 - 3 L 2 4 - 1
= 3.
- Determine the 86th percentile of Z. Calculate the sample standard deviation s for the list x = {0, 0, 4, 8}. avg = 12/4 = 3
s
x
=
H 0 - 3 L 2 +H 0 - 3 L 2 +H 4 - 3 L 2 +H 8 - 3 L 2 4 - 1
= 3.
s
4 x + 9 = † 4 † sx = 4 (3.82971)
- We've selected random samples of peo- ple with or without medication, the score being x = blood pressure decrease over a 5 minute period. Assume large populations. x with med = 12.3 s with med = 3.2 n = 60 x without med = 3.7 s with med = 1.2 n = 90 Give the 95% CI for mwith med - mwithout med.
- We've selected random samples of peo- ple with or without medication, the score being x = blood pressure decrease over a 5 minute period. Assume large populations. x with med = 12.3 s with med = 3.2 n = 60 x without med = 3.7 s with med = 1.2 n = 90 Give the 95% CI for mwith med - mwithout med. (12.3 - 3.7) ± 1.
- 2 60
+
2 90