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Solutions to homework problems related to constructing confidence intervals for means and proportions using the t-distribution. The examples given involve calculating the confidence intervals for the mean difference, mean, and proportion, as well as determining the required sample size to achieve a desired margin of error.
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Setup: n = 86 y¯ = 60. 43 s = 3. 06 a) SE¯y = √sn = √^3.^0686 = .33 mm
b)
a) Since we are using s instead of σ we will use the T-distribution instead of a z-score.
95% CI = ¯y ± t(df )α/ 2
√s n
y ¯ = 28. 7 s = 4. 6 n = 6 SE¯y = 4 √.^66 = 1. 9 t(6 − 1). 05 / 2 = t(5). 025 = 2. 571
95%CI = 28. 7 ± (2.571)(1.9) = (23. 8 , 33 .6) ⇒ 23. 8 < μ < 33. 6 μg/ml
We are highly confident (95% confidence) that the true mean of the blood serum concen- tration of Gentamicin (μg/m) in three-year-old female Suffolk sheep is between 23.8 and 33.6 μg/m.
b) The population mean μ is the blood serum concentration of Gentamicin (1.5 hours after injection of 10 mg/kg body weight) in healthy three-year-old female Suffolk sheep.
c) No. The 95% refers to the percentage (in a meta-experiment) of confidence intervals that would contain μ. since the width of a confidence interval depends on n, the percentage of observations contained in the confidence interval also depends on n, and would be very small if n were large.
We want a 95% CI for the mean difference, so only the numbers in the right column will be used. 95% CI = ¯y ± t(df )α/ 2
√s n
a)¯y = 13. 0 s = 12. 4 n = 10 SEy¯ = 12 √ 10.^4 = 3. 92 t(10 − 1). 05 / 2 = t(9). 025 =
95%CI = 13 ± (2.262)(3.92) = (4. 1 , 21 .9) ⇒ 4. 1 < μ < 21. 9 pg/ml
b) We are 95% confident that the average drop in HBE levels from January to May in the population of all participants in physical fitness programs like the one in the study is between 4.1 and 21.9 pg/ml.
Construct a CI for the proportion p n = 959 pˆ =. 157 For pˆ a) ˆp ± z α 2 (SEpˆ) z α 2 = z. 102 = z. 05 = 1. 645
SEpˆ =
ˆp(1−pˆ) n =
.157(1−.157) 959 =^.^0117
. 157 ± (1.645)(.0177) ⇒. 1378 < p <. 1762
b) The confidence interval from part (a) is a confidence interval for the probability of in- terference with the pacemaker for a specific type of cellular phone tested. We are 90% confident that the true population proportion for cell phone interference with pacemakers is between 13.78% and 17.62%.
M Enew = M E 2 old =.^1992 =. 0995 ⇒ .0995 = (2)
.√ 238 n
n = (2)(. 0995 .238) ⇒ n = ( (2)(.238)
. 0995
To have a margin of error half the size of the one found in part (b) or smaller a sample size of 23 or larger is needed.
a) ¯y = 2. 275 s =. 238 n = 8 SEy¯ = √sn = √. 2238. 275 =. 084
b) Since we are using s instead of σ we will use the T-distribution instead of a z-score.
98% CI = ¯y ± t(df )α/ 2 SE¯y t(8 − 1). 02 / 2 = t(7). 01 = 2. 998
c) We are 98% confident that the average diameter of the stem of a wheat plant three weeks after flowering is between 2.02 and 2.53 mm.
d) We want a sample size, n, that will give us a margin of error (ME) half the size of the one found in part (b).
M E = tα/ 2
√s n
By scanning the .01 column (98% CI) of the t-table you can see that
the t-statistics are approximately 2.5.
M Eold = (2.998)(0.84) =. 252 ⇒
M Enew = M E 2 old =.^2522 =. 126 ⇒ .126 = (2.5)
.√ 238 n
n = (2.5)(. 126 .238) ⇒ n = ( (^) (2.5)(.238)
. 126
To have a margin of error half the size of the one found in part (b) or smaller a sample size of 36 or larger is needed.
a) ¯y = (^1) n
i=1 yi^ = 62.^767 s^ =
1 n− 1
i=1(yi^ −^ y¯)^2 = 1.^01127 SEy¯ = √sn = 1.^01127 √ 6 =. 4128
b) 90% CI = ¯y ± t(df )α/ 2 SE¯y t(6 − 1). 1 / 2 = t(5). 05 = 2. 015
For p˜
SEp˜ =
˜p(1−p˜) n+z^2 α 2
.45(.55) n+2^2 6.^02 ⇒^
. 45 ∗. 55 . 022 −^4 6 614.^75
615 or more students need to be sampled to have a standard of error less than 2 percentage points.
For pˆ
SEpˆ =
ˆp(1−pˆ) n ⇒
.45(.55) n 6.^02 ⇒^
. 45 ∗. 55 . 022 6 618.^75
619 or more students need to be sampled to have a standard of error less than 2 percentage points.
