Download Confidence intervals for a population mean

Answers for Confidence intervals for a population mean
Assume that all samples are normally distributed for the questions on this handout.
Remember that if the standard deviation has been calculated on a small sample you
need to use the t-distribution.
1. An analysis was made of peat soils from a number of sites which had similar
vegetation. The total phosphate in the soil was as follows:
(mg/100g dry weight)
39.3 46.6 51.7 46.0 68.3 58.0
Calculate the mean and s.d. of the sample and hence find a 95% confidence
interval for the mean phosphate content of the soil.
Explain clearly in words what this confidence interval means. WRITTEN
answer!!
x  5165
.
  10.27
95% confidence interval 51.65  2.571  4.193
Note
t5,0.025 = 2.571
95% of the times when we do this type of calculation we will be correct when we claim
that the calculated interval includes the true mean.
2. Using specimens from 10 children, determination of the %calcium content of sound
teeth gave the following:
36.39 36.19 34.20 35.15 35.47
35.22 36.11 35.63 36.63 35.59
(i) Find 95% and 99% confidence intervals for the mean %calcium of the teeth.
95% interval
3.5658  0.511
99% interval
3.5658  0.734
For checking
x  35658
.
  0.7137
s.e. = 0.2257
(ii) Important question. Explain why you cannot find a 100% confidence
interval.
Because we are assuming an underlying normal distribution and we can never state a
range for 100% of a normal distribution. (Curve never quite touches the horizontal
axis.)
3. The mean indirect bilirubin level of 16 four-day old infants was found to be 5.98
mg/100cc. The s.d. was calculated to be 3.5 mg/100cc.
Find 90%, 95% and 99% confidence intervals for the mean bilirubin level of the
population.
Confidence intervals are
5.981.53
5.981.86
5.982.58
(90%)
(95%)
(99%)
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4. A sample of 100 apparently normal adult males, aged 25, had a mean systolic blood
pressure of 125. Assuming that the s.d. of the sample is 15 find
(i) a 90% C.I. for the population mean
(ii) a 95% C.I. for the population mean.
90% confidence interval 125  2.5.
95% confidence interval 125  2.9
For this it is not necessary to use the t – distribution. The normal distribution can be
used as the sample size is greater than 30. The z – values can be found either by using
Table 5 and looking up the 2.5% point for 95% interval etc or you can use the bottom
row of Table 10 which is marked as having an infinite number of degrees of freedom.
(Written as  )
General point about the precision of the quoted mean and s.d. (s.e.)
The usual rules are that you give the mean to one more decimal place than the data.
The standard deviation and the standard error should also be quoted to one more
decimal place than the original. BUT this is as a final answer. You need at least one
extra figure of precision while doing the calculations.
For example, in question 1 above you work to 3 d.p. at least but you quote answers
rounded to 2 d.p. at the end.
If you do not work to greater precision than your final answer you can get a very
inaccurate answer.
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