You intend to estimate a population mean μ from the following sample.
41.7 | 43.7 | 37.2 | 30.5 |
34.1 | 21.4 | 41.5 | 40 |
28.4 | 14.4 | 35.2 | 30 |
42.3 | 40.1 | 41 | 24.2 |
24.5 | 47.9 | 18.9 | 35.1 |
34.7 | 32.9 | 31.8 | 43.9 |
35.9 | 18.5 | 22.9 | 50 |
42.9 | 27.2 | 26.8 | 28.2 |
26.6 | 32.3 | 29.2 | 20.7 |
44.3 | 37.3 | 35.6 | 20.2 |
19.9 | 34.9 | 25.1 | 30.2 |
Find the 95% confidence interval. Enter your answer as a tri-linear
inequality accurate to two decimal place (because the sample data
are reported accurate to one decimal place).
Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.
sample size ( n ) = 44, c=95%
now calculate sample mean( ) ans sample standard deviation (s) for given data
we get
sample mean( ) = 32.3659
sample standard deviation (s) = 8.7678
formula for confidence interval is
here tc is the t critical value for c=95% with df= n-1 = 44-1= 43
using t table we get
tc = 2.017
32.3659 − 2.666 < < 32.3659 + 2.666
29.70 < < 35.03
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