1. A random sample of 15 fence posts from a garden center was found to have a mean length of 1.84 m. The population standard deviation for these posts is taken to be 17cm.
(i) Use the t-distribution to find a symmetric 95% confidence interval for the mean length of such posts.
(ii) Interpret the 95% confidence interval found in part (i) above.
The formula for confidence interval is
"m{\\scriptscriptstyle *}=m\u00b1Cr*\\frac {\\sigma} {\\sqrt {n}}" , where "m{\\scriptscriptstyle *}" - true mean value, m - mean value obtained from data, Cr - critical value depends on confidence level, "\\sigma" - population standard deviation(sample if population is unknown), n - sample size
In the given case we have to use two-sided t-value with n - 1 = 14 degrees of freedom as critical value(most likely it is said due to the small sample size), then
"P(T(14)>Cr)={\\frac {1+0.95} 2}=0.975\\implies Cr=2.145" . So,
"m\u00b1Cr*\\frac {\\sigma} {\\sqrt {n}}=1.84\u00b12.145*{\\frac {0.17} {\\sqrt {15}}}=1.84\u00b10.094"
The 95% confidence interval for population mean is (1.84 - 0.094, 1.84 + 0.094) = (1.746, 1.934)
(ii) The obtained result means that, based on the given data, we can estimate with 0.95 probability that the population mean lies between 1.746 and 1.934
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