In [49]:

    

source("../R/functions.R")


    

In [50]:

    

obs = observation(n = 20, S = 379.7231, USS = 7209.525)


    

In [51]:

    

printSingleObservation(obs)


    

    






$ f = n - 1 = 19 $






    






$ SSD = USS - S^2 = 0.0433663194999099 $






    





Estimeret middelværdi






    






$ \mu \leftarrow \bar{x}. = \frac{S}{n} = \frac{379.7231}{20} = 18.986155 \sim\sim N(\mu, \frac{\sigma^2}{n}) $






    






$ StdError = \sqrt{s^2 / n} = \sqrt{0.00228243786841631 / 20} = 0.0106827849094146 $






    





95% konfidensinterval for $\mu$






    






$ c_{95}(\mu) = \bar{x.} \mp t_{0.975}(f) StdError = 18.986155 \mp 0.0223593257834748 = [18.9637956742165; 19.0085143257835] $






    





Estimeret varians






    






$ \sigma^2 \leftarrow s^2 = \frac{SSD}{f} = \frac{0.0433663194999099}{19} = 0.00228243786841631 \sim\sim \sigma^2 \chi^2 (n-1) / (n - 1) $






    





Estimeret spredning






    






$ \sigma \leftarrow s = \sqrt{s^2} = 0.0477748664929198 $






    





Konfidensinterval for $\sigma^2$






    






$ c_{95}(\sigma^2) = [\frac{f s^2}{\chi^2_{0.975}(f)}, \frac{f s^2}{\chi^2_{0.025}(f)}] = [0.0013200379894682, 0.00486905510000587] $

In [48]:

    

printTestMeanHypothesis(obs, 19)


    

    





Tester hypotese om middelværdi






    





Vi laver en $t$-test.






    






$ H_0: \mu = \mu_0 = 19 $






    





$ t $-teststørrelsen bliver






    






$ t(x) = \frac{\bar{x}. - \mu_0}{\sqrt{s^2 / n}} = \frac{18.986155 - 19}{\sqrt{0.00228243786841631 / 20}} = -1.29601036783943 $






    





testsandsynligheden bliver






    






$ p_{obs}(x) = 2(1 - F_{t(n-1)}(| t(x) |)) = F_{f(19)}(|t(-1.29601036783943)|)) = 1.78950517670509 $






    





Da $p_{obs}(x)$ er større end $0.05$ kan hypotesen ikke forkastes.

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