Anjan wishes to test whether his mean journey time is different on Tuesdays. He chooses a random sample of 30 Tuesdays and finds that his mean journey time for these 30 Tuesdays is 40.2 minutes. Assume that the standard deviation for his journey time on Tuesdays is 6.9 minutes. The time, in minutes, for Anjan's journey to work on Mondays has mean $$\(38.4\)$$ and standard deviation $$\(6.9\)$$ Carry out the test at the $$\(10 \%\)$$ significance level. ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................
Exam No：9709_w20_qp_63 Year：2020 Question No：6(b)(ii)
Answer:
\(\mathrm{H}_{0}\) : Population mean journey time (or \(\mu\) ) \(=38.4\)
\(\mathrm{H}_{1}\) : Population mean journey time (or \(\left.\mu\right) \neq 38.4\)
\(\frac{40.2-38.4}{\frac{6.9}{\sqrt{30}}}\)
\(=1.429\)
' \(1.429\) ' \(<1.645\)
There is no evidence that mean journey time has changed.
Alternative method for question 6(b)(ii) - critical values method
\(\mathrm{H}_{0}\) : Population mean journey time \((\) or \(\mu)=38.4\)
\(\mathrm{H}_{1}\) : Population mean journey time \((\) or \(\mu) \neq 38.4\)
\(38+1.645\left(\frac{6.9}{\sqrt{30}}\right)\)
\(=40.47\)
\(40.2<40.47\)
There is no evidence that mean journey time has changed.
\(\mathrm{H}_{1}\) : Population mean journey time (or \(\left.\mu\right) \neq 38.4\)
\(\frac{40.2-38.4}{\frac{6.9}{\sqrt{30}}}\)
\(=1.429\)
' \(1.429\) ' \(<1.645\)
There is no evidence that mean journey time has changed.
Alternative method for question 6(b)(ii) - critical values method
\(\mathrm{H}_{0}\) : Population mean journey time \((\) or \(\mu)=38.4\)
\(\mathrm{H}_{1}\) : Population mean journey time \((\) or \(\mu) \neq 38.4\)
\(38+1.645\left(\frac{6.9}{\sqrt{30}}\right)\)
\(=40.47\)
\(40.2<40.47\)
There is no evidence that mean journey time has changed.
Knowledge points:
6.5.3 formulate hypotheses and carry out a hypothesis test concerning the population mean in cases where the population is normally distributed with known variance or where a large sample is used
Solution:
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