The length of bird songs, measured in seconds, are normally distributed. A rogue biologist (Biologist A)t claims the mean duration for American bird song is µ = 211.7. Another biologist (Biologist B) decides to test this and takes a sample of 12 bird songs and collects their durations. He retrieves the following data (measured in seconds):215.2, 223.5, 205.4, 233.3,200.2, 240.5, 251.6, 231.6,235.3,198.6 ,213.7, 226.9. Perform a hypothesis test of the rogue biologists claim (Biologist A). Test at the 0.05 significance level. In your answer include both critical values and the p-values. Report p-values to the nearest ten-thousandth and critical values to the nearest hundredth.
Answer)
Null hypothesis Ho : u = 211.7
Alternate hypothesis Ha : u not equal to 211.7
As the population s.d is unknown, we will use t distribution to conduct the test.
Sample mean for the data is = 222.9833
S.d = 16.6332
Test statistics is = (sample mean - claimed mean)/(s.d/√n)
Test statistics = (222.9833-211.7)/(16.6332/√12)
t = 2.35
Degrees of freedom is = n-1, 11
For df 11 and test statistics 2.35
P-Value from t distribution is = 0.038487 = 0.0385
As the P-value is < 0.05 (given significance level)
We reject the null hypothesis
So, we do not have enough evidence to support the claim that the mean duration for American bird song is µ = 211
The length of bird songs, measured in seconds, are normally distributed. A rogue biologist (Biologist A)t...