In: Statistics and Probability
a) As we are testing here whether the rating has changed from 5.9, therefore the null and the alternative hypothesis here are given as:
The test statistic here is computed as:
For n - 1 = 24 degrees of freedom, the p-value here is obtained
from the t distribution tables here as:
p = 2P( t24 < -1.72) = 2*0.0488 = 0.0976
As the p-value here is 0.0976 > 0.05 which is the level of significance, therefore the test is not significant here and we cannot reject the null hypothesis here.
Therefore we dont have sufficient evidence here that the mean has changed from 5.9.
b) for 24 degrees of freedom, we have from t distribution
tables:
P( t24 < 2.064) = 0.975
Therefore, due to symmetry, we have ehere:
P( -2.064 < t24 < 2.064) = 0.95
Now the confidence interval here is computed as:
This is the required 95% confidence interval for the population mean rating here.