For letter a
[tex]z= \frac{x-u}{SD \sqrt{n} } \\ z= \frac{541.4-525}{100 \sqrt{100} } \\ z=1.64[/tex]
Looking up z=1.64, its equivalent is 0.4495
∈=0.5+0.4495
∈=0.9495
As 0.95>0.9495, we cannot reject the null hypothesis, and the result is not significant.
b.)[tex]Z= \frac{(mean-u)SD}{ \sqrt{n} } \\ Z= \frac{(541.5-525)100}{ \sqrt{100} } \\ Z=1.65[/tex]
Looking it up, P(Z>1.65)=0.0505. The result is not significant.
c. The p values are larger than the accepted level of 0.05, which is why the null is rejected. As we cannot reject the null, it is to be accepted that the mean score is 525. This then requires more statistical testing to be done to prove the null.