Question

1. Find the P-value for the hypothesis test with the standardized test statistic z. Decide whether to reject H0 for the level of significance α.      ( Marks 2)

Right-tailed test

z = 0.52

α = 0.06

Answer 1

The value of the standardized test statistic z = 0.52 and test is right tailed test

The level of significance is = 0.06

We find P-value of above standardized test statistic z = 0.52 using following Excel function

=1 - NORMSDIST(z)

Here, z = 0.52

=1 - NORMSDIST(0.52) then press Enter we get P-value is as below

P-value = 0.301532 (Round up to 2 decimal places)

P-value = 0.30

We comparing the above P-value with given and take decision about whether to reject or do not reject Ho using following condition.

1) P-value    then we reject Ho

2) P-value    then we do not reject Ho

We comparing the values we calculate as above and the given level of significance is = 0.06

P-value = 0.30    = 0.06

We do not reject Ho at given level of significance is = 0.06