Question

A professor administers a standardized test that has a µ = 50 and σ = 6. Her class of 31 students (n = 31) has an average score of 55.

What is the probability of the results if the null is true (the p-value)? Report the proportion of the area under the normal curve to the right of the z observed.

Answer 1

Solution :

= 50

=55

=6

n = 31

This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :    = 50

H_a : > 50

Test statistic = z

= ( - ) / / n

= (55-50) / 6 / 31

= 4.64

P(z > 4.64) = 1 - P(z < 4.64 ) = 1 - 1

P-value = 0