Question

A school is interested in whether its student's scores on a standardized test exceed the
national average. The national average is 40 (μ = 40) and the standard deviation is 8 (cr =
8). To determine this, the school officials administer the test to a sample of 16 (n = 16)
students and obtain a mean score of 44 (M = 44). Perform a z-test to determine whether

the students in the school differ significantly from the national average. Perform a two-
tailed test. Compute the obtained z score.

Answer 1

Solution :-

Given :-

Maen ( u ) = 40

Sample Mean ( X ) = 44

Sample Size ( n ) = 16

Stand Devi ( ) = 8

Significance Level ( ) = 0.05

Hypothesis :-

Ho : u = 40

H1 : u 40

( This is two tailed test )

Test Statistic :-

Z stat = 2

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Z critical :- By using Z table at ( ) = 0.05

Z critical ( Zc ) = 1.96

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P -Value :-

by using p -Value Approach,

P ( Z > 2 ) = 1 - P ( Z < 2 )

P ( Z > 2 ) = 1- 0.9772

P ( Z > 2 ) = 0.0228

For Two tailed test,

P -value = 2 * 0.0228

P -value = 0.0456

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Decision :- Since P -Value ( 0.0456) < significance level ( 0.05 )

Reject H0

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Conclusion :-

Therefore, there is enough evidence to claim that the population mean μ is different than 40, at the 0.05 significance level.