A random sample of 70 eighth students’ scores on a national mathematics assessment test has a mean score of 295 with a standard deviation of 45. This test result prompts a state school administrator to declare that the mean score for the state’s eight graders on this exam is more than 290. At α = 0.15, is there enough evidence to support the administrator’s claim? Complement parts (a) through (e).
(a) State the claim mathematically and identify 
and 
.
(b) Find the standardized test statistic z, and its corresponding area.
(c) Find the P-value.
(d) Decide whether to reject or fail to reject the null hypothesis.
(e) Interpret your decision in the content of the original claim.
Solution
(a) State the claim mathematically and identify and .
A null hypothesis is a statistical hypothesis that contain a statement of equality, such as ≤, = , ≥. This alternative hypothesis is the complement of the null hypothesis. It is a statement that must be true if the null hypothesis is false, and it contains a statement of strict inequality, such as > , ≠, or <. The claim is what you believe.
The hypothesis and the claim are shown below.
:µ ≤ 290
: µ > 290 (claim)
(b) Find the standardized test statistic z, and its corresponding area.
The standardized test statistic is
Where 
is the sample mean, µ is the hypothesized mean value, σ is the population standard deviation and n is the sample size. If n ≥ 30 you can use σ 
, the sample standard
The population standard deviation σ is unknown but because n ≥ 30, use s in place of σ .
Substitute given value = 295 , µ = 290 , σ = 45, and n = 70 to find the test statistic.
Substitute.
z 
0.93
Recall that a standard normal distribution table given the area to the left of a z-score.
Determine the area to the left of z = 0.93, rounding to three decimal places.
Area = 0.824
(c) Find the P-value.
Since contain the greater than inequality symbol (>), this is a right-tailed test. In a right-tailed test the P-value is the area to the right of the test statistic. Subtract 0.824 from 1 to find the area to the right of z = 0.93. This area us the P-value.
P-value = 1 – 0.824 = 0.176
(d) Decide whether to reject or fail to reject the null hypothesis.
If the P-value is less than the level of significance α, reject the null hypothesis. If the P-value is greater then the level of significance α, fail to reject the null hypothesis .
For this problem, α = 0.15 and P = 0.176. Fall to reject because 0.176 > 0.15.
(e) Interpret your decision in the content of the original claim.
The original claim that the mean score for the state’s eighth graders on the exam is more then 290 is the alternative hypothesis, . If we reject the null hypothesis, there is enough evidence to support the claim. If we fail to reject the null hypothesis, there is not enough evidence to support the claim.
Thus, at the 15% significance level, there is not enough evidence to support the administrator’s claim that the mean score for the state’s eights graders on the exam is more than 290.