Question 1200159: A school conducts an interest aptitude test for final students, the teacher think that the average student aptitude test score is 470 with a known standard deviation value of 10, so the school takes a sample of the talent interest test value data from 9 students as follows:
Student A Score 440
Student B Score 460
Student C Score 485
Student D Score 509
Student E Score 496
Student F Score 477
Student G Score 457
Student H Score 465
Student I Score 495
The hypothesis put forward by the management department is as follows:
H0: μ = 470
H1: μ # 470
By using a = 5%, testing in two tails (two sides), test this opinion and give the conclusion!
Note:
Do this problem using manual formulas in a structured manner, don't use calculations from excel, minitab, or the like.
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Given information:
mu = 470 = mean
sigma = 10 = standard deviation
Data set = {440, 460, 485, 509, 496, 477, 457, 465, 495}
n = 9 = sample size
alpha = 0.05 = significance level
Hypotheses:
H0: μ = 470
H1: μ # 470
This is a two tailed test due to the "not equal" sign in the alternative hypothesis.
Let's calculate the sample mean xbar.
xbar = (sum of the scores)/(sample size)
xbar = (440+460+485+509+496+477+457+465+495)/(9)
xbar = (4284)/(9)
xbar = 476
Compute the test statistic
z = (xbar - mu)/(sigma/sqrt(n))
z = (476 - 470)/(10/sqrt(9))
z = 1.8
Take notice how I'm using the Z distribution instead of T distribution.
This is because the value of sigma is known.
Use a Z table like this one
https://www.ztable.net/
such tables are found in the back of your stats textbook.
For exams, your teacher would likely give you a reference sheet.
Use that table to find that
P(Z < 1.80) = 0.96407
This value is found in the row that starts with +1.8, and it is in the column that has 0 at the very top.
Scroll down to the section titled "How to Read The Z Table" on that page for more information and an example.
Since P(Z < 1.80) = 0.96407, this means the area to the left of z = 1.80 is roughly 0.96407
The area to the right of this z score is...
P(Z > 1.80) = 1 - P(Z < 1.80)
P(Z > 1.80) = 1 - 0.96407
P(Z > 1.80) = 0.03593
We double this value since we're doing a two tailed test.
0.03593 doubles to 0.07186 which is the approximate p-value.
Rule: if the p-value is smaller than alpha, then reject the null.
We have
alpha = 0.05
p-value = 0.07186
and can see that the p-value is not smaller than alpha, so we fail to reject the null. This means we have no choice but to "accept" the null until more information comes along to disprove it.
Conclusion: Accept the null hypothesis that μ = 470 is the case.
Interpretation: The mean test score appears to be 470. The teacher's claim appears to be correct.
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