Answer:
The interval is [910.053; 959.946]
p-value 0.00596
Decision: Reject null hypothesis.
Step-by-step explanation:
Hello!
You need to make a 95% Confidence Interval for the population mean of scholarship examination scores for the freshman.
It is known to be μ= 900 and the assistant dean wants to test if it changed.
The study variable is:
X: -scholarship examination score of one applicant.
population variance is known as σ²= (180)²
Assuming that the variable has a normal distribution the formula for the interval is:
X[bar] ± [tex]Z_{1-\alpha /2}[/tex]*[tex]\frac{S}{\sqrt{n} }[/tex]
935 ± 1.96*[tex]\frac{180}{\sqrt{200} }[/tex]
The interval is [910.053; 959.946]
To test if the examination scores have changed the hypothesis is:
H₀: μ = 900
H₁: μ ≠ 900
α: 0.05
To use a Confidence Interval the following conditions should be met:
1) Both the test and the interval should be made for the same parameter.
2) The hypothesis has to be two_tailed
3) Confidence level 1 - α and significance level α should be complementary.
To make the decision you have to see if the value given to the population mean in the null hypothesis is contained or not by the interval.
If the value is contained by the interval, you do not reject the null hypothesis.
If the value is not contained by the interval, then the decision is to reject the null hypothesis.
Since 900 is not contained by the 95% Confidence interval [910.05; 959.95], the decision is to reject the null hypothesis. This means that the scholarship examination scores of freshman applications have changed.
To calculate the p-value you have to know the value of the statstic under the null hypothesis:
Z= [tex]\frac{935-900}{\frac{180}{\sqrt{200} } }[/tex]= 2.749 ≅ 2.75
p-value is
P(Z<2.75) + P(Z>2.75)= P(Z<2.75) + (1 - P(Z<2.75))= 0.00298+ (1 - 0.9702= 0.00596
I hope it helps!
