Answer:
p-value: 1.000
There is enough evidence at the 1% level of significance to suggest that the proportions are not equal.
Step-by-step explanation:
We will be conducting a difference of 2 proportions hypothesis test
The hypothesis for this test is:
H_0: p1 - p2=0
H_a: p1 - p2 ≠0
(p1 ) = 252/300 = 0.84
(p2) = 195/300 = 0.65
This is a 2 tailed test with a significance level of 1%. So our critical values are: z > 2.575 and z < -2.575
See the attached photo for the calculations for this test
Answer:41/180
Step-by-step explanation:
It equals 8 1/8 because 5 + 3 = 8 plus the 1/8
Answer:
The minimum score required for admission is 21.9.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 20.6, \sigma = 5.2](https://tex.z-dn.net/?f=%5Cmu%20%3D%2020.6%2C%20%5Csigma%20%3D%205.2)
A university plans to admit students whose scores are in the top 40%. What is the minimum score required for admission?
Top 40%, so at least 100-40 = 60th percentile. The 60th percentile is the value of X when Z has a pvalue of 0.6. So it is X when Z = 0.255. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.255 = \frac{X - 20.6}{5.2}](https://tex.z-dn.net/?f=0.255%20%3D%20%5Cfrac%7BX%20-%2020.6%7D%7B5.2%7D)
![X - 20.6 = 0.255*5.2](https://tex.z-dn.net/?f=X%20-%2020.6%20%3D%200.255%2A5.2)
![X = 21.9](https://tex.z-dn.net/?f=X%20%3D%2021.9)
The minimum score required for admission is 21.9.