The College Student Journal (December 1992) investigated differences in traditional and nontraditional students, where nontradit
1 answer:
Answer:
There is a 0.13% probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65.
Step-by-step explanation:
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by
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. Subtracting 1 by the pvalue, we This p-value is the probability that the value of the measure is greater than X.
For this problem, we have that:
Based on the study results, we can assume the population mean and standard deviation for the GPA of nontraditional students is and
.
We have a sample of 100 students, so we need to find the standard deviation of the sample, to use in the place of in the z score formula.
.
What is the probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65?
This is 1 subtracted by the pvalue of Z when . So
A zscore of 3 has a pvalue of 0.9987.
So, there is a 1-0.9987 = 0.0013 = 0.13% probability that the random sample of 100 nontraditional students have a mean GPA greater than 3.65.
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Answer:
a) Null hypothesis:
Alternative hypothesis:
And we have this property
The degrees of freedom for the numerator on this case is given by where k =3 represent the number of groups.
The degrees of freedom for the denominator on this case is given by .
And the total degrees of freedom would be
The mean squares between groups are given by:
And the mean squares within are:
And the F statistic is given by:
And the p value is given by:
So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.
b)
The degrees of freedom are given by:
The confidence level is 99% so then and
and the critical value would be:
The confidence interval would be given by:
Step-by-step explanation:
Previous concepts
Analysis of variance (ANOVA) "is used to analyze the differences among group means in a sample".
The sum of squares "is the sum of the square of variation, where variation is defined as the spread between each individual value and the grand mean"
Part a
Null hypothesis:
Alternative hypothesis:
If we assume that we have groups and on each group from
we have
individuals on each group we can define the following formulas of variation:
And we have this property
The degrees of freedom for the numerator on this case is given by where k =3 represent the number of groups.
The degrees of freedom for the denominator on this case is given by .
And the total degrees of freedom would be
The mean squares between groups are given by:
And the mean squares within are:
And the F statistic is given by:
And the p value is given by:
So then since the p value is lower then the significance level we have enough evidence to reject the null hypothesis and we conclude that we have at least on mean different between the 3 groups.
Part b
For this case the confidence interval for the difference woud be given by:
The degrees of freedom are given by:
The confidence level is 99% so then and
and the critical value would be:
The confidence interval would be given by:
Answer:
150
Step-by-step explanation:
90-60 is 30
180-30 is150
