In: Statistics and Probability
An administrator has hypothesized that student major preference within the College of Science differs in the current academic year compared to previous years. Using a sample of undeclared undergraduates, intended majors were recorded and compared to the mean percentages computed from several prior years (see table below). Complete the table, provide the requested information, and perform a chi-square goodness-of-fit test. Use an alpha of .05.
Major |
Reported |
Mean % |
Expected |
Chi Square (round to two decimal places) |
Biology |
90 |
40 |
||
Chemistry |
25 |
15 |
||
Environmental Science |
12 |
5 |
||
Physics |
19 |
10 |
||
Psychology |
54 |
30 |
||
N = |
200 |
100 |
χ2obt =
df =
χ2cv =
Concerning the above question, type your results and interpretation in an APA formatted paragraph.
ANSWER::
Chi square test for goodness of fit:
Expected count=200*pi
Major | fo | pi | fe | Chi square=(fo-fe)^2/fe |
Biology | 90 | 0.4 | 80 | 1.250 |
Chemistry | 25 | 0.15 | 30 | 0.833 |
Environmental | 12 | 0.05 | 10 | 0.400 |
Physics | 19 | 0.1 | 20 | 0.050 |
Psychology | 54 | 0.3 | 60 | 0.600 |
Total | 200 | 1 | 200 | 3.133 |
Test statistic:
Chi square=3.133
df=5-1=4
p-value=CHIDIST(3.133,4)=0.5358
Fail to reject the null hypothesis.
not significant.
There is not sufficient evidence to conclude that major preference within the college of science differs in the current academic year compared to previous years.
(OR) TRY THIS ANSWER
Expected Values are Calculated by
Expected Value =N*pi
now
Major | Reported | Mean % | Expected |
Biology | 90 | 40 | 80 |
Chemistry | 25 | 15 | 30 |
Environmental Science | 12 | 5 | 10 |
Physics | 19 | 10 | 20 |
Psychology | 54 | 30 | 60 |
N | 200 |
Now chi square statistics is given by
Chi square DF=r-1=5-1=4
Now P Value is given by
P-Value =P( Chi square >3.133)=0.54
P-Value is very high so we failed to reject H0, therefore, there is no evidence to reject the claim that intended major proportions is the same as that of the prior year .
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