The table below shows the number of students

Achatesopo8

Achatesopo8

Answered question

2022-04-11

The table below shows the number of students absent from a college on particular days in a semester.
 Day  Sun  Mon  Tue  Wed  Thu  Total  Frequency 1008087103110480
Use chi-square goodness of fit test to test the claim that the absence is distributed equally among the days at 0.05 level of significance.

Answer & Explanation

cab65699m

cab65699m

Beginner2022-04-12Added 14 answers

Hypothesis testing is conducted for the parameters of the population like population mean, population standard deviation and so on. It is to check whether he sample values conforms to the pre stated parameter values of the original distribution or not. The hypothesis for the single population mean is conducted using the z or t test. The selection of the test depends upon the value of population parameter that is being tested by conducting the hypothesis.
There are three types of chi-square tests:
1. Chi-square test of independence
2. Chi-square goodness of fit
3. Chi-square test of homogeneity
The formula to compute the chi-square goodness of fit is:
χ2=(O-E)2)E
Here,
χ2=Chi-square test statistic
O=Obsereved frequency
E=Expected frequency
The formula to compute the expected frequency is:
Expected frequency=n×p
Here,
n=Sum of frequency
p=Proportion
The hypotheses are:
H0:p1=p2=p3=p4=p5=15
H1: At least one pi is not equal
The calculation is done as:
 OE(OE)(OE)2E 10096.0004.0000.167 8096.00016.0002.667 8796.0009.0000.844 10396.0007.0000.510 11096.00014.0002.042Sum480480.0000.0006.229
The test statistic is:
χ2=((O-E)2)E)
=0.167+2.667+....+2.042
=6.229
The degree of freedom is:
df=Number of categories-1
=5-1
=4
The p-value is 0.1826.
Here, P-value is greater than the level of significance (0.05).
Hence, the decision is fail to reject null hypothesis
Conclusion:
Thus, there is sufficient evidence to support the claim that the absence is distributed equally among the days.

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