1. A sample of 100 exper
1. A sample of 100expert public health researchers were surveyed and their opinionabout the possibility of the CO-Vid 19 spiking in the fall wasaccessed. They were asked to predict if the spike wouldbe much higher than in the spring, about the same as in thespring, a very small spike or no spike at all. Of the100, 40 reported much higher than in the spring spike, 30 reportedabout the same as in the spring, 20 a very small spike and 10 nospike at all. The researcher wanted to know if theseobtained results deviated significantly from what would be expectedif there were no differences among the expectancies.
a. State the null andalternative hypotheses
b. Perform chi-squaregoodness of fit test
c. What is the criticalvalue of chi-square for this test
d. What is yourdecision with respect to the nullhypothesise. Do theseresults deviate significantly from what would be expected bychance? Describe this deviation
Answer:
a)
Hypothesis:
Null Hypothesis Ho: There is NO significant difference betweenthe observed and the expected value.
Alternative Hypothesis Ha: There is a significant differencebetween the observed and the expected value.
b)
Chi-square test of goodness of fit : The formula is givenbeolw;
As per below calculation table , we get Chi-square statisticvalue = 20 …….(1)
| Predict the spike in the spring | |||||
| High | Same | Very small | No | Total | |
| Observed (O) | 40 | 30 | 20 | 10 | 100 |
| Expected (E) | 25 | 25 | 25 | 25 | 100 |
| (O-E)^2 | 225 | 25 | 25 | 225 | |
| (O-E)^2 / E | 9 | 1 | 1 | 9 | 20 |
c)
Chi-square critical value for degrees of freedom (df) [ 4-1=3 ]and alpha(0.05) = 7.815 …..(2)
( Refer below Chi-square table)
d)
From the above (1) and (2) ; we get;
Since Chi-square statistic value = 20 > Chi-square criticalvalue=7.815
Decision : Reject Ho
Conclusion : The obtained results deviated significantly fromwhat would be expected if there were no differences among theexpectancies.
### End of answers
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