question archive You are conducting a multinomial hypothesis test (α = 0
You are conducting a multinomial hypothesis test (α = 0
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You are conducting a multinomial hypothesis test (α = 0.05) for the claim that all 5 categories are equally likely to be selected. Complete the table.Category Observed Expected Frequency Frequency A 6 B 13 C 23 D 24 E 12 Report All Answers Accurate To Three Decimal Places. But Retain Unrounded Numbers For Future Calculations.
What is the chi-square test-statistic for this data? (Report answer accurate to three decimal places, and remember to use the unrounded Pearson residuals in your calculations.)
χ2=
What are the degrees of freedom for this test?
d.f.=
What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =
The p-value is...
- less than (or equal to) α
- greater than α
This test statistic leads to a decision to...
- reject the null
- accept the null
- fail to reject the null
- accept the alternative
As such, the final conclusion is that...
- There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
- There is not sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
- The sample data support the claim that all 5 categories are equally likely to be selected.
- There is not sufficient sample evidence to support the claim that all 5 categories are equally likely to be selected.
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Answer:
expected value =15.6 (for all categories)
χ2 =15.205
d.f.= 4
p-value =0.0043
- The p-value less than α
- This test statistic leads to a decision to reject the null
- final conclusion is that There is sufficient evidence to warrant rejection of the claim that all 5 categories are equally likely to be selected.
Step-by-step explanation
Since for equality of all 5 categories the expected value will be n/5 =78/5 =15.6
Category Observed Expected (O - E)²/E A 6 15.6 5.9077 B 13 15.6 0.4333 C 23 15.6 3.5103 D 24 15.6 4.5231 E 12 15.6 0.8308 sum 78 78 15.2051
χ2= ∑(O - E)² / E =15.2051
p-value =P(x2 >15.2051) at k-1=4 degrees of freedom
= 0.0043
