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