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alexandr402 [8]
4 years ago
10

Identify the CONCLUSION of a hypothesis test of the following claim and sample data: Claim: "The average annual household income

in Warren County is $47,500." A random sample of 86 households from this county is obtained, and they have an average annual income of $48,061 with a standard deviation of $2,351. Test the claim at the 0.02 significance level.
Mathematics
1 answer:
Hatshy [7]4 years ago
7 0

Complete Question

The options for the above question is  

a There is not sufficient evidence to warrant rejection of the claim.    

b  There is sufficient evidence to warrant rejection of the claim.

c There is sufficient evidence to support the claim.    

d There is not sufficient evidence to support the claim.

Answer:

Option A is correct

Step-by-step explanation:

From the question we are told that

     The  population mean is  \mu  =$47,500

     The sample size is n  =  86

      The sample  mean is  \= x  =$48,061

      The standard deviation is \sigma  =$2,351

       The  level of significance is  \alpha  =  0.02

The null hypothesis  is  

     H_o  :  \mu =$47,500

 The  alternative hypothesis is

      H_a  :  \mu \ne $47,500

The critical value of  \alpha from the t-Distribution table is  Z_{\frac{\alpha }{2} } =  2.326

Now the test statistics is mathematically evaluated as

       t  =  \frac{\= x -  \mu }{ \frac{\sigma }{\sqrt{n} }  }

substituting values

      t  =  \frac{48061  -  47500 }{\frac{2351}{\sqrt{86} }  }

      t  = 2.21

Now from the values  obtained we can see that

        Z_{\frac{\alpha }{2} }  >  t

hence we fail to reject the null hypothesis

Hence there is not sufficient evidence to warrant rejection of the claim

     

       

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Answer:

The imaginary part is 0

Step-by-step explanation:

The number given is:

x=(\cos(12)+i\sin(12)+ \cos(48)+ i\sin(48))^6

First, we can expand this power using the binomial theorem:

(a+b)^k=\sum_{j=0}^{k}\binom{k}{j}a^{k-j}b^{j}

After that, we can apply De Moivre's theorem to expand each summand:(\cos(a)+i\sin(a))^k=\cos(ka)+i\sin(ka)

The final step is to find the common factor of i in the last expansion. Now:

x^6=((\cos(12)+i\sin(12))+(\cos(48)+ i\sin(48)))^6

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=(\cos(72)+i\sin(72))+6(\cos(60)+i\sin(60))(\cos(48)+ i\sin(48))+15(\cos(48)+i\sin(48))(\cos(96)+ i\sin(96))+20(\cos(36)+i\sin(36))(\cos(144)+ i\sin(144))+15(\cos(24)+i\sin(24))(\cos(192)+ i\sin(192))+6(\cos(12)+i\sin(12))(\cos(240)+ i\sin(240))+(\cos(288)+ i\sin(288))

The last part is to multiply these factors and extract the imaginary part. This computation gives:

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Im x^6=\sin 72+6cos 60\sin 48+6\sin 60\cos 48+15\cos 96\sin 48+15\sin 96\cos 48+20\cos 36\sin 144+20\sin 36\cos 144+15\cos 24\sin 192+15\sin 24\cos 192+6\cos 12\sin 240+6\sin 12\cos 240+\sin 288

(It is not necessary to do a lengthy computation: the summands of the imaginary part are the products sin(a)cos(b) and cos(a)sin(b) as they involve exactly one i factor)

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