The mean income per person in the United States is $38,000, and the distribution

Question

The mean income per person in the United States is $38,000, and the distribution of incomes follows a normal distribution. A random sample of 13 residents of Wilmington, Delaware, had a mean of $48,000 with a standard deviation of $9,200. At the 0.10 level of significance, is that enough evidence to conclude that residents of Wilmington. Delaware, have more income than the national average? State the null hypothesis and the alternate hypothesis. State the decision rule for 0.100 significance level (Round your answer to 3 decimal places.) Compute the value of the test statistic (Round your answer to 2 decimal places.) Is there enough evidence to substantiate that residents of Wilmington. Delaware have more income than the national average at the 0.100 significance level?

Explanation / Answer

a)

Formulating the null and alternative hypotheses,              
              
Ho:   u   <=   38000  
Ha:    u   >   38000   [ANSWER]

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b)
              
As we can see, this is a    right   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    12          
tcrit =    +   1.356217334      

Hence, reject Ho when t > 1.356 [ANSWER]

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c)
              
Getting the test statistic, as              
              
X = sample mean =    48000          
uo = hypothesized mean =    38000          
n = sample size =    13          
s = standard deviation =    9200          
              
Thus, t = (X - uo) * sqrt(n) / s =    3.919077473   [ANSWER]

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d)      
              
As t > 1.356, we   REJECT THE NULL HYPOTHESIS.  

Hence,

[[REJECT]] Ho. There is [[SIGNIFICANT]] evidence... [ANSWER]         

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