Suppose you do not know the population mean fee charged to H&R Block customers last year. Instead, suppose you take a sample of size n-8 and find a sample mean of 350. Assume that the distribution for fees is normally distributed with a sample standard deviation of $100. i. Before conducting the survey, suppose you believed based on your previous observations, your best guess for population standard deviation of fee charged to H&R Block is $50. With this assumption in mind, What should your sample size n approximately be if you want:

Margin-of-Error of to be 2 % and confidence level to be 95 %?
Margin-of-Error of to be 4% and confidence level to be 95%?
Margin-of-Error of to be 4 % and confidence level to be 99%?

ii. 90% confidence interval for the population mean of fees H&R Block.

a. Calculate the margin of error (MOE) of x using a 10% significance level.
b. Calculate the 90 % confidence interval.
c. Suppose an analyst belief that the population mean fee is equal to $185. Using a 90% confidence level. can we conclude the analyst is right? Why or why not?

i a

i b

i c

iia

iib

    \

iic

    There is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval

Step-by-step explanation:

From the question we are told that

     The sample size is

      The sample mean is      

      The sample standard deviation is  

Considering question i

    i a

         At    

given that the confidence level is 95%  =  0.95

         the level of significance would be  

The critical value of   from the normal distribution table is  

So  the sample size is mathematically evaluated as

=>        

=>        

 i b

  At      and  confidence level  = 95%  =>     =>  

=>          

=>          

 i c

       At       confidence level  = 99%  =>    

The critical value of   from the normal distribution table is  

=>    

=>    

=>    

Considering ii

Given that the level of significance is  

Then the critical value  of   from the normal distribution table is  

Generally the margin of error is mathematically represented as

substituting values

Generally the 90% confidence interval is mathematically evaluated as

=>      

=>    

So the interpretation is that there is 90% confidence that the mean  fee charged to H&R Block customers last year is in the interval .So there is insufficient evidence to conclude that the analyst is right because the population mean fee by the analyst does not fall within the confidence interval.

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step-by-step explanation: