Question

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?

Answer 1

Answer:

i $\to$ a

    $n = 96040000$

i $\to$ b

    $n_1 =24010000$

i $\to$ c

    $n_2 =41602500$

ii$\to$a

     $E = 58.16$

ii$\to$b

    $291.84 < \mu < 408.16$\

ii$\to$c

    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 $n = 8$

      The sample mean is  $\= x = \ 350$    

      The sample standard deviation is  $\ 100$

Considering question i

    i $\to$ a

         At   $E = 0.02$  

given that the confidence level is 95%  =  0.95

         the level of significance would be  $\alpha =1-0.95 = 0.05$

The critical value of  $\frac{\alpha }{2}$ from the normal distribution table is  

        $Z_{\frac{ \alpha }{2} } = 1.96$

So  the sample size is mathematically evaluated as

            $n = [ \frac{Z_{\frac{\alpha }{2} } * \sigma }{E} ]^2$

=>        $n =[ \frac{ 1.96 * 100}{ 0.02} ]^2$

=>         $n = 96040000$

 i $\to$ b

  At  $E_1 = 0.04$    and  confidence level  = 95%  =>  $\alpha_1 = 0.05$   =>  $Z_{\frac{\alpha_1 }{2} } = 1.96$

             $n_1 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_1} ]^2$

=>           $n_1 =[ \frac{ 1.96 * 100}{ 0.04} ]^2$

=>           $n_1 =24010000$

 i $\to$ c

       At   $E_2 = 0.04$     confidence level  = 99%  =>    $\alpha_2 = 0.01$

The critical value of  $\frac{\alpha_2 }{2}$ from the normal distribution table is  

        $Z_{\frac{ \alpha_2 }{2} } = 2.58$

=>    $n_2 = [ \frac{Z_{\frac{\alpha_2 }{2} } * \sigma }{E_2} ]^2$

=>    $n_2 =[ \frac{ 2.58 * 100}{ 0.04} ]^2$

=>    $n_2 =41602500$

Considering ii

Given that the level of significance is  $\alpha = 0.10$

Then the critical value  of  $\frac{\alpha }{2}$ from the normal distribution table is  

           $Z_{\frac{\alpha }{2} } = 1.645$

Generally the margin of error is mathematically represented as

          $E = Z_{\frac{\alpha }{2} } * \frac{\sigma }{\sqrt{n} }$

substituting values

         $E = 1.645 * \frac{100 }{\sqrt{8} }$

         $E = 58.16$

Generally the 90% confidence interval is mathematically evaluated as

         $\= x - E < \mu < \= x + E$

=>      $350 - 58.16 < \mu < 350 + 58.16$

=>     $291.84 < \mu < 408.16$

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.