Question

5 The lowest and highest observations in a population are 22 and 60, respectively. What is the minimum sample size required to estimate with 95% confidence if the desired margin of error is E-19? What happens to nif you decide to estimate with 99% confidence? (You may find it useful to reference the table. Round Intermediate calculations to at least 4 decimal places and value to 3 decimal places. Round up your answers to the nearest whole number) 11.11 points Confidence Level 95% ebook

Answer 1

Highest observation = 60

Lowest observation = 22

Range = 60 - 22 = 38

By range rule of standard deviation, we have a rough estimate of standard deviation equal to one - fourth of the range.

So, standard deviation = $\sigma$ ~ 38/4 = 9.5

Now, margin of errror = E = 1.9

For confidence level 95%, we have Critical value of z (for two tailed test at 0.05 level of significance) = = 1.96 (can be obtained from the z table by finding the z corresponding to the area close to 0.05/2)

<0.05/2

So, sample size for mean for 95% confidence will be -

$n = (\frac{z_{0.05/2}* \sigma}{E})^{2}$

$= (\frac{1.96* 9.5}{1.9})^{2}$

= 96.04

~ 96

For 99% confidence level, the critical value of z (for two tailed test at 0.01 level of significance) is $z_{0.01/2}$ = 2.58 (can be obtained from the z table by finding the z corresponding to the area close to 0.01/2)

So, sample size =

$n = (\frac{z_{0.01/2}* \sigma}{E})^{2}$

  $= (\frac{2.58* 9.5}{1.9})^{2}$

= 166.4

~ 166