Answered

Evaluate [tex]\( E = z^\ \textless \ em\ \textgreater \ \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \)[/tex] for [tex]\( z^\ \textless \ /em\ \textgreater \ = 1.10, \sigma_1 = 0.77, \sigma_2 = 0.71, n_1 = 15 \)[/tex], and [tex]\( n_2 = 60 \)[/tex].

[tex]\[ E = \square \][/tex]

(Round to two decimal places as needed.)



Answer :

GinnyAnswer
Sure, let's evaluate the given expression step-by-step:

We start with the given formula:
[tex]\[ E = z^ \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} \][/tex]

Given values:
- [tex]\(z^ = 1.10\)[/tex]
- [tex]\(\sigma_1 = 0.77\)[/tex]
- [tex]\(\sigma_2 = 0.71\)[/tex]
- [tex]\(n_1 = 15\)[/tex]
- [tex]\(n_2 = 60\)[/tex]

Step 1: Calculate [tex]\(\frac{\sigma_1^2}{n_1}\)[/tex]
[tex]\[ \sigma_1^2 = 0.77^2 = 0.5929 \][/tex]
[tex]\[ \frac{\sigma_1^2}{n_1} = \frac{0.5929}{15} \approx 0.039527 \][/tex]

Step 2: Calculate [tex]\(\frac{\sigma_2^2}{n_2}\)[/tex]
[tex]\[ \sigma_2^2 = 0.71^2 = 0.5041 \][/tex]
[tex]\[ \frac{\sigma_2^2}{n_2} = \frac{0.5041}{60} \approx 0.0084016667 \][/tex]

Step 3: Add the two variances to get the total variance term
[tex]\[ \frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2} \approx 0.039527 + 0.0084016667 = 0.04792833333333334 \][/tex]

Step 4: Calculate the square root of the variance term
[tex]\[ \sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}} = \sqrt{0.04792833333333334} \approx 0.21892540586540735 \][/tex]

Step 5: Multiply by [tex]\(z^*\)[/tex]
[tex]\[ E = 1.10 \times 0.21892540586540735 \approx 0.2408179464519481 \][/tex]

Step 6: Round the result to two decimal places
[tex]\[ E \approx 0.24 \][/tex]

Therefore, the value of [tex]\(E\)[/tex], rounded to two decimal places, is:
[tex]\[ E = 0.24 \][/tex]

Other Questions