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In a survey of 200 employees, 65% agreed that the management of the cafeteria should be changed. If the confidence interval is 95%, the margin of error for this survey is +/- 6.75% and the maximum number of employees who could have agreed is about?

a. 144 b. 140 c. 136

Respuesta :

Answer:

Option A) 144

Step-by-step explanation:

We are given the following in the question:

Sample size, n = 200

[tex]\hat{p} = 65\% = 0.65[/tex]

Margin of error =

[tex]\pm 6.75\% = \pm 0.0675[/tex]

Confidence interval:

[tex]\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

[tex]\hat{p}\pm \text{Margin of error}[/tex]

Putting values:

[tex](0.65 \pm 0.0675)\\=(0.5825,0.7175)[/tex]

Thus, maximum proportion is

[tex]0.7175 = 71.75\%[/tex]

Maximum number of employees who could have agreed =

[tex]71.75\% \times 200\\\\=\dfrac{71.75}{100}\times 200\\\\= 143.5[/tex]

This, approximately maximum 144 employees agreed that the management of the cafeteria should be changed.

Option A) 144

Answer:

The answers are: in the first one +/- 6.75% and the second one: 144.

Step-by-step explanation:

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