Mathematics: 1. A survey shows that 15% of nurses are being underpaid in a hospital that employs 200 nurses. (i) Construct a 90% confidence interval to represent the proportion of underpaid nurses in the hospital. Give an interpretation of this interval. [4] (ii) Suppose that in another hospital, 12% of the 120 nurses employed are underpaid. Construct a 90% confidence interval for the difference in proportions of nurses being underpaid at both hospitals. [4]

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1. A survey shows that 15% of nurses are being underpaid

jbernaaaaaal

30.01.2022

(i) Interval for proportion of underpaid nurses: 0.15 ± 0.0415.

(ii) Interval for difference in proportion: 0.03 ± 0.0641

Step-by-step explanation: Confidence Interval for a representation of the proportion is calculated by:

$p_{hat}$ ± z. $\sqrt{\frac{p_{hat}(1-p_{hat})}{n} }$

(i) $p_{hat}$ is the proportion, in this case, of underpaid nurses: $p_{hat}$ = 0.15

Confidence Interval is 90%, so z = 1.645

The interval is:

0.15 ± 1.645. $\sqrt{\frac{0.15(0.85)}{200} }$

0.15 ± 1.645*0.0252

0.15 ± 0.0415

Which means that, in 90% of times, the proportion of underpaid nurses wil be between 0.1085 and 0.1915.

(ii) For the difference in proportions:

$p_{hat_1}-p_{hat_2}$ ± z. $\sqrt{\frac{p_{hat_1}(1-p_{hat_1})}{n_{1}}+\frac{p_{hat_2}(1-p_{hat_2})}{n_{2}} }$

The proportion for the second hospital is $p_{hat_2}$ = 0.12 and, since it is the same confidence interval, z = 1.645.

Calculating:

(0.15-0.12) ± 1.645. $\sqrt{\frac{0.15(0.85)}{200} + \frac{0.12(0.88)}{120} }$

0.03 ± 1.645*0.0389

0.03 ± 0.0641

The 90% confidence interval for the difference in proportions of nurses being underpaid is between 0.0341 and 0.0941

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