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Q: Suppose a new treatment for a certain disease is given to a sample of 200...

Q: Suppose a new treatment for a certain disease is given to a sample of 200 patients. The treatment was successful for 164 of the patients. Assume that these patients are representative of the population of individuals who have this disease. Calculate a 98% confidence interval for the proportion successfully treated. (Round the answers to three decimal places.)

A: ___ to ___

Expert Solution

Solution :

Given that,

n = 200

x = 164

Point estimate = sample proportion = = x / n = 164 / 200 = 0.82

1 - = 1 - 0.82 = 0.18

At 98% confidence level the z is ,

= 1 - 98% = 1 - 0.98 = 0.02

/ 2 = 0.02 / 2 = 0.01

Z_/2 = Z_0.01 = 2.326

Margin of error = E = Z_{/ 2} * $\sqrt$ (( * (1 - )) / n)

= 1.96 * ( $\sqrt$ ((0.82 * 0.18) /200 )

= 0.063

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.82 - 0.063 < p < 0.82 + 0.063

0.757< p < 0.883

The 98% confidence interval for the population proportion p is : 0.757 to 0.883

milcah answered 3 months ago

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