Mathematics: A researcher wants to estimate the proportion of depressed individuals taking a new anti-depressant drug who find relief. A random sample of individuals who had been taking the drug is questioned; of them found relief from depression. Based upon this, compute a confidence interval for the proportion of all depressed individuals taking the drug who find relief. Then find the lower limit and upper limit of the confidence interval.

2 AStep-by-step explanation:...

A researcher wants to estimate the proportion of

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23.07.2021

The answer is "(0.7322,0.8234)".

Step-by-step explanation:

please find the complete question in the attached file.

The sample has 225 persons, thus n = 225. Of these 175 people, the sample proportion was decreased through depression

$\to \hat{p}=\frac{175}{225}=0.7778$

So z-value is 1.645 for an interval of 90%. The 90% trust interval will be

$\to \hat{p}\pm 1.645\sqrt{\frac{\hat{p}\left ( 1-\hat{p} \right )}{n}}=0.7778\pm 1.645\sqrt{\frac{0.7778\left ( 1-0.7778 \right )}{225}}$

$=0.7778\pm 1.645\cdot 0.0277\\\\=0.7778\pm 0.0456$

The confidence interval therefore is 90% (0.7322,0.8234).

The 90% confidence interval's lower limit is 0.7322.

The 90% confidence interval is 0.8234 at its upper limit.

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