# In a sample of 375 college seniors, 307 responded positively when asked if they have spring...

## Question:

In a sample of 375 college seniors, 307 responded positively when asked if they have spring fever. Based upon this, compute a {eq}90\% {/eq} confidence interval for the proportion of all college seniors who have spring fever.

## Confidence Intervals

The confidence intervals, with a certain level of confidence, are estimated from observed statistics of the sampling distribution. The confidence level is simply a numerical measure as to how likely the intervals will contain the value of the unknown parameter.

Given Information

There is a total of 375 collage seniors and 307 of them responded positively about the spring fever.

{eq}X \sim Bin\left( {375,p} \right){/eq}

The pivotal quantity:

{eq}\begin{align*} \dfrac{{X - np}}{{\sqrt {n\hat p\left( {1 - \hat p} \right)} }} \sim Normal\left( {0,1} \right)\\ \dfrac{{\dfrac{X}{n} - p}}{{\sqrt {\dfrac{{\hat p\left( {1 - \hat p} \right)}}{n}} }} \sim Normal\left( {0,1} \right) \end{align*}{/eq}

Where,

{eq}\begin{align*} \hat p &= \dfrac{X}{n}\\ &= \dfrac{{307}}{{375}}\\ &= 0.81867 \end{align*} {/eq}

The confidence interval for proportions are calculated as follows:

{eq}P\left( { - 1.6449 < \dfrac{{\hat p - p}}{{\sqrt {\dfrac{{\hat p\left( {1 - \hat p} \right)}}{n}} }} < 1.6449} \right) = 0.90 {/eq}

{eq}\begin{align*} &p \in \left[ {\hat p \pm {Z_{\alpha = 0.05}}\sqrt {\dfrac{{\hat p\left( {1 - \hat p} \right)}}{n}} } \right]\\ &p \in \left[ {0.81867 - 1.6449\sqrt {\dfrac{{0.81867 \times 0.18133}}{{375}}} ,0.81867 + \sqrt {\dfrac{{0.81867 \times 0.18133}}{{375}}} } \right]\\ &p \in \left[ {0.78594,0.85140} \right] \end{align*} {/eq}

Hence, the 90% confidence interval for the population proportion is (0.78594, 0.85140).