Question

In a random sample of $80$ teenagers, the average number of texts handled in a day is $50$. The $96\%$ confidence interval for the mean number of texts handled by teens daily is given as $46$ to $54$

a) What is the standard deviation of the sample?

b) If the number of samples were doubled, by what factor would the confidence interval change? (keeping the same confidence level)

Answer 1

Step-1: Observing the given data:

Random sample space $n=80$

Average number $\overline{x}=50$

Confident level $=96\%$

Upper limit of mean $\overline{x}+E=54$$\Rightarrow 50+E=54\Rightarrow E=4$

Lower limit of the mean $\overline{x}-E=46$$\Rightarrow 50-E=46\Rightarrow E=4$

Step-2: Finding the value of standard deviation:

We know that,

$E=Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$ where $\sigma$ is standard deviation

First find $Z_{\frac{\alpha }{2}}$

Confident level$=96\%$ So, $\alpha =1-\frac{96}{100}=\frac{4}{100}=0.04$

$\Rightarrow \frac{\alpha }{2}=\frac{0.04}{2}=0.02$

In a $z$ Score table we check the value closest to $1-0.02=0.98$ is $0.9798$ that is corresponding to whole number $2$ and decimal number $0.05$.

So, the value of $Z_{\frac{\alpha }{2}}=2.05$

Substitute $E=4;$ $Z_{\frac{\alpha }{2}}=2.05$; $n=80$ in $E=Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$

$$\begin{gathered} \Rightarrow 4=2.05\times \frac{\sigma }{\sqrt{80}} \\ \Rightarrow \sigma =\frac{4\times \sqrt{80}}{2.05} \\ =\frac{4\times 8.9443}{2.05} \\ =\frac{35.7771}{2.05} \\ =17.4522 \end{gathered}$$

Hence, the standard deviation of the given $80$ samples is $17.4522$

Step-3: If the number of samples were doubled, by what factor would the confidence interval change:

The new sample space $n=160$

$\sigma =17.4522$

we know that,

$E=Z_{\frac{\alpha }{2}}\frac{\sigma }{\sqrt{n}}$

$$\begin{gathered} \Rightarrow E=2.05\times \frac{17.4522}{\sqrt{160}}\left[\because Z_{\frac{\alpha }{2}}=2.05;\sigma =17.4522;n=160\right] \\ =\frac{35.777}{\sqrt{160}} \\ =\frac{35.777}{12.649} \\ =2.828 \end{gathered}$$

We know that,

Upper limit of mean $\overline{x}+E$

$=50+2.828=52.828$

Lower limit of the mean $\overline{x}-E$

$=50-2.828=47.172$

Therefore, the confidence interval is $\left[47.172,52.828\right]$.

Hence, the confidence interval change because of sample space.

Hence,

(a) The standard deviation of the given $80$ samples is $17.4522$.

(b) The confidence interval change because of sample space.