While interpreting various results from a set of data, a researcher needs to know how sure is he while dealing with the data. Confidence interval is a range within which most plausible values would occur. To calculate confidence interval, one needs to set confidence level as 90%, 95%, or 99% etc. Most commonly used confidence level is 95%. Confidence interval represents a particular interval within which the data is 95% (or whatever the confidence level chosen) sure or certain for a particular outcome. The formula for confidence interval is given below:

and

Where,

$n$ = Number of terms

$x$ = Sample Mean

$s$ = Standard Deviation

$z_{\frac{\alpha }{2}}$ = Value corresponding to $\frac{\alpha }{2}$ in $z$ table

$t_{\frac{\alpha }{2}}$ = Value corresponding to $\frac{\alpha }{2}$ in $t$ table

$\alpha$ = $1$ - $\frac{confidence\ level}{100}$.

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Few problems based on confidence interval are given below: ### Solved Examples

**Question 1: **Average score obtained by 10 students in a test is 25. Standard deviation of the sample is 8. Find the confidence interval at 95% confidence level?

** Solution: **

n = 10

Mean x = 25

Standard deviation σ = 8

Confidence level = 95% = 0.95

$\alpha$ =1 - $\frac{confidence\ level}{100}$

$\alpha$ =1- $\frac{95}{100}$ = 0.05

Formula for confidence interval:

$Confidence\ interval=x\pm t_{\frac{\alpha }{2}}\left ( \frac{\sigma }{\sqrt{n}} \right )$

$Confidence\ interval=25\pm t_{\frac{0.05}{2}}\left ( \frac{8}{\sqrt{10}} \right )$

= $25\pm 2.09302\left ( \frac{8}{\sqrt{10}} \right )$

**Question 2: **A survey was performed on 50 families asking how many members are there in the family. The mean and standard deviation of the survey was 6 and 5 respectively. If confidence level is 95%, what would be the confidence interval of the survey?

** Solution: **

n = 50

Mean x = 6

Standard deviation σ = 5

Confidence level = 95% = 0.95

$\alpha$ =1 - $\frac{confidence\ level}{100}$

$\alpha$ =1 - $\frac{95}{100}$ = 0.05

Formula for confidence interval:

$Confidence\ interval$ = $x \pm z_{\frac{\alpha }{2}}\left ( \frac{\sigma }{\sqrt{n}} \right )$

$Confidence\ interval$ = $6 \pm z_{\frac{0.05}{2}}\left ( \frac{5}{\sqrt{10}} \right )$

= $6\pm 0.51 \left ( \frac{5}{\sqrt{10}} \right )$

= 4.579 to 7.421

n = 10

Mean x = 25

Standard deviation σ = 8

Confidence level = 95% = 0.95

$\alpha$ =1 - $\frac{confidence\ level}{100}$

$\alpha$ =1- $\frac{95}{100}$ = 0.05

Formula for confidence interval:

$Confidence\ interval=x\pm t_{\frac{\alpha }{2}}\left ( \frac{\sigma }{\sqrt{n}} \right )$

$Confidence\ interval=25\pm t_{\frac{0.05}{2}}\left ( \frac{8}{\sqrt{10}} \right )$

= $25\pm 2.09302\left ( \frac{8}{\sqrt{10}} \right )$

=19.28 to 30.72

n = 50

Mean x = 6

Standard deviation σ = 5

Confidence level = 95% = 0.95

$\alpha$ =1 - $\frac{confidence\ level}{100}$

$\alpha$ =1 - $\frac{95}{100}$ = 0.05

Formula for confidence interval:

$Confidence\ interval$ = $x \pm z_{\frac{\alpha }{2}}\left ( \frac{\sigma }{\sqrt{n}} \right )$

$Confidence\ interval$ = $6 \pm z_{\frac{0.05}{2}}\left ( \frac{5}{\sqrt{10}} \right )$

= $6\pm 0.51 \left ( \frac{5}{\sqrt{10}} \right )$

= 4.579 to 7.421