# With 90% confidence, for the sample mean 362.00, the sample standard deviation 11.80, and the...

## Question:

With {eq}90 \% {/eq} confidence, for the sample mean {eq}362.00 {/eq}, the sample standard deviation {eq}11.80 {/eq}, and the sample size {eq}35 {/eq}, what is the upper confidence limit?

## Confidence Interval:

In the confidence interval, there are two confidence limits or bounds, namely, the upper limit and the lower limit. The confidence interval is constructed by using the randomly selected independent data values at a specified confidence level.

Given Information:

• The size of the sample (n) is 35.
• The confidence level (CL) is 0.90.
• The sample mean {eq}\left( {\bar x} \right) {/eq} is 362.00.
• The sample standard deviation (s) is 11.80.

The significance level {eq}\left( \alpha \right) {/eq} is given by:

{eq}\begin{align*} \alpha &= 1 - CL\\ &= 1 - 0.90\\ &= 0.10 \end{align*} {/eq}

Since the sample size is greater than 30, so we will use z-critical value.

At the significance level 0.10, the two tailed critical value {eq}\left( {{z^ * }} \right) {/eq} obtained from the standard normal table is +/- 1.645.

The upper confidence limit for the true mean can be calculated as:

{eq}\begin{align*} \bar x + \left( {{z^ * } \times \dfrac{s}{{\sqrt n }}} \right) &= 362.00 + \left( {1.645 \times \dfrac{{11.80}}{{\sqrt {35} }}} \right)\\ &= 362.00 + 3.28\\ &= 365.28 \end{align*} {/eq}

Therefore, the upper confidence limit is 365.28. 