where [latex]\bar{x}[/latex] is the sample mean and the standard error used is the standard error of the sample mean, [latex]\frac{s}{\sqrt{n}}[/latex].
The [latex]t[/latex]-critical value in the confidence interval will depend on the sample size (degrees of freedom for the [latex]t[/latex]-distribution: [latex]df=n-1[/latex]) and the confidence level.
Using the statistical tool below, find the [latex]t\text{-critical value}[/latex] for a:
a) [latex]95\%[/latex] confidence interval with a random sample of 45.
b) [latex]90\%[/latex] confidence interval with a random sample of 45.
If [latex]n=45[/latex], then [latex]df = 45-1=44[/latex].
Using the “Find Percentile/Quartile” tab with a [latex]df=44[/latex] and a “Two-Tailed” type of percentile, we found that:
a) [latex]t\text{-critical value}[/latex] for a [latex]95\%[/latex] is [latex]\pm 2.015[/latex]
b) [latex]t\text{-critical value}[/latex] for a [latex]90\%[/latex] is [latex]\pm 1.68[/latex]
A group of engineers developed a new design for a steel cable. They need to estimate the amount of weight the cable can hold. The weight limit will be reported on cable packaging. The engineers take a random sample of [latex]45[/latex] cables and apply weights to each of them until they break. The mean breaking weight for the [latex]45[/latex] cables is [latex]768.2[/latex] lb. The standard deviation of the breaking weight for the sample is [latex]15.1[/latex] lb. What should the engineers report as the mean amount of weight held by this type of cable?
Let’s use these sample statistics to construct a [latex]95\%[/latex] confidence interval for the mean breaking weight of this type of cable.
[latex]\text{estimate }\pm \text{ margin of error}[/latex]
We are [latex]95\%[/latex] confident that the mean breaking weight for all cables of this type is between [latex]763.7[/latex] lb and [latex]772.7[/latex] lb.