Question

It is important that face masks used by firefighters be able to withstand high temperatures because firefighters commonly work in temperatures of 200-500°F. In a test of one type of mask, 12 of 60 masks had lenses pop out at 250°. Construct a 90% upper confidence limit for the true proportion of masks of this type whose lenses would pop out at 250°. (Round your answers to four decimal places.)

Answer 1

Answer:

$(-\infty, 0.2661)$

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The population proportion have the following distribution

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$

n=60 represent the sample size  

X=12 represent the number of masks that had lenses pop out at 250°

$\hat p =\frac{12}{60}=0.2$ represent the estimated proportion of masks that  had lenses pop out at 250°

p represent the population proportion of masks that  had lenses pop out at 250°

Confidence =0.9 or 90%

$\alpha=0.1$ represent the significance level

Confidence interval

On this case we want a interval on this form : $(-\infty,\hat p +z_{\alpha}\sqrt{\frac{\hat p (1-\hat p)}{n}})$

So the critical value would be on this case $z_{\alpha}=1.28$ and we can use the following excel code to find it: "=NORM.INV(1-0.1,0,1)"

We found the lower limit like this:

$0.2 +1.28\sqrt{\frac{0.2 (1-0.2)}{60}}=0.2661$

And the interval would be: $(-\infyt, 0.2661)$