Question

Twenty-two concrete blocks were sampled and tested for crushing strength in order to estimate the proportion that were sufficiently strong for a certain application. Eighteen of the 22 blocks were sufficiently strong. Use the small-sample method to construct a 95% confidence interval for the proportion of blocks that are sufficiently strong.

Answer 1

Answer:

The 95% confidence interval with the normla method would be given by (0.657;0.979)

The 95% confidence interval witht he small sample method would be given by (0.784;0.852)

Step-by-step explanation:

1) Notation and definitions

$X=18$ number of blocks that were sufficiently strong

$n=22$ random sample taken

$\hat p=\frac{18}{22}=0.818$ estimated proportion of blocks that were sufficiently strong

$p$ true population proportion of blocks that were sufficiently strong

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

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

2) Confidence interval (Normal method)

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$. And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.657$

$0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22}}=0.979$

The 95% confidence interval with the normla method would be given by (0.657;0.979)

3) Confidence interval (Small sample method)

The confidence interval for the mean is given by the following formula:  

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n} \frac{N-n}{N-1}}$

But we need to know the total size for the population, and on this case we don't know but let's assumed that N=2000 for example .

If we replace the values obtained we got:

$0.818 - 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.784$

$0.818 + 1.96\sqrt{\frac{0.818(1-0.818)}{22} \frac{2000-22}{2000-1}}=0.852$

The 95% confidence interval witht he small sample method would be given by (0.784;0.852)