Question

A civil engineer is analyzing the compressive strength of concrete. Compressive strength is approximately normally distributed with variance of 1000 (psi)2. A random sample of 12 specimens has a mean compressive strength of 3250 psi. (a) Construct a 95% confidence interval on mean compressive strength. (b) Suppose that the manufacturer of the concrete claims the average compressive strength is 3270 psi. Based on your answer in part (a), what would you say about this claim

Answer 1

Answer:

a) $3250-1.96\frac{31.623}{\sqrt{18}}=3232.108$    

$3250+1.96\frac{31.623}{\sqrt{18}}=3267.892$    

So on this case the 95% confidence interval would be given by (3232.108;3267.892)    

b) For this case since the value of 3270 is higher than the upper limit for the confidence interval so we can conclude that the true mean is not higher than 3270 at 5% of signficance. We can reject the claim at the significance level of 5%.

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".

$\bar X= 3250$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

$\sigma^2 =1000$represent the sample standard variance

$s = \sqrt{1000}$ represent the sample deviation

n=12 represent the sample size  

Part a

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

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$

Now we have everything in order to replace into formula (1):

$3250-1.96\frac{31.623}{\sqrt{18}}=3232.108$    

$3250+1.96\frac{31.623}{\sqrt{18}}=3267.892$    

So on this case the 95% confidence interval would be given by (3232.108;3267.892)    

Part b

For this case since the value of 3270 is higher than the upper limit for the confidence interval so we can conclude that the true mean is not higher than 3270 at 5% of signficance. We can reject the claim at the significance level of 5%.