Answer to Question #337569 in Statistics and Probability for Yen

Question

Val wanted to know the average shearing strength, in pounds (lbs.), of a particular kind of rivet sold in a hardware store. He tested 20 rivets as samples and got the following results.

518

490

513

598

510

532

512

455

500

512

501

487

498

496

500

498

515

520

497

502

Construct a confidence interval for the population mean using 99% confidence.

Answer 1

$$\bar{x}=\dfrac{1}{20}( 518+490+513+598+510$$

$$+532+512+455+500+512+501$$

$$+487+498+496+500+498+515$$

$$+520+497+502)=507.7$$

$$s^2=\dfrac{1}{20-1}\displaystyle\sum_{i=1}^{20}(x_i-507.7)^2=\dfrac{13280.2}{19}$$

$$s=\sqrt{\dfrac{13280.2}{19}}\approx26.4378$$

The critical value for $\alpha = 0.01$ and $df = n-1 = 19$ degrees of freedom is $t_c = z_{1-\alpha/2; n-1} = 2.860935.$

The corresponding confidence interval is computed as shown below:

$$CI=(\bar{x}-t_c\times\dfrac{s}{\sqrt{n}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$=(507.7-2.860935\times\dfrac{26.4378}{\sqrt{20}}, \bar{x}+t_c\times\dfrac{s}{\sqrt{n}})$$

$$507.7+2.860935\times\dfrac{26.4378}{\sqrt{20}})$$

$$=(490.7871, 524.6129)$$

Therefore, based on the data provided, the 99% confidence interval for the population mean is $490.7871< \mu < 524.6129,$ which indicates that we are 99% confident that the true population mean $\mu$ is contained by the interval $(490.7871, 524.6129).$