Answer to Question #156860 in Statistics and Probability for SASWAT

Question

The average breaking strength of steel rods is specified to be 18.5 thousand lbs. for this a sample of 14 rods was tested. The mean and standard deviation obtained were 17.85 and 1.955, respectively. Test the significance of the deviation by using 5% level of significance. (t=2.16 at 5%. d.f = 13) O

Answer 1

Given, $n=14, \sigma=1.955, \bar{x}=17.85, \mu=18.5, \alpha=0.05$

Hypothesis is:

$H_O:\mu=18.5$ Vs $H_1:\mu\ne18.5$

$\Z=(\bar{x}-\mu)/(\sigma| \sqrt n)=(17.85-18.5)/(1.955 | \sqrt 14)=-1.244$

P-value at 0.05 and $\Z=1.244$ corresponds to 0.2135

Therefore, $2P(\Z$ $\text{\textless}-1.244)=2\times 0.10749=0.2150$

$P-value=0.2150$

Since $P-value(0.2150) \gt 0.05$ , we failed to reject $H_O$

In conclusion from the above, it is clear that the average breaking strength of steel rods is 18.5.