Question

A company manufactures batteries that the CEO claims will last an average of 350 hours under normal use. A researcher randomly selected 20 batteries from the production line and tested these batteries. The tested batteries had a mean life span of 320 hours with a standard deviation of 50 hours. Do you agree with CEO?

Answer 1

H₀ $\mu=350$

H_a $\mu<350$

$n=20, \bar {X}=320 s=50$

Since population standard deviation is unknown, $\bar{X}$ follows t_n-1 distribution with mean $\mu$ and variance $\frac{s}{\sqrt{n}}$

$t-value=\frac{\bar{X}-\mu}{\frac{s}{\sqrt{n}}}$

$=\frac{320-350}{\frac{50}{\sqrt{20}}}=-2.68328$

The critical region is on the left, at 5% level of significance, t_0.05,19 is$-1.729133$

The critical value can be obtained from =T.INV(0.05,19) Excel formula.

Taking absolute values, t-value is greater than the critical value; thus, we reject the bill hypothesis and conclude that the batteries average lifespan is less than 350. The CEO's claim is not supported.

Similarly, t_0.95,19 can be used, and from t- table or =T.INV(0.95,19), the critical value is 1.729132812, and by taking absolute value of t- value, similar conclusion is arrived at.