Question

If 5 pieces of certain ribbon selected at random have mean breaking strength

of 169.5 pounds with standard deviation 5.7, do they confirm to the

specification mean breaking strength of 180 pounds.

Use t଴.଴ଵ = 3.747 with 4 d.o.f.

Answer 1

The following null and alternative hypotheses need to be tested:

$H_0:\mu=180$

$H_a:\mu<180$

This corresponds to a left-tailed test, for which a t-test for one mean, with unknown population standard deviation, using the sample standard deviation, will be used.

Based on the information provided, the significance level is $\alpha = 0.01,$ $df=n-1=4$ degrees of freedom, and the critical value for a left-tailed test is $t_c = -3.747.$

The rejection region for this left-tailed test is $R = \{t: t < -3.747\}.$

The t-statistic is computed as follows:

$$t=\dfrac{\bar{x}-\mu}{s/\sqrt{n}}=\dfrac{169.5-180}{5.7/\sqrt{5}}\approx-4.119$$

Since it is observed that $t = -4.119 <-3.747= t_c,$ it is then concluded that the null hypothesis is rejected.

Using the P-value approach:

The p-value for left-tailed, $df=4$ degrees of fredom, $t=-4.119$ is $p = 0.007312,$ and since  $p = 0.007312 < 0.01=\alpha,$ it is concluded that the null hypothesis is rejected.

Therefore, there is enough evidence to claim that the population mean $\mu$

is less than 180, at the $\alpha = 0.01$ significance level.