Question

A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is 135 to test this claim, you randomly selected a sample of 32 systems and find evidence to reject the manufacturer's be 135.7 with a standard deviation of 3.3 at a=0.10, do you have enough evidence to reject the manufacturer's claim?

Answer 1

Hypothesized Population Mean $\mu=135$

Population Standard Deviation $\sigma=3.3$

Sample Size $n=32$

Sample Mean $\bar{x}=135.7$

Significance Level $\alpha=0.10$

Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

$H_0: \mu=135$

$H_1: \mu\not=135$

This corresponds to two-tailed test, for which a z-test for one mean, with known population standard deviation will be used.

Based on the information provided, the significance level is $\alpha=0.10,$ and the critical value for two-tailed test is $z_c=1.6449.$

The rejection region for this two-tailed test is $R=\{z:|>1.6449\}.$

The $z$ - statistic is computed as follows:

$$z=\dfrac{\bar{x}-\mu}{\sigma/\sqrt{n}}=\dfrac{135.7-135}{3.3/\sqrt{32}}\approx1.19994$$

Since it is observed that $|z|=1.19994<1.6449=z_c,$ it is then concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than $135,$ at the $\alpha=0.01$ significance level.

Using the P-value approach: The p-value for two-tailed, the significance level $\alpha=0.10, z=1.19994,$ is $p=2P(Z>1.19994)=0.230163,$ and since $p=0.230163>0.10=\alpha,$ it is concluded that the null hypothesis is not rejected.

Therefore, there is not enough evidence to claim that the population mean $\mu$ is different than $135,$ at the $\alpha=0.10$ significance level.