A manufacturer of sprinkler systems designed for fire protection claims that the average activating temperature is at least 135°F. To test this claim, you randomly select a sample of 32 systems and find the mean activation temperature to be 133°F. Assume the population standard deviation is 3.3°F. At a = 0.10, do you have enough evidence to reject the manufacturer’s claim?

If we compare the p value and the significance level given for example [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significantly lower than 135 so the claim not makes sense

Step-by-step explanation:

Data given and notation

[tex]\bar X=133[/tex] represent the sample mean

[tex]\sigma=3.3[/tex] represent the standard deviation for the population

[tex]n=32[/tex] sample size

[tex]\mu_o =135[/tex] represent the value that we want to test

[tex]\alpha[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to determine if the true mena is at least 135, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 135[/tex]

Alternative hypothesis:[tex]\mu < 135[/tex]

We know the population deviation, so for this case is better apply a z test to compare the actual mean to the reference value, and the statistic is given by:

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".

Calculate the statistic

We can replace in formula (1) the info given like this:

[tex]z=\frac{133-135}{\frac{3.3}{\sqrt{32}}}=-3.428[/tex]

Calculate the P-value

Since is a one-side lower test the p value would be:

[tex]p_v =P(z<-3.428)=0.0003[/tex]

Conclusion

If we compare the p value and the significance level given for example [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we reject the null hypothesis, and the true mean is significantly lower than 135 so the claim not makes sense