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Sonja [21]
3 years ago
8

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?
Mathematics
1 answer:
nirvana33 [79]3 years ago
3 0

Answer:

z=\frac{133-135}{\frac{3.3}{\sqrt{32}}}=-3.428      

p_v =P(z  

If we compare the p value and the significance level given for example \alpha=0.1 we see that p_v 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      

\bar X=133 represent the sample mean

\sigma=3.3 represent the standard deviation for the population      

n=32 sample size      

\mu_o =135 represent the value that we want to test    

\alpha represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)      

p_v 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:\mu \geq 135      

Alternative hypothesis:\mu < 135      

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:      

z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}} (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:      

z=\frac{133-135}{\frac{3.3}{\sqrt{32}}}=-3.428      

Calculate the P-value      

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

p_v =P(z  

Conclusion      

If we compare the p value and the significance level given for example \alpha=0.1 we see that p_v 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      

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