Question

1. When operating properly, a chemical plant has a daily production rate that is normally distributed with a mean value of 880 tons/day and a standard deviation of 21 tons/day. During an analysis period, the output is measured with random sampling for 50 days and the mean output was found to be 871 tons/day. With a 95% confidence level, determine if the plant is operating properly.

Answer 1

Answer:

$z=\frac{871-880}{\frac{21}{\sqrt{50}}}=-3.03$  

$p_v =2*P(z$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 880 tons/day at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=871$ represent the sample mean    

$\sigma=21$ represent the population standard deviation for the sample  

$n=50$ sample size  

$\mu_o =880$ represent the value that we want to test  

$\alpha=0.05$ 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 check if the mean differs from 880, the system of hypothesis would be:  

Null hypothesis:$\mu =880$  

Alternative hypothesis:$\mu \neq 880$  

Since we know the population deviation, 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:  

$z=\frac{871-880}{\frac{21}{\sqrt{50}}}=-3.03$  

P-value  

Since is a two-sided test the p value would be:  

$p_v =2*P(z$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is different from 880 tons/day at 5% of signficance.