Question

The viscosity of a liquid detergent is supposed to average 800 centistokes at 25° C. A random sample of 16 batches of detergent is collected, and the average viscosity is 812. Suppose we know that the standard deviation of viscosity is o = 25 centistokes: (a) State the hypotheses that should be tested, (b) Test these hypotheses using a = 0.05. What are your conclusions?, (c) What is the P-value for the test?,(d) Find a 95 percent confidence interval on the mean.

Answer 1

Answer:

a) Null hypothesis:$\mu = 800$        

Alternative hypothesis:$\mu \neq 800$  

b) $z=\frac{812-800}{\frac{25}{\sqrt{16}}}=1.92$    

c) $p_v =2*P(z>1.92)=0.0548$    

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not different from the value of 800.

d) $812- 1.96 \frac{25}{\sqrt{16}}=799.75$

$812+ 1.96 \frac{25}{\sqrt{16}}=824.25$

Since the confidence interval contains the value 800 this result agrees with the result obtained in the hypothesis test we fail to reject the null hypothesis that the true mean is 800

Step-by-step explanation:

Data given and notation        

$\bar X=812$ represent the mean for the sample    

$\sigma=25$ represent the standard deviation for the population      

$n=16$ sample size        

$\mu_o =800$ 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)    

a) State the null and alternative hypotheses.        

We need to conduct a hypothesis in order to determine if the true mean is different from 800 centistokes:      

Null hypothesis:$\mu = 800$        

Alternative hypothesis:$\mu \neq 800$        

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".    

b) Calculate the statistic        

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

$z=\frac{812-800}{\frac{25}{\sqrt{16}}}=1.92$        

c) Calculate the P-value        

Since is a two tailed test the p value would be:        

$p_v =2*P(z>1.92)=0.0548$    

Conclusion        

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we FAIL to reject the null hypothesis, and the the actual mean is not different from the value of 800.

d) Confidence interval

The confidence interval for the mean is given by the following formula:  

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$   (1)  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.

$z_{\alpha/2}=1.96$

And replacing we got:

$812- 1.96 \frac{25}{\sqrt{16}}=799.75$

$812+ 1.96 \frac{25}{\sqrt{16}}=824.25$

Since the confidence interval contains the value 800 this result agrees with the result obtained in the hypothesis test we fail to reject the null hypothesis that the true mean is 800