Question

Pressing pills A drug manufacturer forms tablets by compressing a granular material that contains the active ingredient and various fillers. The hardness of a sample from each batch of tablets produced is measured to control the compression process. The target value for the hardness is m = 11.5. The hardness data for a random sample of 20 tablets are 11.627 11.613 11.493 11.602 11.360 11.374 11.592 11.458 11.552 11.463 11.383 11.715 11.485 11.509 11.429 11.477 11.570 11.623 11.472 11.531 Is there convincing evidence at the 5% level that the mean hardness of the tablets differs from the target value?

Answer 1

Answer:

$t=\frac{11.516-11.5}{\frac{0.0950}{\sqrt{20}}}=0.753$    

$p_v =2*P(t_{(19)}>0.753)=0.461$  

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 have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly different from 11.5 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

Data given: 11.627 11.613 11.493 11.602 11.360 11.374 11.592 11.458 11.552 11.463 11.383 11.715 11.485 11.509 11.429 11.477 11.570 11.623 11.472 11.531

We can calculate the sample mean and deviation with these formulas:

$\bar X = \frac{\sum_{i=1}^n X_i}{n}$

$s = \sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=11.516$ represent the sample mean

$s=0.0950$ represent the sample standard deviation  

$n=20$ sample size  

$\mu_o =11.5$ represent the value that we want to test

$\alpha=0.05$ represent the significance level for the hypothesis test.  

t 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 is different from 11.5, the system of hypothesis would be:  

Null hypothesis:$\mu = 11.5$  

Alternative hypothesis:$\mu \neq 11.5$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-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:  

$t=\frac{11.516-11.5}{\frac{0.0950}{\sqrt{20}}}=0.753$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=20-1=19$  

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

$p_v =2*P(t_{(19)}>0.753)=0.461$  

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 have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly different from 11.5 at 5% of signficance.