Question

The percent, X , of shrinkage o n drying for a certain type of plastic clay has an average shrinkage percentage :, where parameter : is unknown. A random sample of 45 specimens from this clay showed an average shrinking percentage of 18.4 and a standard deviation of 2.2.
Required:
a. Estimate at 5% level of significance whether the true average shrinkage percentage U: is greater than 17.5 and write your conclusion.
b. Report the p-value.

Answer 1

Answer:

a) $t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744$    

The degrees of freedom are given by:

$df=n-1=45-1=44$  

The critical value for this case is $t_{\alpha}=1.68$ since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

b) $p_v =P(t_{(44)}>2.744)=0.0044$  

Step-by-step explanation:

Information given

$\bar X=18.4$ represent the sample mean

$s=2.2$ represent the sample standard deviation

$n=45$ sample size  

$\mu_o =17.5$ represent the value to verify

$\alpha=0.05$ represent the significance level

t would represent the statistic

$p_v$ represent the p value

Part a

We want to test if the true mean is higher than 17.5, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 17.5$  

Alternative hypothesis:$\mu > 17.5$  

The statistic is given by:

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

And replacing we got:

$t=\frac{18.4-17.5}{\frac{2.2}{\sqrt{45}}}=2.744$    

The degrees of freedom are given by:

$df=n-1=45-1=44$  

The critical value for this case is $t_{\alpha}=1.68$ since the calculated value is higher than the critical we have enough evidence to reject the null hypothesis and we can conclude that the true mean is significantly higher than 18.4

Part b

The p value would be given by:

$p_v =P(t_{(44)}>2.744)=0.0044$