A new curing process developed for a certain type of cement results in a mean compressive strength of 5000 kilograms per square centimeter with a standard deviation of 100 kilograms. To test the hypothesis that µ = 5000 against the alternative that µ < 5000, a random sample of 25 pieces of cement is tested. The critical region is defined to be x<4970. Find the probability of committing a type I error when H0 is true.

The critical region on this case is [tex]\bar X <4970[/tex] so then if the value of [tex]\bar X \geq 4970[/tex] we fail to reject the null hypothesis. In other case we reject the null hypothesis

Null and alternative hypotheses to be tested

We need to conduct a hypothesis in order to determine if the true mean is 5000, the system of hypothesis would be:

Null hypothesis:[tex]\mu = 5000[/tex]

Alternative hypothesis:[tex]\mu \neq 5000[/tex]

Let's define the random variable X ="The compressive strength".

We know from the Central Limit Theorem that the distribution for the sample mean is given by:

[tex]\bar X \sim N(\mu , \frac{\sigma}{\sqrt{n}})[/tex]

Find the probability of committing a type I error when H0 is true.

The definition for type of error I is reject the null hypothesis when actually is true, and is defined as [tex]\alpha[/tex] the significance level.

So we can define [tex]\alpha[/tex] like this:

[tex]\alpha= P(Error I)= P(\bar X <4970, when,\mu = 5000)[/tex]

And in order to find this probability we can use the Z score given by this formula:

[tex]Z=\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]

And the value for the probability of error I is givn by:

[tex]\alpha= P(\bar X <4970) =P(\frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}<\frac{4970 -5000}{\frac{100}{\sqrt{25}}})=P(Z<-1.5)=0.0668[/tex]