A drug company is considering marketing a new local anesthetic. The effective time of the anesthetic the drug company is currently producing has a normal distribution with an mean of 7.4 minutes with a standard deviation of 1.2 minutes. The chemistry of the new anesthetic is such that the effective time should be normally distributed with the same standard deviation, but the mean effective time may be lower. If it is lower, the drug company will market the new anesthetic; otherwise, they will continue to produce the older one. A sample of size 36 results in a sample mean of 7.1. A hypothesis test will be done at the 10% level of significance to help make the decision.

Comparing the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the true mean is significantly lower than 7.4 minutes at 10% of significance

Step-by-step explanation:

Data given and notation

[tex]\bar X=7.1[/tex] represent the sample mean

[tex]\sigma=1.2[/tex] represent the population standard deviation

[tex]n=36[/tex] sample size

[tex]\mu_o =7.4[/tex] represent the value that we want to test

[tex]\alpha=0.1[/tex] represent the significance level for the hypothesis test.

z would represent the statistic (variable of interest)

[tex]p_v[/tex] 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 true mean i lower than 7.4, the system of hypothesis would be:

Null hypothesis:[tex]\mu \geq 7.4[/tex]

Alternative hypothesis:[tex]\mu < 7.4[/tex]

The statistic is given by

[tex]z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}[/tex] (1)

Calculate the statistic

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

[tex]z=\frac{7.1-7.4}{\frac{1.2}{\sqrt{36}}}=-1.5[/tex]

P-value

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

[tex]p_v =P(z<-1.5)=0.0668[/tex]

Conclusion

Comparing the p value and the significance level given [tex]\alpha=0.1[/tex] we see that [tex]p_v<\alpha[/tex] so we can conclude that we have enough evidence to reject the null hypothesis, and the true mean is significantly lower than 7.4 minutes at 10% of significance