Question

An advertisement claims that Fasto Stomach Calm will provide relief from indigestion in less than 10 minutes. For a test of the claim, 35 randomly selected individuals were given the product; the average time until relief was 9.25 minutes. From past studies, the standard deviation of the population is known to be 2 minutes. Can you conclude that the claim is justified? Find the P-value and let a = 0.05.

Answer 1

Answer:

$z=\frac{9.25-10}{\frac{2}{\sqrt{35}}}=-2.219$  

$p_v =P(z$  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 10 minutes.    

Step-by-step explanation:

Data given and notation  

$\bar X=9.25$ represent the sample mean  

$\sigma=2$ represent the population standard deviation

$n=35$ sample size  

$\mu_o =10$ represent the value that we want to test  

$\alpha=0.05$ 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)  

State the null and alternative hypotheses.  

We need to conduct a hypothesis in order to check if the mean is less than 10 minutes, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 10$  

Alternative hypothesis:$\mu < 10$  

Since we know the population deviation, 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".  

Calculate the statistic  

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

$z=\frac{9.25-10}{\frac{2}{\sqrt{35}}}=-2.219$  

P-value  

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

$p_v =P(z$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, so we can conclude that the true mean is significantly lower than 10 minutes.