Question

A soft-drink machine at a steakhouse is regulated so that the amount of drink dispensed is normally distributed with a mean of 200 milliliters and a standard deviation of 15 milliliters. The soft-drink machine is checked by taking a sample of 9 drinks and computing the average amount of drink dispensed, which is 203 milliliters. The steakhouse manager is concerned that the amount of soft-drink dispensed is more than 200 milliliters. At a 5% significance level, What is the alternative hypothesis associated with the steakhouse manager’s concern?

Answer 1

Answer:

$z=\frac{203-200}{\frac{15}{\sqrt{9}}}=0.6$    

$p_v =P(z>0.6)=0.274$  

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 to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=203$ represent the mean height for the sample  

$\sigma=15$ represent the population standard deviation

$n=9$ sample size  

$\mu_o =200$ 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 higher than 200, the system of hypothesis would be:  

Null hypothesis:$\mu \leq 200$  

Alternative hypothesis:$\mu > 20$  

If we analyze the size for the sample is < 30 but we  know the population deviation so 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:  

$z=\frac{203-200}{\frac{15}{\sqrt{9}}}=0.6$    

P-value

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

$p_v =P(z>0.6)=0.274$  

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 to reject the null hypothesis, so we can't conclude that the true mean is not significantly higher than 200 at 5% of signficance.