Question

The price of a certain combo meal at different franchises of a national fast food company varies from​ $5.00 to ​$17.33 and has a known standard deviation of ​$2.18. A sample of 27 students in an online course that includes students across the country stated that their average price is ​$5.50. The students have also stated that they are generally unwilling to pay more than ​$6.25 for this meal. Formulate and conduct a hypothesis test to determine if you can conclude that the population mean is less than ​$6.25. Use a level of significance of 0.01. Is there sufficient evidence at the 0.01 level of significance that the population mean is less than ​$6.25​?

Answer 1

Answer:

$z=\frac{5.50-6.25}{\frac{2.18}{\sqrt{27}}}=-1.788$    

$p_v =P(Z<-1.788)=0.042$  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly lower than 6.25 at 1% of signficance.  

Explanation:

Data given and notation  

$\bar X=5.60$ represent the sample mean

$\sigma=2.18$ represent the sample population deviation

$n=27$ sample size  

$\mu_o =6.25$ represent the value that we want to test

$\alpha=0.01$ 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 less than 6.25, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 6.25$  

Alternative hypothesis:$\mu < 6.25$  

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:  

$z=\frac{5.50-6.25}{\frac{2.18}{\sqrt{27}}}=-1.788$    

P-value

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

$p_v =P(Z<-1.788)=0.042$  

Conclusion  

If we compare the p value and the significance level given $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to fail reject the null hypothesis, so we can't conclude that the true mean is significantly lower than 6.25 at 1% of signficance.