Question

According to sleep researchers, if you are between the ages of 12 and 18 years old, you need 9 hours of sleep to be fully functional. A simple random sample of 28 students was chosen from a large high school, and these students were asked how much sleep they got the previous night. The mean of the responses was 7.9 hours, with a standard deviation of 2.1 hours. If we are interested in whether students at this high school are getting too little sleep, which of the following represents the appropriate null and alternative hypotheses? (a) H0:μ=7.9 and Ha:μ<7.9(b) H0:μ=7.9 and Ha:μ≠7.9(c) H0:μ=9 and Ha:μ≠9(d) H0:μ=9 and Ha:μ<9(e) H0:μ≤9 and Ha:μ≥9

Answer 1

Answer:

d) H0:μ=9 and Ha:μ<9

$t=\frac{7.9-9}{\frac{2.1}{\sqrt{28}}}=-2.77$    

$p_v =P(t_{(27)}$  

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is significantly lower than 9 hours at 5% of signficance.  

Step-by-step explanation:

Data given and notation  

$\bar X=7.9$ represent the sample mean

$s=2.1$ represent the sample standard deviation

$n=28$ sample size  

$\mu_o =9$ represent the value that we want to test

$\alpha$ 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 students at this high school are getting too little sleep, the system of hypothesis would be:  

Null hypothesis:$\mu \geq 68$  

Alternative hypothesis:$\mu < 68$  

If we analyze the size for the sample is < 30 and we don't know the population deviation so is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:  

$t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}}$  (1)  

t-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".  

The best option for this case would be:

d) H0:μ=9 and Ha:μ<9

Calculate the statistic

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

$t=\frac{7.9-9}{\frac{2.1}{\sqrt{28}}}=-2.77$    

P-value

The first step is calculate the degrees of freedom, on this case:  

$df=n-1=28-1=27$  

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

$p_v =P(t_{(27)}$  

Conclusion  

If we compare the p value and the significance level assumed $\alpha=0.05$ we see that $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, we can conclude that the true mean is significantly lower than 9 hours at 5% of signficance.