Question

Consider the computer output below. Fill in the missing information. Round your answers to two decimal places (e.g. 98.76). Test of mu = 100 vs not = 100 Variable N Mean StDev SE Mean 95% CI (Lower) 95% CI (Upper) T X 19 98.77 4.77 Enter your answer; SE Mean Enter your answer; 95% CI (Lower) Enter your answer; 95% CI (Upper) Enter your answer; T (a) How many degrees of freedom are there on the t-statistic? Enter your answer in accordance to the item a) of the question statement (b) What is your conclusion if ? Choose your answer in accordance to the item b) of the question statement(c) What is your conclusion if the hypothesis is versus ? Choose your answer in accordance to the item c) of the question statement

Answer 1

Answer:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$      

The 95% confidence interval would be given by (96.625;100.915)  

a) $df=n-1= 19-1= 18$

b) $p_v =2*P(t_{18}$      

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

c) The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$      

But again since the p value is higher than the significance level we fail to reject the null hypothesis.

Step-by-step explanation:

Previous concepts and data given

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

$\bar X=98.77$ represent the sample mean    

$s=4.77$ represent the sample standard deviation  

n=19 represent the sample selected  

$\alpha$ significance level    

State the null and alternative hypotheses.    

We need to conduct a hypothesis in order to check if we have significant difference on the mean of 100, the system of hypothesis would be:    

Null hypothesis:$\mu = 100$    

Alternative hypothesis:$\mu \neq 100$    

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we can calculate the Standard error for the mean like this:

$SE_{Mean}=\frac{s}{\sqrt{n}}=\frac{4.77}{\sqrt{19}}=1.094$

The confidence interval for the mean is given by the following formula:  

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)  

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-NORM.INV(0.025,0,1)".And we see that $z_{\alpha/2}=1.96$  

Now we have everything in order to replace into formula (1):  

$98.77-1.96\frac{4.77}{\sqrt{19}}=96.625$  

$98.77+1.96\frac{4.77}{\sqrt{19}}=100.915$  

So on this case the 95% confidence interval would be given by (96.625;100.915)  

Part a

The degree of freedom are given by:

$df=n-1= 19-1= 18$

Part b

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".    

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

$t=\frac{98.77-100}{\frac{4.77}{\sqrt{19}}}=-1.124$      

Then since is a two sided test the p value would be:    

$p_v =2*P(t_{18}$      

If we compare the p value and a significance level for example $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.

Part c

If the system of hypothesis on this case are:

Null hypothesis:$\mu = 100$    

Alternative hypothesis:$\mu > 100$  

The only thing that changes is the p value and would be given by:

$p_v =P(t_{18}>-1.124)=0.862$      

But again since the p value is higher than the significance level we fail to reject the null hypothesis.