Respuesta :
Answer:
[tex]17-2.262\frac{1.9}{\sqrt{10}}=15.641[/tex]
[tex]17+2.262\frac{1.9}{\sqrt{10}}=18.359[/tex]
So on this case the 95% confidence interval would be given by (15.641;18.359)
And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm
Step-by-step explanation:
Previous concepts
A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".
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".
[tex]\bar X=17[/tex] represent the sample mean
[tex]\mu[/tex] population mean (variable of interest)
s=1.9 represent the sample standard deviation
n=10 represent the sample size
Confidence interval
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex] (1)
In order to calculate the critical value [tex]t_{\alpha/2}[/tex] we need to find first the degrees of freedom, given by:
[tex]df=n-1=10-1=9[/tex]
Since the Confidence is 0.95 or 95%, the value of [tex]\alpha=0.05[/tex] and [tex]\alpha/2 =0.025[/tex], and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,9)".And we see that [tex]t_{\alpha/2}=2.262[/tex]
Now we have everything in order to replace into formula (1):
[tex]17-2.262\frac{1.9}{\sqrt{10}}=15.641[/tex]
[tex]17+2.262\frac{1.9}{\sqrt{10}}=18.359[/tex]
So on this case the 95% confidence interval would be given by (15.641;18.359)
And since the lower limit for the confidence interval is higher than 15 we can conclude that at 5% of significance the true mean is higher than 15 cm
