Respuesta :
Answer:
The 95% confidence interval would be given by (29.780;32.020)
Step-by-step explanation:
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=30.9[/tex] represent the sample mean for the sample
[tex]\mu[/tex] population mean (variable of interest)
s=7.5 represent the sample standard deviation
n=174 represent the sample size
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=174-1=173[/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 tabel to find the critical value. The excel command would be: "=-T.INV(0.025,173)".And we see that [tex]t_{\alpha/2}=1.97[/tex], this value is similar to the obtained with the normal standard distribution since the sample size is large to approximate the t distribution with the normal distribution.
Now we have everything in order to replace into formula (1):
[tex]30.9-1.97\frac{7.5}{\sqrt{174}}=29.780[/tex]
[tex]30.9+1.97\frac{7.5}{\sqrt{174}}=32.020[/tex]
So on this case the 95% confidence interval would be given by (29.780;32.020)
The value 29.6 is not contained on the interval calculated.
