He amounts (in ounces) of juice in eight randomly selected juice bottles are: 15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5 construct a 98% confidence interval for the mean amount of juice in all such bottles. assume an approximate normal distribution.

Respuesta :

Given the sample: 15.8, 15.6, 15.1, 15.2, 15.1, 15.5, 15.9, 15.5
mean=15.46
var=0.09410714
standard deviation=0.30676887~0.307
the 98% confidence interval will be:
+/-(0.307/sqrt(8))
=+/-0.1085
thus our answer will be:
15.46+/-0.1085
=15.5685 or15.3515

Answer:

[tex]15.46-2.998\frac{0.307}{\sqrt{8}}=15.13[/tex]    

[tex]15.46+2.998\frac{0.307}{\sqrt{8}}=15.79[/tex]    

So on this case the 98% confidence interval would be given by (15.13;15.79)  

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[/tex] represent the sample mean for the sample  

[tex]\mu[/tex] population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size  

Solution to the problem

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 mean and the sample deviation we can use the following formulas:  

[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex] (2)  

[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex] (3)  

The mean calculated for this case is [tex]\bar X=15.46[/tex]

The sample deviation calculated [tex]s=0.307[/tex]

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=8-1=7[/tex]

Since the Confidence is 0.98 or 98%, the value of [tex]\alpha=0.02[/tex] and [tex]\alpha/2 =0.01[/tex], and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.01,7)".And we see that [tex]t_{\alpha/2}=2.998[/tex]

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

[tex]15.46-2.998\frac{0.307}{\sqrt{8}}=15.13[/tex]    

[tex]15.46+2.998\frac{0.307}{\sqrt{8}}=15.79[/tex]    

So on this case the 98% confidence interval would be given by (15.13;15.79)