Answer:
The required 97.5% confidence interval is
[tex]\text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\\text {CI} = 15.5 \pm 2.8412\cdot \frac{0.31}{\sqrt{8} } \\\\\text {CI} = 15.5 \pm 2.8412\cdot 0.1096\\\\\text {CI} = 15.5 \pm 0.311\\\\\text {CI} = 15.5 - 0.311, \: 15.5 + 0.311\\\\\text {CI} = (15.19, \: 15.81)\\\\[/tex]
Therefore, we are 97.5% confident that the actual mean amount of juice in all such bottles is within the range of 15.19 to 15.81 ounces
.
Step-by-step explanation:
The 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
Let us first compute the mean and standard deviation of the given data.
Using Excel,
=AVERAGE(number1, number2,....)
The mean is found to be
[tex]\bar{x} = 15.5[/tex]
=STDEV(number1, number2,....)
The standard deviation is found to be
[tex]s = 0.31[/tex]
The confidence interval for the mean amount of juice in all such bottles is given by
[tex]$ \text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) $\\\\[/tex]
Where [tex]\bar{x}[/tex] is the sample mean, n is the samplesize, s is the sample standard deviation and [tex]t_{\alpha/2}[/tex] is the t-score corresponding to a 97.5% confidence level.
The t-score corresponding to a 97.5% confidence level is
Significance level = α = 1 - 0.975 = 0.025/2 = 0.0125
Degree of freedom = n - 1 = 8 - 1 = 7
From the t-table at α = 0.0125 and DoF = 7
t-score = 2.8412
So the required 97.5% confidence interval is
[tex]\text {CI} = \bar{x} \pm t_{\alpha/2}(\frac{s}{\sqrt{n} } ) \\\\\text {CI} = 15.5 \pm 2.8412\cdot \frac{0.31}{\sqrt{8} } \\\\\text {CI} = 15.5 \pm 2.8412\cdot 0.1096\\\\\text {CI} = 15.5 \pm 0.311\\\\\text {CI} = 15.5 - 0.311, \: 15.5 + 0.311\\\\\text {CI} = (15.19, \: 15.81)\\\\[/tex]
Therefore, we are 97.5% confident that the actual mean amount of juice in all such bottles is within the range of 15.19 to 15.81 ounces.
