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. Construct a 97.5% confidence interval for the mean amount of juice in all such bottles. Assume an approximate Normal distribution.

The required 97.5% confidence interval is

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

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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

=STDEV(number1, number2,....)

The standard deviation is found to be

The confidence interval for the mean amount of juice in all such bottles is given by

Where is the sample mean, n is the samplesize, s is the sample standard deviation and 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

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.

is this part of a question or something

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step-by-step explanation: