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10. A researcher is interested in estimating the mean alcohol content for a particular brand of cough medicine. They randomly sample 50 bottles of this brand and find a sample mean of 8.6 (percent alcohol). Let u denote the mean alcohol content (in percent) for the population of all bottles of the brand under study. The population standard deviation is 2.2. a.) Construct a 95% confidence interval for u​

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Answer:

To construct a 95% confidence interval for the mean alcohol content of the population of all bottles of the brand under study, you can use the following formula:

Sample mean +/- (t-value * standard error)

The t-value can be found using a t-distribution table or a computer program. It is based on the desired confidence level (in this case, 95%) and the degrees of freedom, which is the sample size minus 1 (in this case, 50-1=49).

The standard error is calculated as:

Standard error = population standard deviation / sqrt(sample size)

Plugging in the given values, you get:

Standard error = 2.2 / sqrt(50) = 0.22

Using a t-distribution table or computer program, you can find that the t-value for a 95% confidence interval with 49 degrees of freedom is 2.01.

Therefore, the 95% confidence interval for the mean alcohol content of the population of all bottles of the brand under study is:

8.6 +/- (2.01 * 0.22)

Which simplifies to:

8.6 +/- 0.44

So the 95% confidence interval for the mean alcohol content of the population is (8.16, 8.94).

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