Ten randomly selected people were asked how long they slept at night. The mean time was 7.1 hours, and the sample standard deviation was 0.78 hour. Find the 95% confidence interval of the mean time. Assume the variable is normally distributed.

A confidence interval can be defined as a range of estimated values that defines the probability that a population parameter would fall or lie within it.

For this exercise, we would use the t-distribution to determine the 95% confidence interval of the mean time because the variance is unknown and shall be replaced by the sample standard deviation.

For the degree of freedom, we have:

Degree of freedom, df = n - 1

Degree of freedom, df = 10 - 1

Degree of freedom, df = 9.

Thus, tα/2 = 2.262.

Assume the variable is normally distributed, the confidence interval would be given by:

X - tα/2(s/√n) < μ < X + tα/2(s/√n)

Substituting the given parameters into the formula, we have;

Confidence interval = 7.1 - 2.262(0.78/√10) < μ < 7.1 + 2.262(0.78/√10).

Confidence interval = 7.1 - 0.56 < μ < 7.1 + 0.56.

Confidence interval = 6.5 < μ < 7.7.

In conclusion, you can be 95% confident that the mean of the ten randomly selected people is between 6.5 and 7.7.

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