Consider a 90% confidence interval for μ. Assume σ is not known. For which sample size n=10 or n=20, is the confidence interval longer?

Answer is 10, I cannot provide an explanation but it was an answer on one of my assignments.

n a recent study, males used a new weight-loss supplement, and all but of them experienced weight loss after two weeks. In the same study, females used the same supplement, and all but of them experienced weight loss after two weeks.

Fill in the blanks of the statement below to make the statement the most reasonable possible.

To determine which sample size results in a longer confidence interval, we can consider the formula used to calculate the confidence interval:

CI = x̄ ± t * (s/√n)

Where:
- CI represents the confidence interval
- x̄ is the sample mean
- t is the critical value from the t-distribution corresponding to the confidence level and degrees of freedom
- s is the sample standard deviation
- n is the sample size

In this case, since σ is not known, we will be using the t-distribution rather than the normal distribution.

Given that the confidence level (1 - α) is 90%, we can find the value of α by subtracting 90% from 100%, resulting in a significance level of 0.1 (α = 0.1).

Now, let's compare the confidence intervals for n = 10 and n = 20.

For n = 10:
- The degrees of freedom (df) will be (n - 1) = (10 - 1) = 9.
- The critical value t for a 90% confidence level and df = 9 can be obtained from a t-table or using statistical software.

For n = 20:
- The degrees of freedom (df) will be (n - 1) = (20 - 1) = 19.
- The critical value t for a 90% confidence level and df = 19 can be obtained from a t-table or using statistical software.

Since the critical value t for n = 10 corresponds to a smaller degrees of freedom (df = 9), it is generally larger than the critical value t for n = 20 with df = 19.

Using the formula above, we can observe that when the critical value t is larger, the confidence interval becomes wider, resulting in a longer interval.

Therefore, the confidence interval for n = 10 will likely be longer than the confidence interval for n = 20.

To determine which sample size, n=10 or n=20, results in a longer confidence interval, we need to understand how the confidence interval is calculated.

A confidence interval is a range of values within which we estimate the population parameter (in this case, the mean, μ) to likely fall. The length of the confidence interval depends on three factors: the level of confidence, the sample size, and the variability of the data.

In this case, we are given that the confidence level is 90%, so there is a 90% chance that the true population mean falls within the interval. However, the variability of the data, represented by the standard deviation σ, is not known. Therefore, we need to use a t-distribution instead of a normal distribution to construct the confidence interval.

The general formula for a confidence interval for the mean when σ is not known is:

CI = x̄ ± t * (s / sqrt(n))

Where:
CI = Confidence Interval
x̄ = Sample Mean
t = t-value from the t-distribution table corresponding to the desired confidence level and degrees of freedom
s = Sample Standard Deviation
n = Sample Size

To determine which sample size results in a longer confidence interval, we need to compare the value of (s / sqrt(n)) for n=10 and n=20. Since we are assuming that the sample standard deviation, s, is constant for both sample sizes, the only difference will be the denominator, sqrt(n).

The square root of 10 is approximately 3.16, and the square root of 20 is approximately 4.47. As sqrt(n) increases, the denominator gets larger, resulting in a smaller value for (s / sqrt(n)). Therefore, as the sample size increases from 10 to 20, the length of the confidence interval will become shorter.

In conclusion, the confidence interval will be longer for n=10 compared to n=20.