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Pretty standard but the unknown variance part is the key detail.
First, identify that the population variance is unknown and the sample size is 64, so use the t-distribution with 63 degrees of freedom. Then, state the confidence interval formula: x̄ ± t_(α/2, df) * (s/√n), and plug in the values to compute the interval. Finally, interpret the interval in the context of the problem.
Pro tip: Mention that with n=64, the t and z critical values are nearly identical (t≈2.00 vs z=1.96), but using t is technically correct and shows statistical rigor. Also, briefly interpret the interval in plain language for a non-technical audience, which is valuable in a healthcare company like CVS Health.
Since the population variance is unknown and the sample size is 64, use the t-distribution with n-1 = 63 degrees of freedom. Note that the sample size is large enough that the t-distribution approximates the normal, but the t is still the correct choice.
The formula for a confidence interval for the population mean with unknown variance is: x̄ ± t_(α/2, df) * (s/√n), where x̄ is the sample mean, s is the sample standard deviation, n is the sample size, and t_(α/2, df) is the critical value from the t-distribution.
For a 95% confidence level, α = 0.05, so α/2 = 0.025. With 63 degrees of freedom, the critical t-value is approximately 1.998 (or 2.00). You can mention that you would look this up in a t-table or compute it using software.
Calculate the standard error: s/√n = 16/√64 = 16/8 = 2. Then margin of error = t * SE ≈ 1.998 * 2 = 3.996. The confidence interval is 102 ± 3.996, which gives (98.004, 105.996).
Interpret the interval in context: We are 95% confident that the true population mean lies between approximately 98.0 and 106.0. This means if we repeated this sampling process many times, 95% of the intervals constructed would contain the true mean.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
This one tripped me up more than it should have.
Use the t-test for Pearson correlation: compute t = r * sqrt((n-2)/(1-r^2)) with df = n-2, then find the two-sided p-value from the t-distribution. Compare p to 0.05 and state the conclusion in context.
Pro tip: Mention that with n=100, the test has reasonable power to detect moderate correlations, but always check assumptions like linearity and normality. Also, note that a non-significant result doesn't prove the null; it just means insufficient evidence.
Define H0: ρ = 0 vs. H1: ρ ≠ 0, and set α = 0.05.
Calculate t = r * sqrt((n-2)/(1-r^2)) using r = 0.25 and n = 100.
df = n - 2 = 98. Find the two-sided p-value from the t-distribution with 98 df.
If p < 0.05, reject H0; otherwise fail to reject. Interpret in context of the variables.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.
First, recognize that the top 2.5% corresponds to a z-score of approximately 1.96. Then, use the formula t = μ + zσ to compute the threshold value. Finally, interpret the result in the context of the process, explaining that only 2.5% of items are expected to exceed this value.
Pro tip: When explaining the context, relate it to business implications, such as identifying high-cost patients or outliers, to demonstrate practical understanding. Also, mention that the z-score for 2.5% is a common critical value in statistics, often used for 95% confidence intervals.
State the mean (μ = 50) and standard deviation (σ = 5) of the normal distribution.
Recall that the top 2.5% corresponds to a z-score of approximately 1.96 (since 97.5% of the distribution lies below it).
Use the formula t = μ + zσ = 50 + 1.96 * 5 = 59.8.
Explain that only 2.5% of items from this process are expected to have values greater than 59.8. In a business context, this could represent the top 2.5% of patients by some metric, helping to identify outliers or high-risk groups.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.