Step 1

The difference between B and C is in the choice of critical value. A typical symmetric confidence interval for a location parameter has the form

$$\text{point estimate}\pm \text{critical value}\times \text{standard error},$$

where in turn

$$\text{standard error}=\frac{\text{standard deviation}}{\sqrt{\text{sample size}}},$$

and

$$\text{critical value}\times \text{standard error}=\text{margin of error}.$$

The choice of critical value is informed by the nature of the sampling distribution. When a normally distributed population has unknown variance, the sampling distribution of the mean is Student's t-distributed; however, when the sample size is large, the difference in critical values is negligible; i.e.,

$$\underset{\nu \to \mathrm{\infty}}{lim}{t}_{\nu ,\alpha}^{\ast}={z}_{\alpha}^{\ast},$$

where ${t}_{\nu ,\alpha}^{\ast}$ is the upper $\alpha $ quantile of the Student's t distribution with $\nu $ degrees of freedom, and ${z}_{\alpha}^{\ast}$ is the upper $\alpha $ quantile of the standard normal distribution. So for scenario B, you will use ${z}_{\alpha /2}^{\ast}$ for a two-sided CI with large sample size as an approximation; for scenario C, you will use ${t}_{\nu ,\alpha /2}^{\ast}$ for the critical value.

Step 2

The difference between A and B lies in the standard deviation. In scenario A, it is presumed to be known, thus the standard error will use the population standard deviation $\sigma $ in the numerator. In scenario B, it is unknown, thus you will use the unbiased estimator of the standard deviation

$$s=\sqrt{\frac{1}{n-1}\sum _{i=1}^{n}({x}_{i}-\overline{x}{)}^{2}}.$$

In practice, the factor of $n-1$ rather than n makes little difference when n is large as in this case. But in scenario C, as $\sigma $ is also unknown and must be estimated from the sample, you must use $n-1$, otherwise your CI will not have the required coverage probability even if you correctly use the t-distribution quantile for the critical value.

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