A random sample was taken from Population1 using the following code.

################
rbeta(150,shape1=1,shape2=1)
################ The sample from Population 1 ############################
0.73, 0.43, 0.8, 0.06, 0.37, 0.79, 0.31, 0.23, 0.28, 0.62, 0.07, 0.35, 0.73, 0.99, 0.13, 0.52, 0.51, 0.17, 0.21, 0.28, 0.18, 0.22, 0.47, 0.98, 0.27, 0.52, 0.56, 0.93, 0.68, 0.34, 0.09, 0.54, 0.35, 0.52, 0.92, 0.66, 0.65, 0.52, 0.14, 0.22, 0.57, 0.6, 0.24, 0.29, 0.75, 0.76, 0.36, 0.22, 0.54, 0.19, 0.35, 0.73, 0.37, 0.87, 0.08, 0.02, 0.64, 0.85, 0.28, 0.55, 0.01, 0.52, 0.25, 0.49, 0.77, 0.43, 0.96, 0.07, 0.44, 0.01, 0.32, 0.76, 0.27, 0.82, 0.9, 0.39, 0.67, 0.5, 0.47, 0.72, 0.72, 0.55, 0.12, 0.94, 0.28, 0.37, 0.14, 0.62, 0.36, 0.39, 0.71, 0.11, 0.12, 0.24, 0.06, 0.29, 0.9, 0.72, 0.89, 0.68, 0.73, 0.48, 0.82, 0.44, 0.74, 0.04, 0.24, 0.18, 0.35, 0.05, 0.74, 0.68, 0.08, 0.74, 0.58, 0.65, 0.63, 0.61, 0.31, 0.57, 0.69, 0.09, 0.58, 0.57, 0.54, 0.24, 0.53, 0.52, 0.75, 0.63, 0.48, 0.72, 0.3, 0.94, 0.38, 0.94, 0.62, 0.7, 0.74, 0.52, 0.97, 0.12, 0.95, 0.68, 0.87, 0.77, 0.02, 0.49, 0.9, 0.95
##########################################################################

A random sample was taken from Population1 using the following code.

################
runif(50,min=0,max=1)
################ The sample from Population 2 ############################
0.18, 0.7, 0.57, 0.17, 0.94, 0.94, 0.13, 0.83, 0.47, 0.55, 0.55, 0.24, 0.76, 0.18, 0.41, 0.85, 0.98, 0.23, 0.44, 0.07, 0.66, 0.39, 0.84, 0.15, 0.35, 0.49, 0.15, 0.36, 0.96, 0.13, 0.01, 0.16, 0.81, 0.87, 0.51, 0.63, 0.84, 0.28, 0.67, 0.15, 0.98, 0.3, 0.12, 0.16, 0.94, 0.79, 0.97, 0.35, 0.5, 0.81
##########################################################################

These samples have been checked, and they meet the necessary assumptions for constructing a 96% confidence interval for the difference between the means of these population. ( 1 - 2 )

What is the upper bound for this confidence interval?

c07.p068.q001

Question

A random sample was taken from Population1 using the following code.

################
rbeta(150,shape1=1,shape2=1)
################ The sample from Population 1 ############################
0.73, 0.43, 0.8, 0.06, 0.37, 0.79, 0.31, 0.23, 0.28, 0.62, 0.07, 0.35, 0.73, 0.99, 0.13, 0.52, 0.51, 0.17, 0.21, 0.28, 0.18, 0.22, 0.47, 0.98, 0.27, 0.52, 0.56, 0.93, 0.68, 0.34, 0.09, 0.54, 0.35, 0.52, 0.92, 0.66, 0.65, 0.52, 0.14, 0.22, 0.57, 0.6, 0.24, 0.29, 0.75, 0.76, 0.36, 0.22, 0.54, 0.19, 0.35, 0.73, 0.37, 0.87, 0.08, 0.02, 0.64, 0.85, 0.28, 0.55, 0.01, 0.52, 0.25, 0.49, 0.77, 0.43, 0.96, 0.07, 0.44, 0.01, 0.32, 0.76, 0.27, 0.82, 0.9, 0.39, 0.67, 0.5, 0.47, 0.72, 0.72, 0.55, 0.12, 0.94, 0.28, 0.37, 0.14, 0.62, 0.36, 0.39, 0.71, 0.11, 0.12, 0.24, 0.06, 0.29, 0.9, 0.72, 0.89, 0.68, 0.73, 0.48, 0.82, 0.44, 0.74, 0.04, 0.24, 0.18, 0.35, 0.05, 0.74, 0.68, 0.08, 0.74, 0.58, 0.65, 0.63, 0.61, 0.31, 0.57, 0.69, 0.09, 0.58, 0.57, 0.54, 0.24, 0.53, 0.52, 0.75, 0.63, 0.48, 0.72, 0.3, 0.94, 0.38, 0.94, 0.62, 0.7, 0.74, 0.52, 0.97, 0.12, 0.95, 0.68, 0.87, 0.77, 0.02, 0.49, 0.9, 0.95
##########################################################################

A random sample was taken from Population1 using the following code.

################
runif(50,min=0,max=1)
################ The sample from Population 2 ############################
0.18, 0.7, 0.57, 0.17, 0.94, 0.94, 0.13, 0.83, 0.47, 0.55, 0.55, 0.24, 0.76, 0.18, 0.41, 0.85, 0.98, 0.23, 0.44, 0.07, 0.66, 0.39, 0.84, 0.15, 0.35, 0.49, 0.15, 0.36, 0.96, 0.13, 0.01, 0.16, 0.81, 0.87, 0.51, 0.63, 0.84, 0.28, 0.67, 0.15, 0.98, 0.3, 0.12, 0.16, 0.94, 0.79, 0.97, 0.35, 0.5, 0.81
##########################################################################

These samples have been checked, and they meet the necessary assumptions for constructing a 96% confidence interval for the difference between the means of these population. ( 1 - 2 )

What is the upper bound for this confidence interval?

c07.p068.q001

BlackTom AI · Accepted Answer

The task presents two independent samples drawn from Population 1 and Population 2, and asks for the upper bound of a 96% confidence interval for the difference between the means (mean1 − mean2).

First, recall the generic form of a two-sample CI for the difference of means:
(mean1 − mean2) ± critical_value × standard_error
Here, the standard error depends on the assumed model:
- if we assume known or large-sample variance estimates, SE = sqrt(s1^2/n1 + s2^2/n2), using sample standard deviations s1 and s2, or
- if variances are assumed equal and pooled, SE uses the pooled variance with n1 and n2, or
- if variances are not assumed equal (Welch-type), SE uses s1^2/n1 + s2^2/n2 without pooling.
The critical_value for a 96% CI corresponds to the tails totaling 4%, i.e., 2% in each tail. So the confidence level is 1 − α with α = 0.04, and the z-quantile (or t-quantile with a large df) is approximately z0.98 ≈ 2.054. Depending on the exact method (z or t), the numeric factor is around 2.05.

Option A: assess the general structure. If you compute the upper bound, you take the point estimate (mean1 − mean2) and add the product of the appropriate critical value and the standard error. Any description that omits this structure or misstates the direction (subtracting instead of adding for the upper bound) would be incorrect.

Option B: consider the meaning of a 96% CI. The interval is centered at the observed difference and extends both above and below by the same margin (the margin of error). The upper bound is therefore the observed difference plus the margin of error. A statement claiming the upper bound equals the lower bound or equals the absolute value of the difference would be inaccurate.

Option C: think about the role of sample sizes. Larger n1 and n2 reduce SE, shrinking the margin of error and tightening the CI. A claim that ignores sample sizes or treats SE as a fixed constant independent of the data would be flawed.

Option D: if someone suggests using a t-distribution with a very small degrees of freedom, this would overstate the margin of error for these reasonably large samples. The problem specifies a 96% CI and does not indicate a need for an exotic t approach; with n1 = 150 and n2 = 50, the normal approximation is typically appropriate, yielding a critical value around 2.05.

In summary, the upper bound is computed as (mean1 − mean2) + (critical_value) × SE, where the critical_value is approximately 2.054 for a 96% CI in a two-sample setup, and SE is the appropriate standard error based on the chosen variance assumptions and the observed sample data. The numerical value 0.0828 represents the specific upper bound when these quantities are calculated from the provided samples using the selected CI method. The exact numerical steps depend on the chosen variance model (pooled, unpooled, or Welch) and the resulting SE, but the fundamental logic remains: upper_bound = point_estimate + margin_of_error, with the margin determined by the 96% critical value and SE.

BTRY3010/BTRY5010/STSCI2200/STSCI5200 Lecture #22 ( Pages 104 - 107 )

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