Which one of the following is NOT a condition for inference comparing two means? both populations are Normally distributed the two sample sizes must be equal populations must be distinct independent simple random samples from two populations

Solve the following problems applying Polya’s Four-Step Problem-Solving strategy.
If six people greet each other at a meeting by shaking hands with one another, how many handshakes take place?

In how many ways can 7 graduate students be assigned to 1 triple and 2 double hotel rooms during a conference

The number of cases of tetanus reported in the US in a single month has a Poisson distribution with a parameter of ${\displaystyle \lambda =4.0.}$ What is the probability that exactly one case of tetanus will be reported?

A group of scientists studied the adhesion of various biofilms to solid surfaces for possible use in environmental technologies. Adhesion assay is conducted by measuring absorbance at A590. If the scientists want the confidence interval to be no wider than 0.55 ${\displaystyle dy\ne -c{m}^2}$ at ${\displaystyle 98\mathrm{\%}}$ confidence interval, how many observations should they take? Assume that the standard deviation is known to be 0.66 ${\displaystyle dy\ne -c{m}^2}$
CHOOSE THE RIGHT ANSWER:
a.39
b.21
c.16
d.25
e.18
f.32

Explain and give a full and correct answer how confusing the two (population and a sample) can lead to incorrect statistical inferences.

The article “Stochastic Modeling for Pavement Warranty Cost Estimation” (J. of Constr. Engr. and Mgmnt., 2009: 352–359) proposes the following model for the distribution of Y = time to pavement failure. Let \(\displaystyle{X}_{{{1}}}\) be the time to failure due to rutting, and \(\displaystyle{X}_{{{2}}}\) be the time to failure due to transverse cracking, these two rvs are assumed independent. Then \(\displaystyle{Y}=\min{\left({X}_{{{1}}},{X}_{{{2}}}\right)}\). The probability of failure due to either one of these distress modes is assumed to be an increasing function of time t. After making certain distributional assumptions, the following form of the cdf for each mode is obtained: \(\displaystyle\Phi{\left[\frac{{{a}+{b}{t}}}{{\left({c}+{\left.{d}{t}\right.}+{e}{t}^{{{2}}}\right)}^{{\frac{{1}}{{2}}}}}\right]}\) where \(\Uparrow \Phi\) is the standard normal cdf. Values of the five parameters a, b, c, d, and e are -25.49, 1.15, 4.45, -1.78, and .171 for cracking and -21.27, .0325, .972, -.00028, and .00022 for rutting. Determine the probability of pavement failure within \(\displaystyle{t}={5}\) years and also \(\displaystyle{t}={10}\) years.

Confidence interval; exponential distribution (normal or student approximation?)
Let's say we have got a sample of size n from an exponential distribution with an unknown mean ${\displaystyle \lambda }$.
We want to construct a confidence interval and so we can compare this:
${\displaystyle \frac{\bar{X}-\lambda }{\sqrt{\frac{S^2}{n}}}}$
to a student t-distribution with ${\displaystyle n-1}$ degrees of freedom.
However, as in the case of the exponential distribution, we know that ${\displaystyle Var\left[X\right]={\left(E\left[X\right]\right)}^2}$, so rather than introducing an estimator for variance, we can simply use one estimator, i.e:
${\displaystyle \frac{\bar{X}-\lambda }{\sqrt{\frac{{\bar{X}}^2}{n}}}.}$
And now the question is:
Do we compare this statistic with normal distribution or again with student t-distribution?