Question - Assessment of an Amateur Swimmer's Performance and Statistical Analysis
Solution:
(a)
To find the probability that Zhou wins a randomly chosen race, we use the area of the overlapping section of two normal distributions.
Let \( Z_Z \) be the standard normal variable for Zhou's time,
\[ Z_Z = \frac{X - 80}{2} \]Let \( Z_T \) be the standard normal variable for Tan's time,
\[ Z_T = \frac{X - 79}{3} \]We need P(\( Z_Z < Z_T \)) which is equivalent to P(\( X_Z < X_T \)).
Let \( D = X_T - X_Z \), where \( D \) follows N(1, \( 2^2 + 3^2 \)) since \( Var(X_T - X_Z) = Var(X_T) + Var(X_Z) \) as they are independent.
Hence, \( D \) ~ N(1, 13).
We can standardize \( D \) to get \( Z_D \) ~ N(0, 1) and find P(\( Z_D > 0 \)) to find the probability that Tan wins:
\[ Z_D = \frac{D - 1}{\sqrt{13}} \] \[ P(Z_D > 0) = P\left(\frac{D - 1}{\sqrt{13}} > \frac{0 - 1}{\sqrt{13}}\right) \] \[ P(Z_D > 0) = P(Z_D > -1/\sqrt{13}) \]Using standard normal tables, find P(\( Z_D > -0.277 \)).
The probability that Zhou wins is the complement of this probability:
\[ P(Zhou\ wins) = 1 - P(Tan\ wins) \] \[ P(Zhou\ wins) = 1 - P(Z_D > -0.277) \](b)
Population mean estimate (unbiased) for Zhou's times (µ̂):
\[ \mû = \bar{X} = \frac{\sum{X}}{n} = \frac{2376.3}{30} \]Population variance estimate for Zhou's times (σ̂²):
\[ \sigmâ^2 = \frac{\sum{X^2} - \frac{(\sum{X})^2}{n}}{n-1} \] \[ \sigmâ^2 = \frac{188653.7 - \frac{(2376.3)^2}{30}}{30 - 1} \](c)
Null Hypothesis \( H_0 \): Zhou's mean time has not reduced, \( \mu = \mu_0 \), where \( \mu_0 \) is the mean time before the exercise regime.
Alternative Hypothesis \( H_1 \): Zhou's mean time has reduced, \( \mu < \mu_0 \).
The test statistic for a left-tailed t-test, since \( n \) is small and population variance is unknown, is:
\[ t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}} \]Use t-tables to find critical value for α = 0.05 and \( n-1 \) degrees of freedom. If \( t < t_{critical} \), reject \( H_0 \).
(d)
Tan should use a 2-tail test because he is trying to determine if his times have either increased or decreased, not just in one direction.
(e)
The two assumptions made by Tan are:
\begin{itemize} \item The sample of recorded times is normally distributed. \item The recorded times are independent of each other. \end{itemize} Please note the actual calculations have not been performed, and the solution steps provided are intended to be general instructions on how to proceed with solving the given problems.CamTutor
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