Example Question - significance level
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Statistical Comparison of New Technique's Mean Time to Failure
<p>The question is asking to perform a hypothesis test to compare the mean times for cracks to appear for two different techniques at a 5% significance level. We will use a t-test for comparing two independent means with the assumption that both populations are normally distributed, and we assume equal variances since \(s_{\bar{x}}\) is the same for both techniques.</p> <p>Null hypothesis, \(H_0\): \(\mu_1 = \mu_2\) (The mean time to failure is the same for both techniques.)</p> <p>Alternative hypothesis, \(H_1\): \(\mu_1 < \mu_2\) (The mean time to failure for the new technique is greater than the current technique.)</p> <p>We calculate the test statistic using the formula:</p> \[ t = \frac{\bar{x}_1 - \bar{x}_2}{s_{\bar{x}} \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}} \] <p>Where \(\bar{x}_1\) is the mean for the current technique, \(\bar{x}_2\) is the mean for the new technique, \(s_{\bar{x}}\) is the standard deviation of the sample means, \(n_1\) is the sample size for the current technique, and \(n_2\) is the sample size for the new technique. Plugging the values from the table:</p> \[ t = \frac{249.1 - 250.1}{3.8 \sqrt{\frac{1}{200} + \frac{1}{100}}} \] \[ t = \frac{-1.0}{3.8 \sqrt{0.01 + 0.005}} \] \[ t = \frac{-1.0}{3.8 \sqrt{0.015}} \] \[ t = \frac{-1.0}{3.8 \cdot 0.122474487} \] \[ t \approx \frac{-1.0}{0.46539} \] \[ t \approx -2.1486 \] <p>The degrees of freedom in this case is the smaller of \(n_1 - 1\) and \(n_2 - 1\), so we use \(n_2 - 1 = 99\).</p> <p>Now we compare the calculated t-value to the critical t-value from the t-distribution table at the 5% significance level with 99 degrees of freedom. If our calculated t-value is less than the critical t-value we reject the null hypothesis.</p> <p>(b) Based on the results of the statistical test, if the null hypothesis is rejected, we can say that the new technique significantly increases the mean time before components crack and need to be replaced. However, if the null hypothesis cannot be rejected, then we cannot support the statement that the new technique significantly increases the time before failure.</p>
Assessment of an Amateur Swimmer's Performance and Statistical Analysis
<p>(a)</p> <p>To find the probability that Zhou wins a randomly chosen race, we use the area of the overlapping section of two normal distributions.</p> <p>Let \( Z_Z \) be the standard normal variable for Zhou's time,</p> \[ Z_Z = \frac{X - 80}{2} \] <p>Let \( Z_T \) be the standard normal variable for Tan's time,</p> \[ Z_T = \frac{X - 79}{3} \] <p>We need P(\( Z_Z < Z_T \)) which is equivalent to P(\( X_Z < X_T \)).</p> <p>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.</p> <p>Hence, \( D \) ~ N(1, 13).</p> <p>We can standardize \( D \) to get \( Z_D \) ~ N(0, 1) and find P(\( Z_D > 0 \)) to find the probability that Tan wins:</p> \[ Z_D = \frac{D - 1}{\sqrt{13}} \] \[ P(Z_D > 0) = P\left(\frac{D - 1}{\sqrt{13}} > \frac{0 - 1}{\sqrt{13}}\right) \] \[ P(Z_D > 0) = P(Z_D > -1/\sqrt{13}) \] <p>Using standard normal tables, find P(\( Z_D > -0.277 \)).</p> <p>The probability that Zhou wins is the complement of this probability:</p> \[ P(Zhou\ wins) = 1 - P(Tan\ wins) \] \[ P(Zhou\ wins) = 1 - P(Z_D > -0.277) \] <p>(b)</p> <p>Population mean estimate (unbiased) for Zhou's times (µ̂):</p> \[ \mû = \bar{X} = \frac{\sum{X}}{n} = \frac{2376.3}{30} \] <p>Population variance estimate for Zhou's times (σ̂²):</p> \[ \sigmâ^2 = \frac{\sum{X^2} - \frac{(\sum{X})^2}{n}}{n-1} \] \[ \sigmâ^2 = \frac{188653.7 - \frac{(2376.3)^2}{30}}{30 - 1} \] <p>(c)</p> <p>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.</p> <p>Alternative Hypothesis \( H_1 \): Zhou's mean time has reduced, \( \mu < \mu_0 \).</p> <p>The test statistic for a left-tailed t-test, since \( n \) is small and population variance is unknown, is:</p> \[ t = \frac{\bar{X} - \mu_0}{s/\sqrt{n}} \] <p>Use t-tables to find critical value for α = 0.05 and \( n-1 \) degrees of freedom. If \( t < t_{critical} \), reject \( H_0 \).</p> <p>(d)</p> <p>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.</p> <p>(e)</p> <p>The two assumptions made by Tan are:</p> \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.
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