STAT 3360 Notes

• Midpoint: $\boxed{ \bar{X} }$;
• Critical Value: $\boxed{ Z_{\alpha/2} }$;
• Margin of Error: $\boxed{ Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} }$;
• Lower Confidence Limit (LCL): $\boxed{ \bar{X} - Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} }$;
• Upper Confidence Limit (UCL): $\boxed{ \bar{X} + Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} }$;
• Confidence Interval: $\boxed{ [\bar{X} - Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}, \bar{X} + Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}}] }$;
• Width of Confidence Interval: $\boxed{ 2 \cdot Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} }$.

• the Z-test for $\boxed{H_0: \mu \le \mu_0 \text{ vs } H_A: \mu > \mu_0}$ (which is equivalent to $\boxed{H_0: \mu = \mu_0 \text{ vs } H_A: \mu > \mu_0}$) can be constructed as follows,
  • Test Statistic: $\boxed{T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{\sigma}}$;
  • Critical Value: $\boxed{Z_{\alpha}}$;
  • Rejection Rule: Reject $H_0$ if $\boxed{T > Z_{\alpha}}$.
• the Z-test for $\boxed{H_0: \mu \ge \mu_0 \text{ vs } H_A: \mu < \mu_0}$ (which is equivalent to $\boxed{H_0: \mu = \mu_0 \text{ vs } H_A: \mu < \mu_0}$) can be constructed as follows,
  • Test Statistic: $\boxed{T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{\sigma}}$;
  • Critical Value: $Z_{1 - \alpha}$, which equals $\boxed{- Z_{\alpha}}$;
  • Rejection Rule: Reject $H_0$ if $T < Z_{1 - \alpha}$, which is equivalent to $\boxed{T < -Z_{\alpha}}$.
• the Z-test for $\boxed{H_0: \mu = \mu_0 \text{ vs } H_A: \mu \neq \mu_0}$ can be constructed as follows,
  • Test Statistic: $\boxed{T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{\sigma}}$;
  • Critical Values: $Z_{1 - \alpha/2}$ (which equals $\boxed{- Z_{\alpha/2}}$) and $\boxed{Z_{\alpha/2}}$;
  • Rejection Rule: Reject $H_0$ if $T > Z_{\alpha/2}$ or $T < - Z_{\alpha/2}$, which is equivalent to $\boxed{|T| > Z_{\alpha/2}}$.

• Let $X$ be the score of a student, then we know
  • $X$ follows a Normal distribution;
  • the population mean of $X$, denoted by $\mu$, is unknown;
  • the population standard deviation of $X$ is $\sigma = 3$;
  • the sample size $n = 16$;
  • the sample mean $\bar{X} = 79$;
  • the sample standard deviation is $s = 3.5$.
• To construct the confidence interval of $\mu$ when population standard deviation $\sigma = 3$ is known,
  • the confidence level is $1 - \alpha = 90\%$, so $\alpha = 0.1$;
  • the midpoint of the confidence interval is $\bar{X} = 79$;
  • the critical value is $Z_{\alpha/2} = Z_{0.1/2} = Z_{0.05} = 1.64$ because $P(Z > 1.64) = 0.05$;
  • the margin of error is $Z_{\alpha/2} \cdot \frac{\sigma}{\sqrt{n}} = 1.64 \cdot \frac{3}{\sqrt{16}} = 1.23$;
  • the lower confidence limit is $79 - 1.23 = 77.77$;
  • the upper confidence limit is $79 + 1.23 = 80.23$;
  • the confidence interval is $[77.77, 80.23]$;
  • the width of the confidence interval is $2 \times 1.23 = 2.46$.

• the confidence level is $1 - \alpha = 90\%$, so $\alpha = 0.1$;
• the critical value is $Z_{\alpha/2} = Z_{0.1/2} = Z_{0.05} = 1.64$ because $P(Z > 1.64) = 0.05$;
• since the width of the confidence interval is $2M$ where $M$ is the margin of error, to let $2M < 2$, we need $M < 1$;
• by formula, to achieve $M<1$, the smallest sample size should be $n = \left( \frac{z_{\alpha/2} \sigma}{M} \right)^2 = \left( \frac{1.64 \cdot 3}{1} \right)^2 = 24.21 \approx 25$.

• Here we want to confirm the claim that $\mu < 80$, so we let it be the alternative hypothesis. That is, we want to test $H_0: \mu \ge 80$ vs $H_A: \mu < 80$ (or equivalently $H_0: \mu = 80$ vs $H_A: \mu < 80$);
• The test statistic is $T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{\sigma} = \frac{\sqrt{16}(79 - 80)}{3} = -1.33$;
• The significance level is $\alpha = 5\%$;
• The critical value is $- Z_{\alpha} = - Z_{0.05} = - 1.64$;
• The rejection rule is: reject $H_0$ if $T < - Z_{\alpha}$;
• Here $T = -1.33 > -1.64 = -Z_{\alpha}$, so we don't have enough evidence to reject $H_0$ and thus don't have enough evidence to say that the population mean is lower than $80$.

• The hypotheses are the same, ie, $H_0: \mu \ge 80$ vs $H_A: \mu < 80$ (or equivalently $H_0: \mu = 80$ vs $H_A: \mu < 80$);
• The test statistic is the same, ie, $T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{\sigma} = \frac{\sqrt{16}(79 - 80)}{3} = -1.33$;
• The significance level now is $\alpha = 10\%$;
• The critical value now is $- Z_{\alpha} = - Z_{0.1} = - 1.28$;
• The rejection rule is the same, ie, reject $H_0$ if $T < - Z_{\alpha}$;
• Here $T = -1.33 < -1.28 = -Z_{\alpha}$, so we have enough evidence to reject $H_0$ and thus have enough evidence to say that the population mean is lower than $80$.

• Comparing Q4 with Q3, we see that at a higher significance level, we are more likely to reject $H_0$. Recall that the significance level is just the probability of Type I error, and Type I error occurs if we reject $H_0$ while $H_0$ is true. With a higher significance level, we are allowed more chance for Type I error and thus are more aggressive in rejecting $H_0$.
• In Q3, we were conservative (prudent) in rejecting $H_0$ by using a lower significance level $\alpha = 5\%$. We didn't reject $H_0$ and thus avoided the risk of Type I error. However, if $H_0$ is not true, then we miss the chance to disprove it (ie, Type II error occurs).
• In Q4, we were aggressive in rejecting $H_0$ by using a high significance level $\alpha = 10\%$. We rejected $H_0$ and supported $H_A$. However, if $H_0$ is true and $H_A$ is not true, then Type I error occurs.
• Therefore, the choice of significance is critical for the decision making, and different significance level can lead to even opposite conclusions for the same data.
• In general, a lower significance level means a conservative (prudent) attitude toward Type I error, which leads to a low chance of rejecting $H_0$. Therefore, at a low significance level,
  • if $H_0$ is rejected, then we believe we have enough evidence to support $H_A$ because we are already prudent;
  • if $H_0$ is not rejected, then we don't say "$H_0$ is true". Instead, we say "we don't have enough evidence to reject $H_0$". When we are conservative, we need very strong evidence for rejecting $H_0$. Therefore, it may happen that $H_0$ is not true but we choose not to reject it because we believe the evidence is not strong enough.
• In general, a high significance level means an aggressive attitude toward Type I error, which leads to a high chance of rejecting $H_0$. Therefore, at a high significance level,
  • even if $H_0$ is rejected, we are still not quite confident in $H_A$ because we are inclined to reject $H_0$ and thus have a high chance of rejecting a true $H_0$;
  • if $H_0$ is not rejected, then we believe there is enough evidence to support $H_0$ (ie, $H_0$ is not rejected even when we are aggressive).

• For example, the number in the red box means $P(t_{15} > 1.753) = 0.05$, so $t_{0.05, 15} = 1.753$.

  

• Midpoint: $\boxed{ \bar{X} }$;
• Critical Value: $\boxed{ t_{\alpha/2, n-1} }$;
• Margin of Error: $\boxed{ t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} }$;
• Lower Confidence Limit (LCL): $\boxed{ \bar{X} - t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} }$;
• Upper Confidence Limit (UCL): $\boxed{ \bar{X} + t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} }$;
• Confidence Interval: $\boxed{ \left[ \bar{X} - t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}}, \bar{X} + t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} \right] }$;
• Width of Confidence Interval: $\boxed{ 2 \cdot t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} }$.

• the t-test for $\boxed{H_0: \mu \le \mu_0 \text{ vs } H_A: \mu > \mu_0}$ (which is equivalent to $\boxed{H_0: \mu = \mu_0 \text{ vs } H_A: \mu > \mu_0}$) can be constructed as follows,
  • Test Statistic: $\boxed{T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{s}}$;
  • Critical Value: $\boxed{t_{\alpha, n-1}}$;
  • Rejection Rule: Reject $H_0$ if $\boxed{T > t_{\alpha, n-1}}$.
• the t-test for $\boxed{H_0: \mu \ge \mu_0 \text{ vs } H_A: \mu < \mu_0}$ (which is equivalent to $\boxed{H_0: \mu = \mu_0 \text{ vs } H_A: \mu < \mu_0}$) can be constructed as follows,
  • Test Statistic: $\boxed{T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{s}}$;
  • Critical Value: $t_{1-\alpha, n-1}$, which equals $\boxed{- t_{\alpha, n-1}}$
  • Rejection Rule: Reject $H_0$ if $T < t_{1-\alpha, n-1}$, which is equivalent to $\boxed{T < - t_{\alpha, n-1}}$
• the t-test for $\boxed{H_0: \mu = \mu_0 \text{ vs } H_A: \mu \neq \mu_0}$ can be constructed as follows,
  • Test Statistic: $\boxed{T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{s}}$;
  • Critical Values: $t_{1-\alpha/2, n-1}$ (which equals $\boxed{- t_{\alpha/2, n-1}}$) and $\boxed{t_{\alpha/2, n-1}}$;
  • Rejection Rule: Reject $H_0$ if $T > t_{\alpha/2, n-1}$ or $T < - t_{\alpha/2, n-1}$, which is equivalent to $\boxed{|T| > t_{\alpha/2, n-1}}$.

• Let $X$ be the score of a student, then we know
  • $X$ follows a Normal distribution;
  • the population mean of $X$, denoted by $\mu$, is unknown;
  • the population standard deviation of $X$, denoted by $\sigma$, is unknown;
  • the sample size $n = 16$;
  • the sample mean $\bar{X} = 79$;
  • the sample standard deviation $s = 3.5$.
• To construct the confidence interval of $\mu$ when $\sigma$ is unknown,
  • the confidence level is $1 - \alpha = 90\%$, so $\alpha = 0.1$;
  • the midpoint of the confidence interval is $\bar{X} = 79$;
  • the critical value is $t_{\alpha/2, n-1} = t_{0.1/2, 16-1} = t_{0.05, 15} = 1.753$;
  • the margin of error is $t_{\alpha/2, n-1} \cdot \frac{s}{\sqrt{n}} = 1.753 \cdot \frac{3.5}{\sqrt{16}} = 1.534$;
  • the lower confidence limit is $79 - 1.534 = 77.466$;
  • the upper confidence limit is $79 + 1.534 = 80.534$;
  • the confidence interval is $[77.466, 80.534]$;
  • the width of the confidence interval is $2 \times 1.534 = 3.068$.

• Here we want to confirm the claim that $\mu < 80$, so we let it be the alternative hypothesis. That is, we want to test $H_0: \mu \ge 80$ vs $H_A: \mu < 80$ (or equivalently $H_0: \mu = 80$ vs $H_A: \mu < 80$);
• The test statistic is $T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{s} = \frac{\sqrt{16}(79 - 80)}{3.5} = -1.143$;
• The significance level is $\alpha = 5\%$;
• The critical value is $- t_{\alpha, n-1} = - t_{0.05, 16-1} = - 1.753$;
• The rejection rule is: reject $H_0$ if $T < - t_{\alpha, n-1}$;
• Here $T = -1.143 > -1.753 = -t_{\alpha, n-1}$, so we don't have enough evidence to reject $H_0$ and thus don't have enough evidence to say that the population mean is lower than $80$.

• The hypotheses are the same, $H_0: \mu \ge 80$ vs $H_A: \mu < 80$ (or equivalently $H_0: \mu = 80$ vs $H_A: \mu < 80$);
• The test statistic is the same, $T = \frac{\sqrt{n}(\bar{X} - \mu_0)}{s} = \frac{\sqrt{16}(79 - 80)}{3.5} = -1.143$;
• The significance level now is $\alpha = 10\%$;
• The critical value now is $-t_{\alpha, n-1} = -t_{0.1, 16-1} = - 1.341$;
• The rejection rule is the same: reject $H_0$ if $T < - t_{\alpha, n-1}$;
• Here $T = -1.143 > -1.341 = -t_{\alpha, n-1}$, so we still don't have enough evidence to reject $H_0$ and thus still don't have enough evidence to say that the population mean is lower than $80$.

• Midpoint: $\boxed{ \hat{p} }$;
• Critical Value: $\boxed{ Z_{\alpha/2} }$;
• Margin of Error: $\boxed{ Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} }$;
• Lower Confidence Limit (LCL): $\boxed{ \hat{p} - Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} }$;
• Upper Confidence Limit (UCL): $\boxed{ \hat{p} + Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} }$;
• Confidence Interval: $\boxed{ \left[ \hat{p} - Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}, \hat{p} + Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \right] }$;
• Width of Confidence Interval: $\boxed{ 2 \cdot Z_{\alpha/2} \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} }$;

At confidence level $1-\alpha$, to achieve a confidence interval for the population proportion $p$ whose margin of error is no wider than $M$, the smallest sample size needed is

If the value of $n$ calculated by the formula is not an integer, then we should round it up in order to ensure the actual margin of error is no wider than $M$.