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Next: Hypothesis tests to compare Up: Statistical Decisions Previous: Hypothesis tests for a

Hypothesis tests for a population mean

For this problem we use the same sampling theory derived for confidence intervals of population means. If the population is approximately normally distributed or if the sample size is large, then the sampling distribution of

\begin{displaymath}
t = \frac{\overline{X} - \mu_0}{s/\sqrt{n}}
\end{displaymath}

is a t-distribution with n-1 degrees of freedom.

  1. $H_0:\ \mu \le \mu_0$
    $H_1:\ \mu > \mu_0$
    The burden of proof is to show that $\mu > \mu_0$, where $\mu_0$ is the reference value for the population mean that is initially assumed to be true. The test statistic for this test is

    \begin{displaymath}
t = \frac{\overline{X} - \mu_0}{\sqrt{s/n}}.
\end{displaymath}

    The p-value is P(T > t), the area to the right of t under the t-density with n-1 d.f.


  2. $H_0:\ \mu \ge \mu_0$
    $H_1:\ \mu < \mu_0$
    The test statistic for this test is

    \begin{displaymath}
t = \frac{\overline{X} - \mu_0}{\sqrt{s/n}}.
\end{displaymath}

    The p-value is P(T < t), the area to the left of t under the t-density with n-1 d.f.

  3. For the two-sided test,
    $H_0:\ \mu = \mu_0$
    $H_1:\ \mu \ne \mu_0$
    the test statistic is |t| and the p-value is the total area under both tails of the t-density,

    \begin{displaymath}
P(T < -\vert t\vert) + P(T > \vert t\vert),
\end{displaymath}

More advanced tools are required to obtain the power function of this test. However, we can approximate the power function if the sample size is large. First consider a one-sided test of the hypotheses:
$H_0:\ \mu \le \mu_0$
$H_1:\ \mu > \mu_0$
with level of significance $\alpha$. We would reject the null hypothesis if the t-statistic,

\begin{displaymath}
T = \frac{\overline{X} - \mu_0}{s/\sqrt{n}},
\end{displaymath}

is greater than or equal to the critical value, $t_{n-1,1-\alpha}$, where this critical value is chosen so that the area to the right of this value under the t-distribution with n-1 degrees of freedom equals $\alpha$. In the large sample case, this critical value is approximately equal to the corresponding value from the standard normal distribution. In R the approximate critical value is obtained by

\begin{displaymath}
t_{n-1,1-\alpha} = qt(1-\alpha,n-1) \approx z_{1-\alpha} = qnorm(1-\alpha).
\end{displaymath}

This is equivalent to rejecting the null hypothesis if

\begin{displaymath}
\overline{X} \ge \mu_0 + z_{1-\alpha}s/\sqrt{n}.
\end{displaymath}

Therefore, the power function for this test is given by

\begin{displaymath}
P_\mu(\overline{X} \ge \mu_0 + z_{1-\alpha}s/\sqrt{n}),
\end{displaymath}

where $\mu$ represents the actual population mean. In the large sample case, we can use the standard normal distribution to approximate the distribution of

\begin{displaymath}
T = \frac{\overline{X} - \mu}{s/\sqrt{n}},
\end{displaymath}

This gives

\begin{eqnarray*}
{\rm Power}(\mu) &=& P_\mu(\overline{X} \ge \mu_0 + z_{1-\alph...
...\
&=& 1 - pnorm(\frac{\mu_0 - \mu}{s/\sqrt{n}} + z_{1-\alpha})
\end{eqnarray*}

Suppose we have a random sample of size n=80 with sample mean 65, sample s.d. 12, and we wish to test
$H_0:\ \mu \le 60$
$H_1:\ \mu > 60$
at 5% level of significance. In R a plot of this power function can be obtained by
xbar = 65
s = 12
n = 80
alpha = .05
mu0 = 60
criticalValue = qnorm(1-alpha)
mu = seq(58,68,length=201)
power = 1 - pnorm(criticalValue + (mu0 - mu)*sqrt(n)/s)
plot(mu,power,type="l",xlab=expression(mu),ylab="Power")
title("Power Function of One-sided T-test")
abline(h=alpha,col="red")
abline(h=.9,col="blue")
points(mu0,alpha,pch=19,col="red")
text(mu0,alpha,paste("Power(60) =",alpha),adj=c(-.1,1.2))
beta = .1
mu1 = max(mu[power <= 1-beta])
points(mu1,1-beta,pch=19,col="blue")
text(mu1,1-beta,paste("Power(",round(mu1,2),") = ",1-beta,sep=""),adj=c(-.1,1.2))
Note that the power function at the null hypothesis is always equal to the level of significance. This plot also shows that if the population mean is 63.9, then the probability this test will reject the null hypothesis is 0.9.


next up previous
Next: Hypothesis tests to compare Up: Statistical Decisions Previous: Hypothesis tests for a
Larry Ammann
2014-12-08