1
$\begingroup$

Consider a random sample of size $n$ from a normal distribution with unknown mean $μ$ and unknown variance $σ^2.$ Suppose the sample mean is $\bar X$ and the sample variance is $S^2.$ We would like to test $H_0:μ=μ_0$ versus $H_1:μ≠μ_0.$

What test statistic should you should use to test $H_0$ versus $H_1?$ I believe that the test statistic should be $T=\frac{X¯-μ_0}{S/\sqrt{n},}$ which follows a t distribution with $n-1$ degrees of freedom. Am I correct?

Suppose we now want to test $H_0:σ^2=σ^2_0$ versus $H_1:σ^2≠σ^2_0.$

Which of these test statistics should we use? I think that we should use $W=\frac{(n−1)S^2}{σ^2_0},$ which follows a $\chi^2( n-1)$ distribution. Am I correct? Thank you for your help.

$\endgroup$
2
  • $\begingroup$ Both of your tests are appropriate. I have discussed the t test in my answer. (I have also shown computer printout for the chi-squared test.) $\endgroup$
    – BruceET
    Nov 29, 2020 at 10:39
  • $\begingroup$ Thank you sir I have just seen this comment. $\endgroup$
    – mathslover
    Nov 29, 2020 at 13:47

1 Answer 1

0
$\begingroup$

For asking questions here, you should learn how to use JaX to format equations. The appropriate t statistic is $T = \frac{\bar X - \mu_0}{S/\sqrt{n}},$ where $n, \bar X,$ and $S$ are the sample size, mean, and standard deviation, respectively.

T test. You want to do a 2-sided test. If $|T| \ge t^*,$ then you reject $H_0: \mu = \mu_0$ against the two-sided alternative $H_0: \mu \ne \mu_0,$ at the 5% level, where values $\pm t^*$ cut probability 0.025 from the upper and lower tails, respecteively, of Student's t distribution with $n - 1$ degrees of freedom. You can find such 'critical values' by using printed tables of the CDFs of t distributions or by using software (a statistical calculator or a computer program). For example, using R statistical software with n = 15 you would get critical values $\pm 2.145,$ as follows:

qt(c(.025,.975), 14)
[1] -2.144787  2.144787

Example: Traditionally, first midterm exams for a course have an average score of $\mu = 100.$ The fifteen students enrolled this semester had scores summarized as follows:

summary(x)
   Min. 1st Qu.  Median    Mean 3rd Qu.    Max. 
  75.84  103.65  107.75  109.17  117.52  131.51 
length(x);  sd(x)
[1] 15             # sample size
[1] 13.72826       # sample SD
stripchart(x, pch="|")

enter image description here

In particular, the sample mean this year is $\bar X = 109.17.$ This is above $\mu = 100,$ but is it 'significantly' different at the 5% level of significance? A two-sided, one-sample t test in R gives the following result.

t.test(x, mu=100)

        One Sample t-test

data:  x
t = 2.588, df = 14, p-value = 0.02147
alternative hypothesis: true mean is not equal to 100
95 percent confidence interval:
 101.5710 116.7759
sample estimates:
mean of x 
 109.1734 

Notice that the test statistic is $T = 2.588,$ so that $|T| > 2.145.$ We reject $H_0: \mu = 100$ in favor of the alternative $H_a: \mu \ne 100.$ Another way to see that $H_0$ is rejected is that the P-value (probability) of a test statistic as far or farther from $0$ then $2.145)$ is $0.02147 \le 0.05 - 5\%.$ The method of finding the P-value is shown separately below.

1 - diff(pt(c(-2.588, 2.588), 14))
[1]  0.02147282

In the figure below the critical values are shown as vertical orange lines. Each cuts probability $0.025$ from its respective tail of $\mathsf{T}(\nu=14).$ The observed value of $T$ is shown as a solid black line. The dotted black line is equally far from $0.$ The total area under the density curve outside these two black lines is the P-value.

enter image description here

R code for the figure:

curve(dt(x, 14), -4, 4, col="blue", ylab="PDF", xlab="t", 
       main="Density of T(14)")
 abline(h=0, col="green2");  abline(v=0, col="green2")
 abline(v = c(-2.144787, 2.144787), col="orange")
 abline(v = 2.588);  abline(v = -2.588, lty="dotted") 

Test for variance. Finally, I will show you a variance test using the chi-squared distribution, and the same data as in my Example above, from a recent release of Minitab statistical software.

Test and CI for One Variance 

Method

Null hypothesis         σ = 15
Alternative hypothesis  σ ≠ 15

The chi-square method is only for the normal distribution.

Statistics

 N  StDev  Variance
15   13.7       188

95% Confidence Intervals

               CI for       CI for
Method          StDev      Variance
Chi-Square  (10.1, 21.7)  (101, 469)

Tests

                 Test
Method      Statistic  DF  P-Value
Chi-Square      11.73  14    0.743

Note: As your first lesson in using JaX, here is what I typed to get the formula in my first paragraph above:

... t statistic is $T = \frac{\bar X - \mu_0}{S/\sqrt{n}},$ where $n, \bar X,$ and $S$ are ....

$\endgroup$
1
  • $\begingroup$ Thank you sir your answer was really informative. However my question was if the test statistics I used in each case are correct given the information above. Again, I want to thank you for your answers. $\endgroup$
    – mathslover
    Nov 29, 2020 at 12:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.