# Determine the test statistic for each case

Cross Validated Asked by mathslover on February 27, 2021

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?

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="|")


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.

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 ....

Answered by BruceET on February 27, 2021