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?

Thank you for your help.

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

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