Student's t-test one sample

Single sample t-test analysis flow

• Calculate conjecture values for the average and variance of the population
• We know that averages and variance are unbiased statistical values for the population

 $$\begin{matrix}\hat{\mu}=\bar{X}\\{\hat{\sigma}}^2=s^2=\frac{\sum{(x_i-\bar{X})^2}}{N-1}\\\end{matrix}$$

• The conjecture value for the variance is not enough, we must guess the value of the variance from the average distribution of the sample. When the population variance is known, we find that:
   $$\sigma_{\bar{X}}^2=\frac{\sigma^2}{N}\Rightarrow\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{N}}$$
• Now we have an estimator for the standard deviation of the sample mean value (standard error). So that when substituting into the previous equation, we get the formula:
   $${\hat{\sigma}}_{\bar{X}}=\frac{\hat{\sigma}}{\sqrt{N}}=\frac{s}{\sqrt{N}}=\frac{\sqrt{\frac{\sum{(x_i-\bar{X})^2}}{N-1}}}{\sqrt{N}}$$
• Now we can create a statistical test similar to the z test of a single sample, where the population variance is suspected from the data sample:
   $$ z\ \ =\ \ \frac{\bar{x}-\mu}{{\hat{\sigma}}_{\bar{x}}}\ \ =\ \ \frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{N}}}$$ the parameter σ replaced with its conjecture value, s, so that:
   $$ t=\frac{\bar{x}-\mu}{\sigma_{\bar{X}}}=\frac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}$$
• Unlike the z distribution, the distribution of t values depends on the degree of freedom (df). For N = $\infty$, t = z
• For a single sample t-test (average value test with a certain value), df = N-1

Example 1 (Two Tailed Test)

Incandescent lamp entrepreneur A says that his lamp can last about 800 hours. Recently, it has been suspected that its service life has changed. To prove this, a study was carried out by testing 50 lamps. It turns out that the average is 792 hours with a standard deviation (s) = 55 hours. Investigate at a significant level of 5% whether the quality of the lamp has changed?

Answer:

1. Step 1: Claim: the durability of the lamp has already changed, symbolically it can be expressed by the μ ≠ 800 vs. μ = 800.
2. Step 2: Of the two equations above, μ ≠ 800 does not contain an element of equality, so it becomes an alternative hypothesis (H₁) and its null hypothesis: μ = 800.
3. H₀: μ = 800
4. H₁: μ ≠ 800
5. Significant level: α = 0.05
6. Determine the statistical test: the sample is taken randomly; σ unknown; n > 30 (normally distributed). Of these conditions, the appropriate statistical test is the t-test.
7. Calculate t_stat and determine t_critical:

 $$ t=\frac{\bar{x}-\mu_{\bar{x}}}{\frac{s}{\sqrt{n}}}=\frac{792-800}{\frac{55}{\sqrt{50}}}=-1.029$$

1. df = n-1 = 50-1 = 49; α = 0.05
2. t_critical = t (_α/2.df) = t_(0.025.49) = 2.01
3. Since |t_stat| < t_critical then H₀ is accepted!

Example 2 (One Tailed Test):

The average time it took for students to manually re-apply at University A in previous semesters was about 45 minutes. A new registration using an information system is being tried in the hope of reducing the registration time for students when compared to the old way. For this reason, a random sample of 10 students who had registered in the next semester was taken using the new system. It turns out that the average time it takes to register is about 35 minutes with a standard deviation of 9.5 minutes. Do you believe in those expectations? Test at a significance level of 5%?

Answer:

1. Step 1: Claim: the time required is less than 45 minutes, symbolically it can be expressed by μ < 45 vs. μ ≥ 45.
2. Step 2: Of the two equations above, μ < 45 does not contain an element of equality, so it becomes an alternative hypothesis (H₁) and its null hypothesis: μ = 45.
3. H₀: μ = 45
4. H₁: μ < 45
5. Significant level : α = 0.05
6. Determine the statistical test: the sample is taken randomly; σ unknown; n < 30; normally distributed. Of the three conditions, the appropriate statistical test is the t-test.
7. Calculate t_stat and determine t_critical:

 $$ t=\frac{\bar{x}-\mu_{\bar{x}}}{\frac{s}{\sqrt{n}}}=\frac{35-45}{\frac{9.5}{\sqrt{10}}}=-3.3$$

1. df = n-1 = 10-1 = 9; α = 0.05
2. t_critical = t (_{α, df}) = t_(0.05.9) = 1,833
3. Because |t_stat| > |t_critical| then H₀ is rejected!
4. Conclusion: From the statistical test above, we feel confident that the time required with the new registration system is indeed less than 45 minutes.

Example 3 (One Tailed Test)

Based on experience in previous years, the average body temperature of newly admitted medical students is believed to be less than 98.6°F. To ensure that the average body temperature of newly entered students remains below that grade, a senior student plans to re-examine the claim. But because of his busy schedule, he only collected data from 12 college students. The average body temperature of the 12 students is as follows:

 98.0 97.5 98.6 98.8 98.0 98.5 98.6 99.4 98.4 98.7 98.6 97.6

To test the claim, he used a significant level of 0.05 stating that the average body temperature did come from the student population with an average of less than 98.6°F.

Answer

Before conducting a hypothesis test, we must first explore the data. Visually, check the presence or absence of outliers as well as whether based on normal histograms and quantile plots we can assume that the data comes from a normally distributed population. From the above data, we obtain statistical values: n = 12, = 98.39, s = 0.535. The average example = 98.39 is indeed smaller than 98.6, but is it enough to determine that the figure is actually smaller than 98.6. Ok, let's do those hypothesis testing steps. In this case, the traditional hypothesis test method will be used by comparing the t-stat value with the critical t-. $\bar{x}$

1. Step 1: Claim: average body temperature is less than 98.6°F, symbolically expressed by μ < 98.6.
2. Step 2: Alternative Hypothesis: μ ≥ 98.6.
3. Step 3: From the two equations above, the μ < 98.6 does not contain an element of equality, so it becomes an alternative hypothesis (H1) and its null hypothesis: μ = 98.6.
4. H0: μ = 98.6
5. H1: μ < 98.6
6. Significant level: α = 0.05
7. Determine the statistical test: the sample is taken randomly; σ unknown; n < 30; normally distributed. Of the three conditions, the appropriate statistical test is the t-test.
8. Calculate t_stat and determine t_critical:

 $$ t=\frac{\bar{x}-\mu_{\bar{x}}}{\frac{s}{\sqrt{n}}}=\frac{96.392-96.8}{\frac{0.535}{\sqrt{12}}}=-1.35$$

1. df = n-1 = 12-1 = 11; α = 0.05
2. t_critica = t(_α,df) = t(_{0.05, 11}) = 1,796
3. Since |t_stat| < t_critical then H0 is accepted!

Calculations with SmartstatXL Excel Add-In

Analysis Results

Interpretation:

Hypothesis:

1. H₀: μ = 98.6
2. H₁: μ < 98.6

Classic Method

t_stat = t_stat = T-Value = -1.349

t_critica = t_table = t_(0.05.9) = -1,796 (obtained from the table value t-student)

Since |t_stat| > |t_critical| 1,349 < 1,796 then H0 is accepted!

That is to say: 95% of us believe that there is not enough evidence to state that the freshman's body temperature is less than 98.6°F. Although the average score was indeed smaller, from the 12 samples taken, it was not strong enough to state that the student body temperature was less than 98.6°F.

Modern Methods:

Tests with modern methods use p-value in determining whether or not a statistical test is significant.

If: P-Value < Significant Level then the significant test or H₀ is rejected

If: P-Value > Significant Level then the non-significant test or H₀ is accepted

In the above case, P-Value = 0.102 > α = 0.05. This indicates that the test is not significant or H₀ is rejected.

That is to say: 95% of us believe that there is not enough evidence to state that the freshman's body temperature is less than 98.6°F. Although the average score was indeed smaller, from the 12 samples taken, it was not strong enough to state that the student body temperature was less than 98.6°F.