Real World Examples

Example 1 : Right-tailed test

An engineer measured the Brinell hardness of 25 pieces of ductile iron that were annealed. The resulting data were:

The engineer hypothesized that the mean Brinell hardness of all such ductile iron pieces is greater than 170. Therefore, he was interested in testing the hypotheses:

$H_0 : \mu = 170$ $H_a : \mu \gt 170$

If we conduct one sample t-test in this example, we will get that the average Brinell hardness of the $n = 25$ pieces of ductile iron was $172.52$ with a standard deviation of 10.31. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 10.31 by the square root of $n = 25$, is $2.06$). The test statistic $t^*$ is $1.22$, and the P-value is $0.117$.

If the engineer set his significance level $\alpha$ at $0.05$ and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic $t^*$ were greater than $1.7109$ (determined using statistical software or a t-table).

Since the engineer's test statistic, $t^* = 1.22$, is not greater than $1.7109$, the engineer fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the $\alpha = 0.05$ level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than $170$.

If the engineer used the P-value approach to conduct his hypothesis test, he would determine the area under a $t_n - 1 = t_{24}$ curve and to the right of the test statistic $t^* = 1.22$.

By statistical software or a t-table, we can get that P-value is $0.117$. Since the P-value, $0.117$, is greater than $\alpha = 0.05$, the engineer fails to reject the null hypothesis. There is insufficient evidence, at the $\alpha = 0.05$ level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than $170$.

Example 2 : Left-tailed test

A biologist was interested in determining whether sunflower seedlings treated with an extract from Vinca minor roots resulted in a lower average height of sunflower seedlings than the standard height of $15.7 cm$. The biologist treated a random sample of $n = 33$ seedlings with the extract and subsequently obtained the following heights:

The biologist's hypotheses are:

$H_0 : \mu = 15.7$ $H_a : \mu \lt 15.7$

If we conduct one sample t-test in this example, we will get that the average height of the $n = 33$ sunflower seedlings was 13.664 with a standard deviation of $2.544$. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation 13.664 by the square root of $n = 33$, is $0.443$). The test statistic $t^*$ is $-4.60$, and the P-value, $0.000$, is to three decimal places.

If the biologist set her significance level $\alpha = 0.05$ and used the critical value approach to conduct her hypothesis test, she would reject the null hypothesis if her test statistic $t^*$ were less than $-1.6939$ (determined using statistical software or a t-table).

Since the biologist's test statistic, $t^* = -4.60$, is less than $-1.6939$, the biologist rejects the null hypothesis. That is, the test statistic falls in the "critical region." There is sufficient evidence, at the $\alpha = 0.05$ level, to conclude that the mean height of all such sunflower seedlings is less than $15.7 cm$.

If the biologist used the P-value approach to conduct her hypothesis test, she would determine the area under a $t_n - 1 = t_{32}$ curve and to the left of the test statistic $t^* = -4.60$.

By statistical software or a t-table, we can get that the P-value is $0.000$, which we take to mean < $0.001$. Since the P-value is less than $0.001$, it is clearly less than $\alpha = 0.05$, and the biologist rejects the null hypothesis. There is sufficient evidence, at the $\alpha = 0.05$ level, to conclude that the mean height of all such sunflower seedlings is less than $15.7 cm$.

Example 3 : Two-tailed test

A manufacturer claims that the thickness of the spearmint gum it produces is $7.5$ one-hundredths of an inch. A quality control specialist regularly checks this claim. On one production run, he took a random sample of $n = 10$ pieces of gum and measured their thickness. He obtained:

The quality control specialist's hypotheses are:

$H_0 : \mu = 7.5$ $H_a : \mu \neq 7.5$

If we conduct one sample t-test in this example, we will get that the average thickness of the $n = 10$ pieces of gums was $7.55$ one-hundredths of an inch with a standard deviation of $0.1027$. (The standard error of the mean "SE Mean", calculated by dividing the standard deviation $0.1027$ by the square root of $n = 10$, is $0.0325$). The test statistic $t^*$ is $1.54$, and the P-value is $0.158$.

If the quality control specialist sets his significance level $\alpha$ at $0.05$ and used the critical value approach to conduct his hypothesis test, he would reject the null hypothesis if his test statistic $t^*$ were less than $-2.2622$ or greater than $2.2622$ (determined using statistical software or a t-table).

Since the quality control specialist's test statistic, $t^* = 1.54$, is not less than $-2.2622$ nor greater than $2.2622$, the quality control specialist fails to reject the null hypothesis. That is, the test statistic does not fall in the "critical region." There is insufficient evidence, at the $\alpha = 0.05$ level, to conclude that the mean thickness of all of the manufacturer's spearmint gum differs from $7.5$ one-hundredths of an inch.

If the quality control specialist used the P-value approach to conduct his hypothesis test, he would determine the area under a $t_n - 1 = t_9$ curve, to the right of $1.54$ and to the left of $-1.54$.

By statistical software or a t-table, we can get that the P-value is $0.158$. Since the P-value, $0.158$, is greater than $\alpha = 0.05$, the quality control specialist fails to reject the null hypothesis. There is insufficient evidence, at the $\alpha = 0.05$ level, to conclude that the mean thickness of all pieces of spearmint gum differs from $7.5$ one-hundredths of an inch.