# S.3.3 Hypothesis Testing Examples

S.3.3 Hypothesis Testing Examples
Brinell Hardness Scores

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

 170 167 174 179 179 187 179 183 179 156 163 156 187 156 167 156 174 170 183 179 174 179 170 159 187

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:

H0 : μ = 170
HA: μ > 170

The engineer entered his data into Minitab and requested that the "one-sample t-test" be conducted for the above hypotheses. He obtained the following output:

#### Descriptive Statistics

N Mean StDev SE Mean 95% Lower Bound
25 172.52 10.31 2.06 168.99

$\mu$: mean of Brinelli

#### Test

Null hypothesis    H₀: $\mu$ = 170
Alternative hypothesis    H₁: $\mu$ > 170

T-Value P-Value
25 172.52

The output tells us 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 α 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 tn - 1 = t24 curve and to the right of the test statistic t* = 1.22:

In the output above, Minitab reports that the 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.

Height of Sunflowers

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:

 11.5 11.8 15.7 16.1 14.1 10.5 9.3 15 11.1 15.2 19 12.8 12.4 19.2 13.5 12.2 13.3 16.5 13.5 14.4 16.7 10.9 13 10.3 15.8 15.1 17.1 13.3 12.4 8.5 14.3 12.9 13.5

The biologist's hypotheses are:

H0 : μ = 15.7
HA: μ < 15.7

The biologist entered her data into Minitab and requested that the "one-sample t-test" be conducted for the above hypotheses. She obtained the following output:

#### Descriptive Statistics

N Mean StDev SE Mean 95% Upper Bound
33 13.664 2.544 0.443 14.414

$\mu$: mean of Height

#### Test

Null hypothesis    H₀: $\mu$ = 15.7
Alternative hypothesis    H₁: $\mu$ < 15.7

T-Value P-Value
-4.60 0.000

The output tells us 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 $\alpha$ at 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):s-3-3

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 α = 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 tn - 1 = t32 curve and to the left of the test statistic t* = -4.60:

In the output above, Minitab reports 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.

Gum Thickness

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:

 7.65 7.6 7.65 7.7 7.55 7.55 7.4 7.4 7.5 7.5

The quality control specialist's hypotheses are:

H0 : μ = 7.5
HA: μ ≠ 7.5

The quality control specialist entered his data into Minitab and requested that the "one-sample t-test" be conducted for the above hypotheses. He obtained the following output:

#### Descriptive Statistics

N Mean StDev SE Mean 95% CI for $\mu$
10 7.550 0.1027 0.0325 (7.4765, 7.6235)

$\mu$: mean of Thickness

#### Test

Null hypothesis    H₀: $\mu$ = 7.5
Alternative hypothesis    H₁: $\mu \ne$ 7.5

T-Value P-Value
1.54 0.158

The output tells us 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 -2.2616 or greater than 2.2616 (determined using statistical software or a t-table):

Since the quality control specialist's test statistic, t* = 1.54, is not less than -2.2616 nor greater than 2.2616, 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 tn - 1 = t9 curve, to the right of 1.54 and to the left of -1.54:

In the output above, Minitab reports 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.

### In closing

In our review of hypothesis tests, we have focused on just one particular hypothesis test, namely that concerning the population mean $\mu$. The important thing to recognize is that the topics discussed here — the general idea of hypothesis tests, errors in hypothesis testing, the critical value approach, and the P-value approach — generally extend to all of the hypothesis tests you will encounter.

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