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## Reject The Null Hypothesis Statement Example

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### Example: Right-tailed test

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

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

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:

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 α = 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 α = 0.05, the engineer fails to reject the null hypothesis. There is insufficient evidence, at the α = 0.05 level, to conclude that the mean Brinell hardness of all such ductile iron pieces is greater than 170.

Note that the engineer obtains the same scientific conclusion regardless of the approach used. This will always be the case.

### Example: 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:

 11.5 11.8 15.7 16.1 14.1 10.5 15.2 19 12.8 12.4 19.2 13.5 16.5 13.5 14.4 16.7 10.9 13.0 15.1 17.1 13.3 12.4 8.5 14.3 12.9 11.1 15 13.3 15.8 13.5 9.3 12.2 10.3

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:

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.

Minitab Note. Minitab will always report P-values to only 3 decimal places. If Minitab reports the P-value as 0.000, it really means that the P-value is 0.000....something. Throughout this course (and your future research!), when you see that Minitab reports the P-value as 0.000, you should report the P-value as being "< 0.001."

If the biologist set her significance level α 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):

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 α = 0.05, and the biologist rejects the null hypothesis. 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.

Note again that the biologist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

### Example: 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:

 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:

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 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 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 qualtiy 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 α = 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 α = 0.05, the quality control specialist fails to reject the null hypothesis. There is insufficient evidence, at the α = 0.05 level, to conclude that the mean thickness of all pieces of spearmint gum differs from 7.5 one-hundredths of an inch.

Note that the quality control specialist obtains the same scientific conclusion regardless of the approach used. This will always be the case.

### 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.

## Method 2 - Using standardized test statistics

One can also use the standardized value z.

Test Statistic, Z = (sample mean – population mean)/(std-dev/sqrt(no. of samples) i.e.

Then, the rejection region becomes

Z= (190 – 180)/(75/sqrt(300)) = 2.309

Our rejection region at 5% significance level is Z> Z0.05 = 1.645

Since Z= 2.309 is greater than 1.645, the null hypothesis can be rejected with the similar conclusion mentioned above.

## Method 3 - P-value calculation

We aim to identify P(sample mean >= 190, when mean = 180)

= P (Z >= (190- 180)/( 75 / sqrt (300))

= P (Z >= 2.309) = 0.0084 = 0.84%

The following table to infer p-value calculations concludes that there is confirmed evidence of average monthly returns being higher than 180.

 p-value Inference less than 1% Confirmed evidence supporting alternative hypothesis between 1% and 5% Strong evidence supporting alternative hypothesis between 5% and 10% Weak evidence supporting alternative hypothesis greater than 10% No evidence supporting alternative hypothesis

Example 2: A new stockbroker (XYZ) claims that his brokerage fees are lower than that of your current stoc broker's (ABC). Data available from an independent research firm indicates that the mean and std-dev of all ABC broker clients are $18 and$6 respectively.

A sample of 100 clients of ABC is taken and brokerage charges are calculated with the new rates of XYZ broker. If the mean of sample is $18.75 and std-dev is same ($6), can any inference be made about the difference in the average brokerage bill between ABC and XYZ broker?

H0: Null Hypothesis: mean = 18

H1: Alternative Hypothesis: mean <> 18 (This is what we want to prove)

Rejection region: Z <= - Z2.5 and Z>=Z2.5 (assuming 5% significance level, split 2.5 each on either side)

Z = (sample mean – mean)/(std-dev/sqrt(no. of samples)

= (18.75 – 18) / (6/(sqrt(100)) = 1.25

This calculated Z value falls between the two limits defined by

- Z2.5 = -1.96 and Z2.5 = 1.96.

This concludes that there is insufficient evidence to infer that there is any difference between the rates of your existing broker and the new broker.

Alternatively, The p-value = P(Z< -1.25)+P(Z >1.25)

= 2 * 0.1056 = 0.2112 = 21.12% which is greater than 0.05 or 5%, leading to the same conclusion.

Graphically, it is represented by the following:

Criticism Points for Hypothetical Testing Method

• Statistical method based on assumptions
• Error prone as detailed in terms of alpha and beta errors
• Interpretation of p-value can be ambigous, leading to confusing results

## The Bottom Line

Hypothesis testing allows a mathematical model to validate a claim or idea with a certain confidence level. However, like majority of statistical tools and models, it is bound by a few limitations. The use of this model for making financial decisions should be considered with a critical eye, keeping all dependencies in mind. Alternate methods like Bayesian Inference are also worth exploring for similar analysis.

For more on practical applications of data to determine risk, see "5 Ways to Measure Mutual Fund Risk."