A soap manufacturing company was distributing a particular brand of a soap through a large number of retail shops. Before a heavy advertisement campaign, the mean sales per week per shop was 140 dozen. After the campaign, a sample of 26 shops was taken and mean sales was found to be 147 dozen with standard deviation 16. Can you consider the advertisement campaign effective ? [Given $t_{25}(0.05) = 2.06$ ]

Q: How do we conclude the campaign is effective?

We compare the computed t value 2.23 with the critical value 2.06 at 25 degrees of freedom and 5% significance. Since 2.23 exceeds 2.06, the difference is statistically significant, we reject the null hypothesis of no effect, and conclude the campaign genuinely raised sales.

Inferential Statistics

t-Test for a Single Mean

Official Answer

Yes, the advertisement campaign can be considered effective at the 5% level.

Hypotheses:

Null $H_0$ : $\mu = 140$ (campaign has no effect).
Alternative $H_1$ : $\mu > 140$ (campaign increased sales).

Given values:

Population mean $\mu = 140$ , sample mean $\bar{x} = 147$ , $s = 16$ , $n = 26$ , degrees of freedom $\nu = 25$ .

Test statistic:

$t = (\bar{x} - \mu) / (s / \sqrt{n}) = (147 - 140) / (16 / \sqrt{26}) = 7 / 3.137 = 2.23$ (approximately).

Decision:

Calculated $|t| = 2.23$ > table value $t_{25}(0.05) = 2.06$ .
So we reject $H_0$ .

Conclusion:

The increase in mean sales is statistically significant, so the advertisement campaign is effective.

t-testsingle meannull hypothesisdegrees of freedomlevel of significancetest statisticreject H0campaign effective

Marking Scheme

10.5 mark: stating $H_0$ : $\mu = 140$ and $H_1$ : $\mu > 140$ .
21.5 marks: correct test statistic $t = (147 - 140)/(16/\sqrt{26}) = 2.23$ (award for correct formula and substitution).
31 mark: comparing $2.23 > 2.06$ , rejecting $H_0$ , and concluding the campaign is effective.

Hint

Use $t = (\bar{x} - \mu)/(s/\sqrt{n})$ with $\nu = n - 1 = 25$ , then compare the computed t with the table value 2.06.

Quick Oral Answer

I test $H_0$ : $\mu = 140$ against $H_1$ : $\mu > 140$ with a one-sample t-test; $t = (147-140)/(16/\sqrt{26}) = 2.23$ , which exceeds the table value 2.06 at 25 degrees of freedom, so I reject $H_0$ and conclude the campaign is effective.

Analysis & Explanation

This is a one-sample t-test because the population standard deviation is unknown and the sample is small ( $n < 30$ ).

Concept:

When sigma is unknown and n is small, the sampling distribution of the mean follows Student's t-distribution with $\nu = n - 1$ degrees of freedom. We compare the standardized deviation of the sample mean from the claimed population mean against the critical t value.

Why one-tailed:

The question asks whether the campaign is 'effective', i.e., whether sales increased, so the alternative is directional ( $\mu > 140$ ) and we use the one-tailed critical value $t_{25}(0.05) = 2.06$ .

Decision rule:

Reject H0 if calculated |t| exceeds the table value. Here $2.23 > 2.06$ , so the result is significant and H0 is rejected.

Exam trap:

Using $\nu = 26$ instead of 25, or forgetting to divide $s$ by $\sqrt{n}$ . Both distort the t value.

Real-world link:

Companies routinely run exactly this A/B style significance test to decide whether marketing spend actually moved sales before scaling a campaign nationally.

Common Mistakes

1Using degrees of freedom $\nu = 26$ instead of $\nu = n - 1 = 25$ .
2Forgetting to divide the standard deviation by $\sqrt{n}$ , giving a wrong t value.
3Concluding 'not effective' by mis-comparing; since $2.23 > 2.06$ the null hypothesis must be rejected.

Interesting Facts

The t-distribution was published in 1908 by William Sealy Gosset under the pen name 'Student' because his employer, the Guinness brewery, barred staff from publishing.

As the sample size grows, the t-distribution approaches the standard normal curve; for n above about 30 the two are nearly identical, which is why the small-sample t-test matters most here ( $n = 26$ ).

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Frequently Asked Questions

Why do we use a t-test and not a z-test here?

The population standard deviation is unknown and the sample size is small ( $n = 26 < 30$ ). Under these conditions the sample mean follows Student's t-distribution with $n - 1 = 25$ degrees of freedom, so the t-test is appropriate rather than the z-test.

How do we conclude the campaign is effective?

We compare the computed t value 2.23 with the critical value 2.06 at 25 degrees of freedom and 5% significance. Since 2.23 exceeds 2.06, the difference is statistically significant, we reject the null hypothesis of no effect, and conclude the campaign genuinely raised sales.