For one-tailed, known mean and standard deviation tests the test statistic to use is the z-score.
Example: A slew of initiatives (increases in excise duty, constraints on pack sizes and restrictions on distribution channels) were introduced in an effort to cut down the consumption of cigarettes. The government’s target was to reduce consumption to less than 120 sticks per smoker, per month.
In a study conducted to assess the success of this initiative, the average consumption of 100 smokers was 28 sticks per week, or 112 sticks per month.
Based on a large-scale consumption study conducted at an earlier date, the standard deviation of consumption of cigarettes was estimated as 40 sticks per month.
From this information, is it possible to gauge whether the government achieved its target?
H0: μ ≥120
HA: μ<120
α = 5%
$$ z = \frac{\bar x-μ}{s/\sqrt n} = \frac{112-120}{40/10} = -2.0 $$p-value = 0.023 < α = 0.05
The p-value (refer Exhibit 33.20) reveals that the probability of obtaining a z-score of −2.0 or lower is approximately 0.023. Based on this we can conclude that there is 97.7% (1 − 0.023) probability that the government’s initiatives succeeded in achieving the target of reducing consumption to less than 120 sticks per month.
In the context of the null hypothesis, there is a probability of 0.023 (2.3%) that the average smoker is now smoking less than 120 sticks per month. As this is significant for the given level of 5%, the null hypothesis is rejected.
Note: A p-value from z-score calculator is provided on this web page.
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