One-Tailed — Known Mean and Standard Deviation: Hypothesis Testing

For one-tailed, known mean and standard deviation tests the test statistic to use is the z-score.

Example: A range of initiatives, including increases in excise duty, constraints on pack sizes, and restrictions on distribution channels, were implemented with the aim of reducing cigarette consumption. The government set a target of decreasing consumption to below 120 sticks per smoker per month.

To evaluate the effectiveness of these measures, a study was conducted, involving 100 smokers. The findings revealed that, on average, each smoker consumed 28 sticks per week, amounting to 112 sticks per month.

To provide further context, an earlier large-scale consumption study estimated the standard deviation of cigarette consumption to be 40 sticks per month.

Based on these findings, can we conclude 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

Exhibit 33.21 Probability that the sample average consumption of cigarettes is less than 120 is 0.023 or 2.3%.

The p-value, as depicted in Exhibit 33.21, indicates that the probability of obtaining a z-score of −2.0 or lower is 0.023. Using this information, we can infer that there is a 97.7% probability (1 − 0.023) that the government’s initiatives have effectively achieved the objective of reducing cigarette consumption to less than 120 sticks per month.

The null hypothesis is rejected because the p-value, 0.023 (2.3%), is less than α, 5%.

Note: A p-value from z-score calculator is provided on this webpage.

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