Two-tailed, known mean and standard deviation tests also use the z-score statistic. We can use the z-score and the normal distribution, provided the population’s standard deviation is known.

*Example:* The mean weight of fresh recruits into the army was 65.8 kg last year.
For a sample of 200 recruits this year, the mean weight is 66.2 kg. Assuming the population standard deviation is 3.2 kg, at 0.05
significance level, can we conclude that the mean weight has changed since last year?

H_{0}: μ=65.8

H_{A}: μ≠65.8

α = 5%

$$ z = \frac{\bar x-μ}{s/\sqrt n} = \frac{66.2-65.8}{3.2/14.14} = 1.77 $$p-value = 0.077 > α = 0.05

The p-value of 0.077, obtained from normal distribution (Exhibit 33.21) for z = 1.77, is not significant for the given level of 5%.

If the actual mean was 65.8 kg, there is a 7.7% probability that the sampled recruits would weigh ≥ 66.2 kg or ≤ 65.4 kg. Since this probability is higher than the significance level of 5%, the null hypothesis is not rejected. We cannot conclude with 95% certainty that the new recruits differ in weight from those recruited last year.

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