Question

Suppose a baker claims that the mean time to prepare a cake is at most 360 minutes. A sample of 25 bakers in randomly selected and it is found that the mean time taken by them to prepare a cake was 270 minutes. Assume also that standard deviation is known to be 120 minutes. Assuming the time to prepare a cake is normally distributed and using a 0.05 level of significance, would you conclude the baker’s claim is true?

Answer 1

$$n = 25, \bar{x} = 270, s = 120, F(t) = 1 - 0.05 = 0.95$$

from the table of Laplace distribution $t = 1.96$

$$s = \sqrt{\frac{n}{n - 1}} \sigma^2$$ $$\sigma = s \sqrt{\frac{n - 1}{n}} = 120 \sqrt{\frac{24}{25}} = 48\sqrt{6}$$ $$\bar{x} - t \frac{\sigma}{\sqrt{t}} < \overline{x_0} < \bar{x} + t \frac{\sigma}{\sqrt{t}}$$ $$\overline{x_0} < 270 + 1.96 \frac{48\sqrt{6}}{\sqrt{24}} = 270 + 1.96 \cdot 24 = 317.4 < 360$$

so the baker's claim is true.