The mean lifetime of light bulbs produced by a company has, in the past, been 1500 hours - Leaving Cert Mathematics - Question 3 - 2015

Using the formula for the z-score: $z = \frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}}}$, we have:

• Sample mean ((\bar{x})): 1475 hours
• Population mean ((\mu)): 1500 hours
• Standard deviation ((s)): 110 hours
• Sample size ((n)): 100

Now substituting the values: $z = \frac{1475 - 1500}{\frac{110}{\sqrt{100}}} = \frac{-25}{11} \approx -2.27$

Now, we compare the calculated z-score of -2.27 with the critical z-values at the 0.05 significance level, which are approximately ±1.96. Since -2.27 is less than -1.96, we reject the null hypothesis. Therefore, there is significant evidence to suggest that the mean lifetime of the bulbs has changed.

Using the z-score tables, we find:

• For z = -2.27, the area to the left is approximately 0.0116 and to the right is approximately 0.9884. Thus, the p-value is: $p = 2 \times P(Z < -2.27) = 2 \times 0.0116 = 0.0232$ So, the p-value is approximately 0.0232.

The p-value of 0.0232 indicates that there is a 2.32% chance of observing a sample mean as extreme as 1475 hours, or more extreme, if the null hypothesis (that the mean lifetime is 1500 hours) were true. Since this p-value is less than the significance level of 0.05, it provides strong evidence against the null hypothesis.