Practice
- A tech company tests a new phone battery. A sample of 30 phones, coming from a normally-distributed population, shows an average battery life of 22.4 hours with a sample standard deviation of 3.5 hours. Construct a 99% confidence interval for the true mean battery life of the new phone model.
Solution
From the problem we get the following information:
- A confidence level of 99%
- Sample Size is \(n=30\)
- The sample mean is \(\bar{x} = 22.4\)
- The sample standard deviation is \(s=3.5\)
First, verify the central limit theorem
- Is the sample random? Yes (stated in the problem)
- Is the sample large enough? Yes (sample size is at least 30)
The Central Limit Theorem holds.
Now, we find the critical value. Since we only have a sample standard deviation, we need to use the t-score with \(DF = n-1 = 29\) degrees of freedom. Doing this, we get \(t_c = 2.756\).
\[E = t_c\frac{s}{\sqrt{n}} = 2.756\frac{3.5}{\sqrt{30}} = 1.76\]
\[\bar{x} + E = 22.4 + 1.76 = 24.16\]
\[\bar{x} - E = 22.4 - 1.76 = 20.64\]
Solution:
We are 99% confident that the true mean for battery life is between 20.64 and 24.16 hours.
Return back to Lesson 18.5