**Q: Motorola wishes to estimate the mean talk times for one of their new phones before the battery must be recharged. In a random sample of 35 phones, the sample mean talk time is 325 minutes (x-bar).**

**a) Why can we say that the sampling distribution of x-bar is approximately normal?**

**b) Construct a 90% confidence interval for the mean talk time of all new Motorola phones, assuming the population standard deviation is 31 minutes.**

Answer:

(Please keep in mind, statistics is not 100% my area of expertise, but I will do my best to provide clear and accurate explanations)

a) The Central Limit Theorem essential tells us that x-bar will be approximately normally distributed for a sufficiently large number of independent random variables (which we are dealing with). So, we can assume that the distributions of the sample means (x-bars) is normally distributed thanks to the CLT. A sample size of 30 or larger is considered “large enough” to assume an approximately normal distribution.

b) We first need to find (or know) the z-value associated with a 90% confidence interval. Our alpha-value is .10 (10%), which makes alpha/2 = .05. So, we need to find the z-value that is associated with a .05 area under the curve (a reverse look-up on a standard normal table of values). The desired z-score is 1.645

In other words, of all the possible x-bar values along the horizontal axis of the normal distribution curve, 90% of them should be within a z-score of 1.645 from the population mean.

A confidence interval can be calculated by using the following formula:

x-bar +/- z*(population standard deviation / sqrt(sample size))

So, a 90% confidence interval, with our given information can be calculated by:

325 +/- 1.645*(31/sqrt(35))

325 +/- 8.6197

Which gives two answers:

325 – 8.6197 = 316.3803

and

325 + 8.6197 = 333.6197

Therefore, we can be 90% confident that the average length of talk time before needing recharge for the population is between 316.3803 and 333.6197 minutes.

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