Question

In a massive attempt to compete with General Electric, the Acme Light Bulb Company issued a newline of bulb. Acme took 100 bulbs from their new line and measured their lifetime. The mean measuredlifetime for the 100 bulbs was 1280 hours and the sample standard deviation for the 100 bulbs was138.32.

(a) The light bulb lifetimes come from an exponential probability distribution. If ? is the meanlifetime of the light bulbs, and the light bulb lifetimes are iid, give the probability distribution ofthe mean lifetime from the sample of 100 light bulbs. Specify the mean and standard deviationof this sample mean lifetime.(b) Construct a 90% confidence interval of the true mean lifetime of Acme

Answer 1

(a) The Normal Distribution,
`\\ \mu_{\overline{X}} = \mu = 1280`, and standard deviation of
`\\ (S)/(√(n)) = 13.832` hours. (b) The 90% confidence interval of the true mean lifetime goes from 1257.04 to 1302.96 hours. hours.

Step-by-step explanation:

We want to estimate the true value of the mean for a population taking a sample from this population. When the sample is greater or equal to 30
`\\ n \geq 30`, no matter which distribution the data come from, the distribution of the sample means will have a normal distribution with mean
`\\ \mu` (the mean of the population) and a standard deviation of
`\\ (\sigma)/(√(n))`. Mathematically:

`\\ \overline{X} \sim N(\mu, (\sigma)/(√(n)))`

Likewise, the variable Z will behave as normal standard distribution
`\\ \mu = 0` and
`\\ \sigma = 1`.

Then

`\\ Z = \frac{\overline{X} - \mu}{(\sigma)/(√(n))}`

`\\ Z \sim N(0, 1)`

Key aspects to solve this question:

1. The sample size is n = 100 bulbs.
2. The mean
   `\\ \mu = 1280` hours.
3. The sample standard deviation
   `\\ S =138.32`

Having this information at hand, we are prepared to solve the questions.

The probability distribution of the mean lifetime from the sample of 100 light bulbs

The light bulb lifetimes come from an exponential probability distribution. However, since the sample size in greater than 30, that is,
`\\ n \geq 30`, the sampling distribution will follow a normal distribution (as previously explained). Then, the probability distribution of the mean lifetime from the sample of 100 light bulbs is the normal distribution.

In this conditions, the mean and standard deviation of this sample mean lifetime are:

The mean of the sample is the same of the population, i. e.,
`\\ \mu_{\overline{X} = \mu`. This is also the mean for the distribution of the sample means. Therefore,
`\\ \mu_{\overline{X}} = \mu = 1280` hours.

The standard deviation in this case is not the population standard deviation (since we do not have this value). Instead, we have the sample standard deviation given
`\\ S = 138.32`. Thus, the standard deviation for the sample is

`\\ (S)/(√(n))`

`\\ (138.32)/(√(100))`

`\\ (138.32)/(10)`

`\\ 13.832` hours.

The 90% confidence interval of the true mean lifetime

With this interval, we need to find two values. We do not have the value for the population standard deviation. The confidence interval is given by the following formula:

Upper limit =
`\\ \mu +t_(c)(S)/(√(n))`

Lower limit =
`\\ \mu -t_(c)(S)/(√(n))`

Where
`\\ t_(c)` is the confidence coefficient for a t-student distribution, with n - 1 degrees of freedom. For a 90% confidence interval is, approximately,
`\\ t_(c) = 1.660`.

We have already determined the values for
`\\ \mu = 1280` hours and
`\\ (S)/(√(n)) = 13.832`hours. Therefore:

Upper limit =
`\\ 1280 +1.660*13.832 = 1302.96` hours.

Lower limit =
`\\ 1280 -1.660*13.832 = 1257.04` hours.

Thus, the 90% confidence interval of the true mean lifetime goes from 1257.04 to 1302.96 hours.