Mathematics, 31.03.2020 03:43 emwemily

Question 1.05 It is known from past experience that the life of traffic light bulbs is normally distributed with a standard deviation of 250 hours. In a study, if you wanted the total width of the two-sided confidence interval on mean to be 120 hours at 90% confidence, how many light bulbs should be selected for the study

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Answer from: andrealch13p3w89f

The required sample size of lightbulbs = 47

Step-by-step explanation:

The total width of the two-sided confidence interval on mean to be 120 hours at 90% confidence.

That is, 60 hours on top and below the mean.

Confidence interval = (Sample mean) ± (Margin of error)

So, from this description that the interval is of width 120 hours of the mean and 60 hours, below and above the mean,

Margin of error = 60 hours.

Margin of Error = (critical value) × (standard deviation of the distribution of sample means)

Critical value for 90% confidence = 1.645

Standard deviation of the distribution of sample means = (standard deviation) ÷ √n

where n = sample size = ?

Standard deviation = 250 hours

60 = 1.645 × (250/√n)

√n = (1.645×250)/60 = 6.854

n = 6.854² = 46.98 = 47 (approximated to the closest whole number)

Hope this Helps!!!

Answer from: gm2

$n=(\frac{1.64(250)}{60})^2 =46.69 \approx 47$

So the answer for this case would be n=47 rounded up to the nearest integer

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

$\bar X$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

$\sigma= 250$ represent the population standard deviation

n represent the sample size

Solution to the problem

For this case the width of the interval is 120 and we have the following relation:

$ME = \frac{width}{2}= \frac{120}{2}=60$

The margin of error is given by this formula:

$ME=z_{\alpha/2}\frac{s}{\sqrt{n}}$ (a)

And on this case we have that ME =60 and we are interested in order to find the value of n, if we solve n from equation (a) we got:

$n=(\frac{z_{\alpha/2} \sigma}{ME})^2$ (b)

The critical value for 90% of confidence interval now can be founded using the normal distribution. And in excel we can use this formula to find it:"=-NORM.INV(0.05;0;1)", and we got $z_{\alpha/2}=1.64$ , replacing into formula (b) we got: