Question

ecn 221 A random sample of 10 miniature Tootsie Rolls was taken from a bag. Each piece was weighed on a very accurate scale. The results in grams were 3.0873.1313.2413.2413.2703.3533.4003.4113.4373.477 Click here for the Excel Data File (a) Construct a 90% confidence interval for the true mean weight. (Round your answers to 4 decimal places.)

Answer 1

`3.3048-1.83(0.132)/(√(10))=3.2284`

`3.3048 +1.83(0.132)/(√(10))=3.3812`

So on this case the 90% confidence interval would be given by (3.2284;3.3812)

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)

s represent the sample standard deviation

n represent the sample size

Data: 3.087 3.131 3.241 3.241 3.270 3.353 3.400 3.411 3.437 3.477

Solution to the problem

The confidence interval for the mean is given by the following formula:

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=3.3048`

The sample deviation calculated
`s=0.132`

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=10-1=9`

Since the Confidence is 0.90 or 90%, the value of
`\alpha=0.1` and
`\alpha/2 =0.05`, and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.05,9)".And we see that
`t_(\alpha/2)=1.83`

Now we have everything in order to replace into formula (1):

`3.3048-1.83(0.132)/(√(10))=3.2284`

`3.3048 +1.83(0.132)/(√(10))=3.3812`

So on this case the 90% confidence interval would be given by (3.2284;3.3812)