Question

Use the given data to construct a confidence interval for the population standard deviation, σσ. Assume that the population has a normal distribution: The football coach randomly selected the following players and timed how long each player took to perform a certain drill. The times (in minutes) were:

[10,10,15,10,11,7,5,5]

Find a 95.0 confidence interval for the population standard deviation σ

A) 2.6194049<σ<6.43192582.6194049<σ<6.4319258
B) 2.2194049<σ<6.83192582.2194049<σ<6.8319258
C) 2.6194049<σ<7.43192582.6194049<σ<7.4319258
D) None of These

Answer 1

The confidence interval for the population standard deviation would be:

C) 2.6194049<σ<7.4319258

Explanation:

1) Data given and notation

s represent the sample standard deviation

`\bar x` represent the sample mean

n=8 the sample size

Confidence=95% or 0.95

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 mean or variance 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.

The Chi Square distribution is the distribution of the sum of squared standard normal deviates.

2) Calculating the confidence interval

The confidence interval for the population variance is given by the following formula:

`((n-1)s^2)/(\chi^2_(\alpha/2)) \leq \sigma \leq ((n-1)s^2)/(\chi^2_(1-\alpha/2))`

On this case we need to find the sample standard deviation with the following formula:

`s=\sqrt{(\sum_(i=1)^8 (x_i -\bar x)^2)/(n-1)}`

And in order to find the sample mean we just need to use this formula:

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

The sample mean obtained on this case is
`\bar x= 9.125` and the deviation s=3.357

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

`df=n-1=8-1=7`

Since the Confidence is 0.95 or 95%, the value of
`\alpha=0.05` and
`\alpha/2 =0.025`, and we can use excel, a calculator or a tabel to find the critical values.

The excel commands would be: "=CHISQ.INV(0.025,7)" "=CHISQ.INV(0.975,7)". so for this case the critical values are:

`\chi^2_(\alpha/2)=16.012`

`\chi^2_(1- \alpha/2)=1.690`

And replacing into the formula for the interval we got:

`((7)(3.357)^2)/(16.012) \leq \sigma ((7)(3.357)^2)/(1.690)`

`4.927 \leq \sigma^2 \leq 46.678`

Now we just take square root on both sides of the interval and we got:

`2.2194049 \leq \sigma \leq 6.8319258`

So the best option would be:

C) 2.6194049<σ<7.4319258