Question

The calories in a random sample of 14 standard-size candy bars have a standard deviation of 21.6. Find the lower limit of the 95% confidence interval for the population standard deviation. Round your answer to the nearest tenth.

Answer 1

The question given to us tells us we have a sample size of 14 candy bars. Because the sample size is below 30, we can use t- distribution.

The standard deviation of the distribution of the candy bars is 21.6.

We are asked to find the lower limit of the population standard deviation given a confidence level of 95%.

In order to solve this problem, we need to follow a few steps.

1. Calculate the alpha level.

2. Use alpha level to find the critical probability

3. Check the critical value on t - distribution table

4. Find the upper and lower limit of the standard deviation.

Now, let us apply these steps.

1. Calculate the alpha level.

`\begin{gathered} \alpha-\text{level}=1-\text{confidence level} \\ \text{confidence level=95\% =}(95)/(100)=0.95 \\ \\ \therefore\alpha=1-0.95=0.05 \end{gathered}`

The alpha level is 0.05.

2. Use alpha level to find the critical probability (p*):

`\begin{gathered} p\cdot=1-(\alpha)/(2) \\ p\cdot=1-(0.05)/(2) \\ \\ \therefore p\cdot=0.975 \end{gathered}`

The critical probability (p*) = 0.975.

3. Check the critical value on t - distribution table:

In order to do this, we need to know the degrees of freedom (df). Degrees of freedom (df) is gotten by:

`\begin{gathered} df=n-1 \\ n=\text{sample size} \\ df=14-1 \\ \therefore df=13 \end{gathered}`

The process for finding the critical value on the t-distribution table is shown below:

The critical value is: 2.16

4. Find the upper and lower limit of the standard deviation:

The formula for finding the limits of the standard deviation is given below:

`\begin{gathered} \pm t_(n-1)*\frac{s}{\sqrt[]{n}} \\ t_(n-1)=\text{critical value} \\ s=\text{ standard deviation} \\ n=\text{sample size} \end{gathered}`

Now, let us calculate the limits:

`\begin{gathered} \pm2.16*\frac{21.6}{\sqrt[]{14}} \\ =\pm2.469\approx\pm12.5\text{ (to nearest tenth)} \\ \\ \therefore\text{lower limit = 21.6 - 12.5=9.1} \end{gathered}`

Therefore, the final answer is:

9.1