Question

Six different national brands of chocolate chip cookies were randomly selected at the supermarket. The grams of fat per serving as as follows: 7; 7; 11; 7; 8; 8. Assume the underlying distribution is approximately normal.Note: If you are using a Student's t-distribution, you may assume that the underlying population is normally distributed. (In general, you must first prove that assumption, though.)Construct a 90% confidence interval for the population mean grams of fat per serving of chocolate chip cookies sold in supermarkets.i. State the confidence interval. (Round you answers to two decimal places.)ii. Sketch the graph.iii. Calculate the error bound. (Round your answer to two decimal places.)

Answer 1

Answer:

i. The 90% confidence interval would be given by (6.72;9.28)

ii. See the figure attached.

iii. $ME=2.02\frac{1.549}{\sqrt{6}}=1.28$

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".

We can calculate the sample mean and the sample deviation with the following formulas:

$\bar X =\frac{\sum_{i=1}^n X_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (X_i -\bar X)^2}{n-1}}$

$\bar X=8$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)

s=1.549 represent the sample standard deviation

n=6 represent the sample size  

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=6-1=5$

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(0.05,5)".And we see that $t_{\alpha/2}=2.102$

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

$8-2.02\frac{1.549}{\sqrt{6}}=6.722$    

$8+2.02\frac{1.549}{\sqrt{6}}=9.278$

So on this case the 90% confidence interval would be given by (6.72;9.28)    

ii. Sketch the graph.

See the figure attached.

.iii. Calculate the error bound. (Round your answer to two decimal places.)

The margin of error is given by:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$

And replacing we got:

$ME=2.02\frac{1.549}{\sqrt{6}}=1.28$