Answer :
Answer:
The 98% confidence interval would be given by (0.245;1.163)
Since the upper value for the confidence interval is higher than 1 we can't conclude that the specification required is meeting at 2% of significance.
Step-by-step explanation:
1) 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".
The data is:
0.51 ,0.69,0.10,0.93,1.31,0.50,0.89
2) Compute the sample mean and sample standard deviation.
In order to calculate the mean and the sample deviation we need to have on mind the following formulas:
[tex]\bar X= \sum_{i=1}^n \frac{x_i}{n}[/tex]
[tex]s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}[/tex]
In Excel the formula is:
=AVERAGE(0.51 ,0.69,0.10,0.93,1.31,0.50,0.89)
On this case the average is [tex]\bar X= 0.704[/tex]
In excel the formula is:
=STDEV.S(0.51 ,0.69,0.10,0.93,1.31,0.50,0.89)
The sample standard deviation obtained was s=0.3867
3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints
In order to find the critical value we need to take in count that our sample size n =7 <30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 98% of confidence, our significance level would be given by [tex]\alpha=1-0.98=0.02[/tex] and [tex]\alpha/2 =0.01[/tex]. The degrees of freedom are given by:
[tex]df=n-1=7-1=6[/tex]
We can find the critical values in excel using the following formulas:
"=T.INV(0.01,6)" for [tex]t_{\alpha/2}=-3.14[/tex]
"=T.INV(1-0.01,6)" for [tex]t_{1-\alpha/2}=3.14[/tex]
The confidence interval for the mean is given by the following formula:
[tex]\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}[/tex]
If we replace we got:
[tex]0.704 - 3.14\frac{0.3867}{\sqrt{7}}=0.245[/tex]
[tex]0.704 + 3.14\frac{0.3867}{\sqrt{7}}=1.163[/tex]
So the 98% confidence interval would be given by (0.245;1.163)
Does it appear that there is too much mercury in tuna? sushi?
Since the upper value for the confidence interval is higher than 1 we can't conclude that the specification required is meeting at 2% of significance.