The expression means 15 minus the product of 5 times the difference of p minus 6. If p = 8, then the value of the expression is
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Answer:
Critical values:
95% confidence interval would be given by (3.646;4.472)
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:
4.65 3.89 2.73 4.35 3.80 4.86 4.33 4.37 4.76 4.05 3.05 3.87
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:
=AVERAGE(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)
On this case the average is
=STDEV.S(4.65,3.89,2.73, 4.35, 3.8, 4.86, 4.33, 4.37, 4.76, 4.05, 3.05, 3.87)
The sample standard deviation obtained was s=0.6503
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 =12 <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 95% of confidence, our significance level would be given by and
. The degrees of freedom are given by:
We can find the critical values in excel using the following formulas:
"=T.INV(0.025,11)" for
"=T.INV(1-0.025,11)" for
The confidence interval for the mean is given by the following formula:
And we can use Excel to calculate the limits for the interval
Lower interval : "=4.059 -2.201*(0.6503/SQRT(12))" =3.646
Upper interval : "=4.059 +2.201*(0.6503/SQRT(12))" =4.472
So the 95% confidence interval would be given by (3.646;4.472)