Question

The grade point averages for 10 randomly selected high school students are listed below. Assume the grade point averages are normally distributed. 2.0 3.2 1.8 2.9 0.9 4.0 3.3 2.9 3.6 0.8 Find a 98% confidence interval for the true mean.

Answer 1

The 98% confidence interval would be given by (1.55;3.53)

A range of values that, with a particular level of confidence, is likely to encompass a population value is called a confidence interval. A population mean is typically stated as a percentage that falls between an upper and lower interval.

The range of values in a confidence interval below and above the sample statistic is called the margin of error.

The 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".

show the sample mean for the given sample.

population mean (the relevant variable)

The sample standard deviation is denoted by s.

n stands for the number of samples.

Resolution of the issue

The following formula produces the mean's confidence interval:

`\bar x` ±
`(t_a)/(2) (s)/(√(n) )` ____________(1)

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar x =`
`\sum_(i =1)^n (x_i)/(n)`_______(2)

`s = \sqrt{(\sum^n_(i=1)(x_i-\bar x))/(n-1) }` ____________(3)

The mean calculated for this case is X = 2.54

The sample deviation calculated s = 1.110

t In order to calculate the critical value
`(t_a)/(2)` we need freedom, given by: to find first the degrees of

df = n-1=10-19

Since the Confidence is 0.98 or 98%, the value of a = 0.02 and a/2 = 0.01, and we can use excel, a calculator or a tabel to find the critical value.

`(t_a)/(2)` = 2.82

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

2.54 - 2.82
`(1.110)/(√(10) )` = 1.55

2.54 + 2.82
`(1.110)/(√(10) )` = 3.53

Answer 2

`2.54-2.82(1.110)/(√(10))=1.55`

`2.54+2.82(1.110)/(√(10))=3.53`

So on this case the 98% confidence interval would be given by (1.55;3.53)

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".

`\bar X` represent the sample mean for the sample

`\mu` population mean (variable of interest)

s represent the sample standard deviation

n represent the sample size

Solution to the problem

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

`\bar X \pm t_(\alpha/2)(s)/(√(n))` (1)

In order to calculate the mean and the sample deviation we can use the following formulas:

`\bar X= \sum_(i=1)^n (x_i)/(n)` (2)

`s=\sqrt{(\sum_(i=1)^n (x_i-\bar X))/(n-1)}` (3)

The mean calculated for this case is
`\bar X=2.54`

The sample deviation calculated
`s=1.110`

In order to calculate the critical value
`t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=10-1=9`

Since the Confidence is 0.98 or 98%, the value of
`\alpha=0.02` and
`\alpha/2 =0.01`, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.01,9)".And we see that
`t_(\alpha/2)=2.82`

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

`2.54-2.82(1.110)/(√(10))=1.55`

`2.54+2.82(1.110)/(√(10))=3.53`

So on this case the 98% confidence interval would be given by (1.55;3.53)