Question

A random sample of 64 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240. If we want to provide a 95% confidence interval for the population mean SAT score, the degrees of freedom for reading the t value is ___.

Answer 1

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

The point estimate of the population mean is
`\hat \mu = \bar X </p><p>In order to calculate the critical value [tex]t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=64-1=63`

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=1400` represent the sample mean

`\mu` population mean (variable of interest)

s=240 represent the sample standard deviation

n=64 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)

The point estimate of the population mean is
`\hat \mu = \bar X </p><p>In order to calculate the critical value [tex]t_(\alpha/2)` we need to find first the degrees of freedom, given by:

`df=n-1=64-1=63`