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

Answer:

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

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

$df=n-1=64-1=63$  

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".  

$\bar X=1400$ 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:  

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

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

$df=n-1=64-1=63$