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 ___.

(1)  

The point estimate of the population mean is we need to find first the degrees of freedom, given by:  

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".  

represent the sample mean  

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:  

(1)  

The point estimate of the population mean is we need to find first the degrees of freedom, given by:  

mark branliest

step-by-step explanation:

.6 is 1/2 of 12

the answer is actually d. i just took the test

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step-by-step explanation: