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A random sample of 120 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 SAT scores, the degrees of freedom for reading the critical values of the "t" value is:_________

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Answer:

1.980

Step-by-step explanation:

Given that a random sample of 120 SAT scores of students applying for merit scholarships showed an average of 1400 with a standard deviation of 240

Here nothing mentioned about the distribution but sample size is very large.

Since population std deviation is not known, only sample std deviation is used.

t test will be more appropriate.

df = degrees of freedom = n-1=119

For two tailed 95% level, t critical value for 119 df would be

1.980

Using this critical value confidence interval would be formed.

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