Thirty randomly selected students took the calculus final. If the sample mean was 95 and the standard deviation was 6.6, construct a 99% confidence interval for the mean score of all students. a) 91.69 < μ < 98.31b) 92.95 < μ < 97.05c) 91.68 < μ < 98.32d) 92.03 < μ < 97.97

c) 91.68 < μ < 98.32

Step-by-step explanation:

Let μ  be the mean score of all students.

Let represents the sample mean score.

As per given , we have

Sample size : n= 30

Sample mean :

Sample standard deviation:

Significance level

Now, Confidence interval of mean when population standard deviation is unknown is given by :-

Degree of freedom = n-1 = 29

Critical two-tailed t-value : [By students' t distribution table]

Confidence interval of mean will be:

i.e. a 99% confidence interval for the mean score of all students : 91.68 < μ < 98.32

So correct option is c) .

dunno

step-by-step explanation:

because they could drop even lower..