The owner of a major Atlanta restaurant wanted to estimate the mean amount spent per customer for dinner meals

The owner of a major Atlanta restaurant wanted to estimate the mean amount spent per customer for dinner meals. A random sample of n=49 customers over a 3 week period revealed a sample average of x=12.60 spent, with a population standard deviation = 2.50. Determine a 95% confidence interval ( rounded to the nearest cent) for the average amount spent per dinner meal of all his customers

a. 12.01 to 13.19
b. 11.76 to 13.44
c. 11.90 to 13.30
d. unable to determine because of insufficient data
e. none of the above answers

A random sample of n=64 children of working mothers showed that they were absent from school an sample average of x=5.3 days per term, with a standard deviation s=1.8 days. Provide a 96% confidence interval for the average number of days absent per term for all students.

a. 5.2321 days to 5.3762 days
b. 4.8151 days to 5.7543 days
c. 4.8387 days to 5.7613 days
d. 4.7722 days to 5.8392 days
e. None of the above.
Can anyone help?

large sample confidence intervals are used to find a region in which we are 100 (1-α)% confident the true value of the parameter is in the interval.

For large sample confidence intervals about the mean you have:

xBar ± z * sx / sqrt(n)

where xBar is the sample mean
z is the zscore for having α% of the data in the tails, i.e., P( Z > |z|) = α
sx is the sample standard deviation
n is the sample size

so here you have:

12.60 ± 1.96 * 2.50 / sqrt(49)

= (11.9, 13.3)

---------------

P( |Z| < 2.053749) = 0.96

5.3 ± 2.053749 * 1.8 / sqrt(64)

= (4.837906, 5.762094)