a. t distribution
b. 98% confidence the population mean number of texts per day is between and texts is 31. 455 and 36.945
c. Two percent will not contain the true population mean number of texts per day.
Confidence interval
Given that,
xbar = 34.2
s = 12.7
n = 119
degrees of freedom, df = n - 1 = 118
a. t distribution
b. CI level is 98%, hence α = 1 - 0.98 = 0.02
α/2 = 0.02/2
= 0.01,
tc = t (α/2, df) = 2.358
CI = [tex](34.2 - 2.358 * \frac{12.7}{\sqrt{119} } , 34.2 + 2.358 *\frac{12.7}{\sqrt{119} } )[/tex]
CI = (31.455 , 36.945)
With 98% confidence the population mean number of texts per day is between 31.455 and 36.945 texts.
(c) Approximately 98 percent of these confidence intervals will contain the true population mean number of texts per day, while approximately 2 percent will not.
To learn more about confidence interval check the given link
https://brainly.com/question/27630001
#SPJ4