Question

Suppose an advertising agency wishes to know the average number of text messages a typical student sends in one month. They take a simple random sample of 25 students and have them give the number of sent texts counted in the previous month’s bill. The sample has a mean of x¯ = 1800 texts and standard deviation of s = 500 texts.

(a) What is the sample distribution, normal or binomial?

(b) How many degrees of freedom are there?

(c) Estimate the mean number of texts a typical student sends using a 95% confidence interval.

Answer 1

a)

As the sample is random

So sampling distribution is normal with mean = 1800, s.d = 500/√25 = 100

b)

df =n-1 = 24

c)

(Xbar +-t * sd/sqrt(n))

CI
xbar	1800.00
sd	500.000
n	25
level of confidence	0.95
alpha	0.05
MOE	206.390
lower	1593.61
upper	2006.39

formulas

CI
xbar	1800
sd	500
n	25
level of confidence	0.95
alpha	=1-B6
MOE	=T.INV.2T(B7,B4-1)*B3/SQRT(B4)
lower	=B2-I12*B3/SQRT(B4)
upper	=B2+I12*B3/SQRT(B4)