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

Answer )

A)

As the sample is random

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

B)

Degrees pf freedom is = n-1 = 24

C)

As the population s.d is unknown we will use t distribution to estimate the interval

Critical value t from t table for 95% confidence level and 24 dof is 2.06

Error = t*s.d/√n = 2.06*500/√25 = 206.39

Confidence interval is given by

(Mean - error, mean + error)

[1593.61, 2006.39].

We are 95% confident that the population mean (μ) falls between 1593.61 and 2006.39.