Answer to Question #158854 in Statistics and Probability for anonymous

Question

Geo-net, a cellular phone company, has collected the following frequency distribution for the length of calls outside its normal customer roaming area:

Length (min.) Frequency

0 65 26

5 6 10 75

10 6 15 139

15 6 20 105

20 6 25 37

25+ 18

400

The sample mean (x–) for this distribution is 14.3 minutes, and the sample standard deviation is

3.7 minutes. Determine whether these data are normally distributed 1a = .052.

Answer 1

H₀: Call lengths outside normal customer roaming areas follow N(μ, σ²)

H₁: Call lengths outside normal customer roaming areas do not follow N(μ, σ²)

The Chi-Square goodness of fit test shall apply for given data, with degrees of freedom equal to one minus the number of categories in which the data is divided. Hence,

df = 6 - 1 = 5

The critical value at significance level of 0.05 from the table is

$χ^2(5, 0.05) = 11.07$

Now let us analyze the given data. We are interested in evaluating the test statistic

$χ^2 = \sum \frac{(O_i-E_i)^2}{E_i}$

Now we need to evaluate the expected values from given data. As number of observations, N=400 so the given mean and standard deviation values can be assumed as the population statistics. Hence, the exected values are calculated using the Z table, by evaluating the proportion of total area under normal distribution in each of the frequency distribution intervals. This tabulation can be done as per below, to obtain the test statistic value, 666.4

In order to calculate the probabilities based on range of Z statistic, the following calculator was helpful

http://davidmlane.com/hyperstat/z_table.html

The test statistic is way off the critical value - 666.4 vs 11.07. So we conclude that given distribution is anything but Normal.