Answer to Question #218026 in Statistics and Probability for Ash

Question

The daily duration of telephone calls received by the enquiry department of a small industry for a randomly selected 11 days over a quarter is as follows:

160, 172, 121, 144, 100, 108, 175, 200, 105, 95, 102 

The manager of industry says that the population variance of the daily duration of calls over the quarter is 1500. The authority thinks that it is over estimated. Justify the claims of the manager at 5% level of significance.

Answer 1

$H_0: \sigma^2 = 1500 \\ H_1: \sigma^2 < 1500 \\ N= 11$

Chi-Square Test for the Variance

Test-statistic:

$T = (N-1)(\frac{s}{\sigma_0})^2 \\ \bar{x}= \frac{160+ 172+ ...+ 95+ 102}{11} = 134.72 \\ s = \sqrt{ \frac{1}{11-1}( (160-134.72)^2+(172-134.72)^2+...+(95-134.72)^2+(102-134.72)^2 ) } = 36.929 \\ \sigma_0 = \sqrt{1500}=38.729 \\ T = (11-1)(\frac{36.929}{38.729})^2 \\ T=9.09 \\ α=0.05$

Lower one-tailed test.

$df=N-1=10 \\ χ^2_{α,10} = 18.307$

Reject H₀ if $T< χ^2_{α,10}$

$T=9.09<χ^2_{α,10}=18.307$

Reject H₀.

There is enough evidence to conclude that that the population variance of the daily duration of calls over the quarter is less than 1500.