Answer to Question #260943 in Statistics and Probability for bri

Question

In 2015, the average duration of long-distance telephone calls from a certain town was 9.4 minutes. A telephone company wants to perform a test to determine whether this average duration of long-distance calls has changed. Twenty calls, originating from the town, was randomly selected and the mean duration was 10.2 minutes with standard deviation 4.8 minutes. a) Using a 1% level of significance, Complete the following: i. Give the null and alternative hypothesis of this test. [2] ii. Determine the critical value(s) of this test. [2] iii. Compute the value of the test statistic. [2] iv. State the decision rule. [1] v. Give your decision based on the available sample evidence. [1] vi. Hence, state your conclusion. [2] b) Construct the 99% confidence interval for the population mean duration of the longdistance calls from the town. [4]

Answer 1

X- duration of telephon call now

H₀: M(X)=9.4 - null hypothesis

H₁:M(X)$\ne9.4$ - alrernative hypothesis

We are usint t-test

Critical value $t_0: P(|T|>t_0)=\alpha \backsim P(T>t_0)=\frac{\alpha}{2}$

where T- is t-statics

$t_0=qt(\frac{0.01}{2},12-1)=qt(0.995,11)=3.106$ -calculation in mathcad

$T=\frac{\overline x-9.4}{\sigma(x) }=\frac{10.2-9.4}{4.8}=\frac{0.8}{4.8}=0.167$

Conclusion: T<$t_{0.995,11}$ therefore null hypothesi M(X)=9.4 is true

Statement about shanging average duration of telephone calls has no 99% confidence.

Let us consider condition $|\frac{\overline x-X}{\sigma(x) }|<t_{0.995,11}$ of valiidity of H₀

From it we have $X \in (10.2-t_{0.995,11}\cdot \sigma(x),10.2+t_{0.995,11}\cdot \sigma(x))=\\(10.2-3.106\cdot 4.8.10.2+3.106\cdot 4.8)=(-4.407,25.109)=\\ (0,25.109 \space min)$

Answer 99% confidence interval for average duration of telephone call is (0.25.109).