Question

Exercises 1–5 refer to the sample data in the following table, which summarizes the last digits of the heights (cm) of 300 randomly selected subjects (from Data Set 1 “Body Data” in Appendix B). Assume that we want to use a 0.05 significance level to test the claim that the data are from a population having the property that the last digits are all equally likely.

Last Digit	0	1	2	3	4	5	6	7	8	9
Frequency	30	35	24	25	35	36	37	27	27	24

What are the null and alternative hypotheses corresponding to the stated claim?

Short answer

The null hypothesis and the alternative hypothesis for the given problem is as follows:

\(\begin{aligned}{l}{H_0}:{p_0} = {p_1} = {p_2} = ... = {p_9}\\{H_1}:{\rm{Atleast}}\;{\rm{one}}\;{\rm{of}}\;{\rm{the}}\;{\rm{probabilities}}\;{\rm{is}}\;{\rm{diiferent}}{\rm{.}}\end{aligned}\)

Step-by-step solution

Given information

The last digits of the heights of a sample of people are tabulated along with their respective frequencies.

Hypotheses

The claim is to test that the sample is chosen from the population with the property that the has last digits of the heights of people are equally likely to occur.

Let\({p_0},{p_1},{p_2},...,{p_9}\)be the probabilities of the last digitof the heights of a sample of people.

The null hypothesis is written as follows:

The probabilities of the last digits of the heights of people are likely to occur equally.

The alternative hypothesis is written as follows:

The probabilities of the last digits of the heights of people are not likely to occur equally.

In terms of notations, the null and alternative hypothesis is:

\(\begin{aligned}{l}{H_0}:{p_0} = {p_1} = {p_2} = ... = {p_9}\\{H_1}:{\rm{Atleast}}\;{\rm{one}}\;{\rm{of}}\;{\rm{the}}\;{\rm{probabilities}}\;{\rm{is}}\;{\rm{diiferent}}{\rm{.}}\end{aligned}\)