Question

Exercises 1-5 refer to the sample data in the following table, which summarizes the last digits of the height cm of 300 randomly selected subjects (from data set 1 " body data"). 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.

1-what are the null and alternative hypotheses corresponding to the stated claim?

2-when testing the claim in Exercise 1, what are the observed and expected frequencies for the last digit of 7?

3-is the hypothesis test left-tailed, right-taled, or two-tailed?

4-If using a 0.05 significance level to test the stated claim, find the number of degrees of freedom.

5-Given that the p-value for the hypothesis test is 0.501, what do you conclude? Does it appear that the height were obtained through measurement or that the subjects reported their heights?

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

Answer 1

1- To est the property of equally likely of the data means all elements have equal chance or in other words the occurance of one will not affect the occurance or non occurance of the other.

It means we need to prove that the observations are independance

To Test the independance we can use the Chi square test will

Null hypothesis H0: There are are no relation between the categorical variables

and the ALternative Hypothesis will be

H1: There is relation between the categorical variables.

2- Since this is a independant discrete data and probability of occurance of each value (p) is 1/10 and non occurance is 9/10

P(X) for each x= 0.1 and Expected frequency is 300X.01 = 30

For the last digit 7 the expected frequency is 30 and observed frequency (given) is 27

3. For an independance test of Chi square if the Critical value is above the calculated value we reject the Null Hypothesis and other wise we accept the null hypothesis > we sill be looking at only one side as whether the Chi square value exceeeds the critical threshold or not hence it will be a one tailed test

4. The degree of freedom of a chi square test is df= (c-1)(r-1) where c is the number of columns and r is the number of rows and n-1 for a single column data  here r=10 and n-1 = 10-1 =9

5. In a Chi square distribution if the p value is greater than the Significance level means there is a strong evidence that we should reject the null hypothesis , here the null hypothsis is that the observations are independent here P=0.501 and Significance level = .05 so we reject the null hypothesis and conclude that there is relation between the oservation or the value of one dpendent on the value of the other > Also the P value is very high here means the relation is very strong .and which indicates that the date obtained maynot from actual measurement.