Question

A random sample of 118 weights of teenagers is obtained and the last digit is recorded. The frequency of each digit is recorded in the following table. If people report their weight, they tend to round. Test the claim that the sample is from a population of weights in which the last digits did NOT occur with the same frequency. Last digit 0 1 2 3 4 5 6 7 8 9 Frequency 20 15 9 8 12 15 6 13 8 12. Repeat the above procedure deleting digits 0 and 5. Do you get different results? What do the results mean?

Answer 1

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: The sample is from a population of weights in which the last digits occur with the same frequency.

Alternative hypothesis: The sample is from a population of weights in which the last digits did NOT occur with the same frequency.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a chi-square goodness of fit test of the null hypothesis.

Analyze sample data. Applying the chi-square goodness of fit test to sample data, we compute the degrees of freedom, the expected frequency counts, and the chi-square test statistic. Based on the chi-square statistic and the degrees of freedom, we determine the P-value.

DF = k - 1 = 3 - 1
D.F =
(E_i) = n * p_i

Expected frequency (O,o-EdiErc Last digit Observed (Ex n x P(x)) 5.698305085 0.86779661 0.66440678 1.223728814 0.003389831 0.86779661 2.850847458 0.122033898 1.223728814 0.003389831 13.52542373 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.8 11.8 118 20 15 0 12 15 6 13 4 6 12 118 Sum


Er,c

X² = 13.525

where DF is the degrees of freedom, k is the number of levels of the categorical variable, n is the number of observations in the sample, E_i is the expected frequency count for level i, O_i is the observed frequency count for level i, and X² is the chi-square test statistic.

The P-value is the probability that a chi-square statistic having 9 degrees of freedom is more extreme than 13.525.

We use the Chi-Square Distribution Calculator to find P(X² > 13.525) = 0.14

Interpret results. Since the P-value (0.14) is greater than the significance level (0.05), we have to failed to reject the null hypothesis.

From the above test we do not sufficient evidence that the results are different.