Question

Randomly selected birth records were obtained and results arelisted below. Use a 0.01 significance level to test the reasonableclaim that births occur with equal frequency on the different daysof the week. How might the apparent lower frequency on Saturday andSunday be explained?

Day Sun Mon Tues Wed Thurs Fri Sat

Births 43 63 59 58 56 57 45

Answer 1

Solution:-

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

Null hypothesis: Births occur with equal frequency on the different days of the week.

Alternative hypothesis: At least one of the proportions in the null hypothesis is false.

Formulate an analysis plan. For this analysis, the significance level is 0.01. 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 = 7 - 1
D.F = 6
(E_i) = n * p_i


X² = 6.17

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 6 degrees of freedom is more extreme than 6.17.

We use the Chi-Square Distribution Calculator to find P(X² > 6.17) = 0.404.

Interpret results. Since the P-value (0.404) is greater than the significance level (0.01), we have to accept the null hypothesis.