I need help with this problem WS#In studying the responses to a multiple-choice test question, the following sample data were obtained. Atthe 0.10 significance level, test the claim that the responses occur with the same frequency.Ho: The responses to the questions occur with the same frequencyHa: The responses to the questions do not occur with the same frequency3Enter the expected frequency rounded to one decimal and the last column rounded to 4 decimals.Observed Expected (0 – E)²Frequency FrequencyEeResponsedABCDE8=80 с$100064592215What is the chi-square test-statistic for this data? Round to 4 decimals.x²What is the p-value? Round to 4 decimals.rL%tgOllуLh&7u* CO8i(9O0р

The chi-square test-statistic is calculated by using the formula : χ2 = Σ[(O–E)2 /E] Where, O = Observed frequency, E = Expected frequency. Results based on p-value : If the p-value is less than or equal to the level of significance (α) then the null hypothesis (H0) is rejected i.e., If p-value ≤ α ⇒ reject the null hypothesis. If the p-value is greater than the level of significance (α) then we fail to reject the null hypothesis (H0 ). If p-value > α ⇒ fail to reject the null hypothesis. According to the question, null and alternative hypotheses are given by, H0 : Responses to the questions occur with same frequency, Ha : Responses to the questions do not occur with same frequency, Expected frequency is calculated using the formula : E = ΣO/n Where, ΣO = Sum of observed frequencies, n = number of groups. From the table provided in the question we have, n = 5. Expected frequency = (8 + 8 + 9 + 22 + 15)/5 Expected frequency = 62/5 Expected frequency = 12.4Using the data provided calculation table is given by, Responses Observed frequency (O) Expected frequency (E) (O–E) (O–E)2 (O–E)2/E A 8 12.4 –4.4 19.36 1.5613 B 8 12.4 –4.4 19.36 1.5613 C 9 12.4 –3.4 11.56 0.9323 D 22 12.4 9.6 92.16 7.4323 E 15 12.4 2.6 6.76 0.5452 Σ(O–E)2/E = 12.0324 (All the values are rounded off as mentioned in the question) ANSWER : The compete table is as follows : Responses Observed frequency Expected frequency (O–E)2 /E A 8 12.4 1.5613 B 8 12.4 1.5613 C 9 12.4 0.9323 D 22 12.4 7.4323 E 15 12.4 0.5452Also, from the last cell of the last column of the calculation table given in step 3 the chi-square test-statistic is given by, χ2 = 12.0324 p-value : For a chi-square distribution it is not possible to find exact p-value using distribution table. Hence, using the Excel to get the exact p-value. Using Excel function :CHISQ.DIST.RT(x, deg_freedom) Where, x = test-statistic, and deg_freedom = degree of freedom Degree of freedom for chi-square is given by, Degree of freedom = n–1 Degree of freedom = 5–1 = 4 Using Excel, p-value =0.017112 Rounding off to 4 decimal places we have, p-value = 0.0171 We know that, level of significance is given α = 0.10 Since, p-value < α i.e., 0.0171 < 0.10. Hence, the null hypothesis is rejected. ANSWER : χ2 = 12.0324 p-value = 0.0171