Problem
4 Chapter 15 chl-squar goodnese of fi (20 points) The following frequeney iable shaws the observod number of people whe teded positive for Covid 19 during the wek of Wrwenher \( 17^{\text {ti }} \) io November \( 19^{\mathrm{k}} 2022 \), in sิan Diego county. 19 Daily Statur Uodate. pdf Assume the weekly sample of 2912 of people infected with Covid 19 in the ecunty of than Diego to be representative of the 5 an Diego population. \begin{tabular}{|l|c|c|c|c|c|c|c|} \hline \( \begin{array}{l}\text { DAY OF } \\ \text { THE } \\ \text { WEEK }\end{array} \) & \( \begin{array}{l}\text { Sunday } \\ (11 / 13 / 22)\end{array} \) & \( \begin{array}{l}\text { Monday } \\ (11 / 14 / 22)\end{array} \) & \( \begin{array}{l}\text { Tuesday } \\ (11 / 15 / 22)\end{array} \) & \( \begin{array}{l}\text { Wednesday } \\ (11 / 16 / 22)\end{array} \) & \( \begin{array}{l}\text { Thursday } \\ (11 / 17 / 22)\end{array} \) & \( \begin{array}{l}\text { Friday } \\ (11 / 1 / 2 / 22)\end{array} \) & \( \begin{array}{l}\text { Saturday } \\ (11 / 19 / 22)\end{array} \) \\ \hline \( \begin{array}{l}\text { NUMBER } \\ \text { OF COVID } \\ \text { CASES }\end{array} \) & 232 & 332 & 395 & 639 & 533 & 450 & 331 \\ \hline \end{tabular} Carry out a chi-gquare goodness of fit to determine if there is convineing evidenee that the proportions of number of Covid-19 cases in San Dicgo county falling into each of the fwe categories are not all the same. Use significance level of \( a-0.05 \) a. Define the proportions pil to p7 b. Hypotheses
Solution
Step 1 ：Step 1: Define proportions p1 to p7: \(p_1, p_2, p_3, p_4, p_5, p_6, p_7\)
Step 2 ：Step 2: State hypotheses: \(H_0: p_1=p_2=p_3=p_4=p_5=p_6=p_7\), \(H_1: \) At least one proportion differs
Step 3 ：Step 3: Calculate expected frequencies: \(E_i = \frac{n}{7} \): \(\frac{2912}{7} \approx 416\)
Step 4 ：Step 4: Compute the chi-square statistic: \(\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i} \)
Step 5 ：Step 5: Find the critical value for \(\alpha = 0.05 \) and degrees of freedom: df = 6, critical value = 12.59
Step 6 ：Step 6: Compare the chi-square statistic to the critical value and make a conclusion
