Please answer using SAS. Please show all steps used to reach answer A researcher claims that the distribution of favorite pizza toppings amongteenagers is as shown below. 200 teenagers are randomly selected and asked what their favoritepizza topping is. The results are shown below. conduct the corresponding hypothesis testing byusing SASTopping Cheese Pepperoni Sausage Mushrooms OnionsFrequency 41% 25% 15%10%9%200 teenagers are randomly selected and asked what their favorite pizza topping is. The resultsare shown below.ToppingCheese Pepperoni Sausage Mushrooms OnionsResults (n = 200) 7852302515Conduct the corresponding hypothesis testing by using SAS. Please include your SAS code andoutput.(a) Set up the appropriate Ho and Ha.(b) Report the test statistic from SAS(c) Report the P-value and make your conclusion at a = .05.

Given information: Topping Cheese Pepperoni Sausage Mushrooms Onions Results(n = 200) 78 52 30 25 15Null hypothesis: H0: The distribution of favourite pizza toppings among teenagers follows the given distribution. Alternate hypothesis: HA: The distribution of favourite pizza toppings among teenagers does not follows the given distribution. For this analysis, the significance level is 0.05. We will conduct a chi-square goodness of fit test of the null hypothesis. Use the given data to form excel table consisting of observed frequency, expected probability, expected frequency and chi square statistics. Results Topping P(x) Expected Or,c-Er,c2Er,c Cheese 78 0.41 82 0.195121951 Pepperoni 52 0.25 50 0.08 Sausage 30 0.15 30 0 Mushrooms 25 0.1 20 1.25 Onions 15 0.09 18 0.5 Sum 200 1 200 2.025121951 Formula for the expected frequency is: E(x) = n*P(x) DF = k - 1 D.F = 5 - 1 D.F = 4Use the table to determine the chi-square statistic, degree of freedom and P-value. χ2=∑(Or,c-Er,c)2Er,c =2.025121951≈2.03 Use the excel formula, "=CHISQ.DIST.RT(2.03,4)" for determining the P-value from the chi-square statistic having degrees of freedom 4. P(χ2>2.03)=0.7302 Since the P-value (0.7302) is greater than the significance level (0.05), we failed to reject the null hypothesis. We failed to reject H0. There is convincing evidence to conclude that distribution of favourite pizza toppings among teenagers follows the given distribution.SAS code: data race; infile "H:\sas\data" dlm=; input topping $; run; Then GOF test can be requested with a "freq" procedure. proc freq data = race; tables race / chisq test p=(41 24 15 10 9); run; SAS output: