News reports indicate that nationally, 45% of Americans are planning"staycations" at home this year instead of a vacation trip. Suppose we take arandom sample of 200 people in our city and find that 113 of them are planning"staycations."Let p = the proportion in our city who are planning staycations this year. A 98%confidence interval for P isa) (0.496, 0.634)b) (0.483, 0.646)c) No confidence interval is necessary, since we know p = 45%.d) (0.480, 0.648)