please help

Q4: Pharmaceutical companies promote their prescription drugs using television advertisements. In a survey of 80 randomly selected viewers, 70 viewers indicated that they had not asked their physician about using the advertised drug. Develop a 95% confidence interval for the proportion of viewers who do discuss a drug seen on TV with their physician. ( 10 ppl did not. Ask) 10/80 calculate Lower case p , What is p? What statistical distribution (Standard Normal, or Student's t) should be used in this scenario? What is the value of z or t? What is the lower confidence limit? What is the upper confidence limit? Is it reasonable to conclude that 25% of viewers discussed an advertised drug with their physician? Why or why not