A survey of 1000 young adults nationwide found that among adults aged 18-29, 35% of women and 30% of men had tattoos. Suppose that these percentages are based on random samples of 500 women and 500 men. Using a 2.5% level of significance, can you conclude that among adults aged 18-29, the proportion of all women who have tattoos exceeds the proportion of all men who have tattoos?

95% confidence interval for adults in age group 18-29 is (-0.007, 0.107)

Step-by-step explanation:

Given  

N1 = n2 = 500

% of males = 0.30

% of females = 0.35

95% confidence interval  

(p2-p1) + z(0.025) sqrt (p1q1/n1 + p2q2/n2)  and (p2-p1) - z(0.025) sqrt (p1q1/n1 + p2q2/n2)  

Substituting the given values, we get –  

(0.35-0.30) + 1.96 sqrt (0.3*0.7/500 + 0.35*0.65/500) and (0.35-0.30) - 1.96 sqrt (0.3*0.7/500 + 0.35*0.65/500)

0.05 + (1.96 *0.029) and 0.05 - (1.96 *0.029)

0.05 + 0.057 and 0.05 – 0.057  

(-0.007, 0.107)

where is the rest of the

step-by-step explanation: