Obtain the required confidence interval for the difference between two population proportions. Assume that independent simple random samples have been selected from the two populations. A survey found that 37 of 74 randomly selected women and 47 of 74 randomly selected men follow a regular exercise program. Find a 95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program. Group of answer choices

The 95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program = ( -0.293, 0.023)

Step-by-step explanation:

The formula for confidence interval for the difference between the proportions is given as:

p1 - p2 ± z × √p1 (1 - p1)/n1 + p2(1 - p2)/n2

From the question

We have two groups.

Group 1 = For women

A survey found that 37 of 74 randomly selected women

p1 = x/n1

n1 = 74

x1 = 37

p1 = 37/74

p1 = 0.5

Group 2 = For Men

A survey found out that 47 of 74 randomly selected men follow a regular exercise program

p2 = x/n1

n2= 74

x2 = 47

p2 = 47/74

p2 = 0.6351351351 ≈ 0.635

z = z score for 95% Confidence Interval = 1.96

The confidence interval for the difference between the proportions is given as:

p1 - p2 ± z × √p1 (1 - p1)/n1 + p2(1 - p2)/n2

0.5 - 0.635 ± 1.96 × √0.5 (1 - 0.5)/74 + 0.635(1 - 0.635)/74

-0.135 ± 1.96 × √(0.5 × 0.5)/74 + (0.635× 0.365)/74

-0.135 ± 1.96 × 0.08068750209663572

-0.135 ± 0.158147504109406

Hence:

= -0.135 - 0.158147504109406

= -0.2931475041

Approximately = -0.293

= -0.135 + 0.158147504109406

= 0.0231475041

Approximately = 0.023

Therefore, the 95% confidence interval for the difference between the proportions of women and men who follow a regular exercise program = ( -0.293, 0.023)

-6 is the correct option from the above given options

hello,

b) 62.47% of the students are between 14 and 18 years old.

step-by-step explanation:

mean = 15 and standard deviation = 2

student between 14 to 18

formula used z-score

for age 14 , z score

using z-table , p(z> -0.5)=0.3085

for age 18 , z score

using z-table , p(z< 1.5)=0.9332

p(-0.5< z< 1.5)=0.9332-0.3085 = 0.6247

thus, 62.47% of the students are between 14 and 18 years old.

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