Question

Here's data on randomly sampled seat belt users in New York City where seat belts are required and in Boston where they are not required.

	Drivers	Wearing Seat Belts
NYC	220	183
Boston	117	68

Let

p1 = proportion of NYC drivers wearing seat belts

p2 = proportion of Boston drivers wearing seat belts

Find

Answer 1

We have to check there is significance between New York City drivers wearing seat belts and in Boston drivers wearing seat belts.

Set hypothesis:

H0: There is no significance difference between NYC drivers wearing seat belts and Boston drivers wearing seat belts (i.e p1 = p2)

Ha: There is significance difference between NYC drivers wearing seat belts and Boston drivers wearing seat belts (i.e p1 ≠ p2)

We are using two sample proportion test,

Test statistic:

Where p1 = proportion of NYC drivers wearing seat belts = 183 / 220 = 0.8318

P2 = proportion of Boston drivers wearing seat belts = 68 / 117 = 0.5812

Calculate SE,

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

So,

p = (p1 * n1 + p2 * n2) / (n1 + n2)

= [(0.8318*220 + 0.5812*117) / (220 + 117)]

p = 0.744807

Therefore,

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }

= sqrt{0.744807*(1-0.744807)*[(1/200) + (1/117)]

SE = 0.049987

Calculate Test statistic,

P-value:

Since we have two-tailed test, the P-value is the probability that the z-score is less than -5.0240 and Greater than 5.0240

Find, P(z<-5.0240) + P(z>5.0240) = <0.0001.

Note: This p-value we have to calculate by using google link as below.

https://www.socscistatistics.com/pvalues/normaldistribution.aspx.

Decision: P-value is less than 0.05 then reject H0.

i.e p1 ≠ p2

Conclusion:

There is significance difference between NYC drivers wearing seat belts and Boston drivers wearing seat belts