Question

Suppose that Adam is a candidate for city council in a large metropolitan area. While campaigning, he claimed that 65% of adult city residents meet or exceed the recommendations of the Center for Disease Control and Prevention (CDC) for physical activity cach week. Jim, a reporter for a local fact-checking website, suspects that Adam pulls numbers out of thin air without doing any research. To test Adam's assertion, Jim surveyed 355 randomly selected adult city residents and found that 214 of them meet or exceed the CDC's recommendations for weekly physical activity, Jim's sample statistics are summarized in the table. The standard error and z-statistic were computed using the null hypothesis Hip = 0.65 where p is the proportion of all adults in the city who meet or exceed the CDC's recommendations for physical activity each week Sample size Sample count Standard z-Statistic error Sample proportion P 0.6028 11 x SE 355 214 0.0253 -1.8638 Do the data summarized in the table provide evidence that Adam's claim was incorrect? To help answer this question, use

Answer 1

To Test :-

H0 : p = 0.65

H1 : p $\neq$ 0.65

Given

n = 355    ;      x = 214

Estimate of sample proprotion will be

= x/n = 214/355 = 0.6028169

P

= 0.6028169

P

Standard Error will be given by

SE = $\sqrt{p(1-p/n)}$

     = $\sqrt{0.65(1-0.65)/355}$    = 0.0253

SE = 0.0253

Test Statistics z :-

z =
P-P SE
= $\frac{0.6028-0.65}{ 0.0253}$ = -1.8638

z = -1.8638

Hence test statistics value i.e z-statistics is -1.8638

To find corresponding p-value

Since alternative hypothesis is of " $\neq$ " type , so this is two tail test , and hence P-value will be given by

P-Value = Pr ( Z < -| z-statistics | ) + Pr ( Z > | z-statistics | )

             = Pr ( Z < -| -1.8638 | ) + Pr ( Z > | -1.8638 | )

P-Value = Pr ( Z < - 1.8638 ) + Pr ( Z > 1.8638 )

Here Z ~ N(0,1) , hence this is symmetric distribution , thus we get

P-Value = Pr ( Z < - 1.8638 ) + Pr ( Z > 1.8638 )   = 2 * Pr ( Z < - 1.8638 )

So P-Value will be given by

P-Value = 2 * Pr ( Z < - 1.8638 )

Now Pr ( Z < - 1.8638 ) can be obtained from statistical book or more accuratley from any software like R/Excel

From R

> 2*pnorm(-1.8638,m=0,s=1)             # 2 * Pr ( Z < - 1.8638 )
[1] 0.0623498

P-Value = 2 * Pr ( Z < - 1.8638 ) = 0.0623498

hence , P-Value is

P-Value : 0.0623                             { Rounding to four decimals }