Question

# A researcher claims that 19% of US adults are smokers. In a random sample of 250...

A researcher claims that 19% of US adults are smokers. In a random sample of 250 US adults 55 were found to be smokers. Use a 0.05 significance level to test the researcher’s claim.

please show step by step details

#### Homework Answers

Answer #1

To test :

H0 : The percentage of US adult smoker is 0.19 , i.e , p = 0.19

H1 :  The percentage of US adult smoker is 0.19 , i.e , p 0.19

Here , n = 250 and alpha = 0.05

The sample proportion , = 55 / 250

= 0.22

The test statistic , z = ( - p) / {p(1-p)/n}

= 0.03 / 0.0248

= 1.21

We know , at alpha = 0.05 , z(critical) = 1.96

Therefore, z < z(critical) and we know that , we fail to reject the null hypothesis when z is less than the critical value.

Therefore , we fail to reject the null hypothesis here and conclude that 19% of US adults are smokers.

Know the answer?
Your Answer:

#### Post as a guest

Your Name:

What's your source?

#### Earn Coins

Coins can be redeemed for fabulous gifts.

##### Not the answer you're looking for?
Ask your own homework help question
ADVERTISEMENT
##### Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

ADVERTISEMENT