Mathematics, 15.04.2020 23:38 angelina0miles

National data in the 1960's showed that about 44% of the adult population had never smoked cigarettes. In 1995 a national health survey interviewed a random sample of 881 adults and found that 52% had never been smokers.

a) Create a 95% confidence interval for the proportion of adults (in 1995) who had never been smokers.
b) Does this provide evidence of a change in behavior among Americans? Using your confidence level, test an appropriate hypothesis and state your conclusion.

Answers: 2

Show answers

Answers

Answer from: shafferakr6

a) $0.52 - 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.487$

$0.52 + 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.553$

The 95% confidence interval would be given by (0.487;0.553)

b) After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Solution to the problem

Part a

In order to find the critical value we need to take in count that we are finding the interval for a proportion, so on this case we need to use the z distribution. Since our interval is at 95% of confidence, our significance level would be given by $\alpha=1-0.95=0.05$ and $\alpha/2 =0.025$ . And the critical value would be given by:

$z_{\alpha/2}=-1.96, z_{1-\alpha/2}=1.96$

The confidence interval for the mean is given by the following formula:

$\hat p \pm z_{\alpha/2}\sqrt{\frac{\hat p (1-\hat p)}{n}}$

If we replace the values obtained we got:

$0.52 - 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.487$

$0.52 + 1.96\sqrt{\frac{0.52(1-0.52)}{881}}=0.553$

The 95% confidence interval would be given by (0.487;0.553)

Part b

After see the confidence interval we see that the lower limits 0.487>0.44 with 0.44 or 44% the estimated proportion for 1960, we can conclude that we have a significant increase in the adult proportion had never smoked cigarettes at 5% of significance.

Answer from: Quest

You would need a laid out graph for this but could you provide a picture of just a graph? i can give stepssolve for y.subtract 4xdivide by -1 (opposites)graph slope intercept formcheck each point to see if it’s on the graph

Answer from: Quest

The answer is

Another question on Mathematics

Mathematics, 21.06.2019 17:00

How do you find the area of a triangle

Answers: 1

Answer

Mathematics, 21.06.2019 20:30

Linda loans her friend $250 with a 5% monthly intrest rate. how much was the whole loan?

Answers: 1

Answer

Mathematics, 21.06.2019 23:30

Aparking garage charges $1.12 for the first hour and 0.50 for each additional hour. what is the maximum length of time tom can park in the garage if we wishes to play no more than $5.62

Answers: 1

Answer

Mathematics, 22.06.2019 00:30

Find the solution on the interval [0, 2pi)

Answers: 1

Answer

You know the right answer?

National data in the 1960's showed that about 44% of the adult population had never smoked cigarette...

Questions