Question

6. Suppose that the sample unemployment rate for those aged 25–60 is 8% based on a...

6. Suppose that the sample unemployment rate for those aged 25–60 is 8% based on a survey of 150 people, while the unemployment rate for those aged 16–24 is 12% based on a survey of 100 people.

(a) Form a 90% confidence interval for the difference between the two population unem- ployment rates.

(b) Construct a test statistic to test the null hypothesis that the two population rates are the same against the alternative hypothesis that the rate is higher in for those aged 16–24, and report the associated P-value.

Homework Answers

Answer #1

= 0.08, n1= 150

= 0.12, n2= 100

a)

c= 90%

formula for confidence interval is

Where Zc is the z critical value for c= 90%

Zc= 1.645

−0.105 < P < 0.025

(−0.105 , 0.025)

Thus we get 90% confidence interval as (−0.105 , 0.025)

b)

Ho: P1 = P2

Ho: P1 <  P2

fformula is

z=-1.05

Test statistics = -1.05

P-Value = P(z < -1.05)

using z table we get

P(z < -1.05) = 0.1465

P-Value =0.1465

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
Similar Questions
Suppose the national unemployment rate is 3.5%. In a survey of 425 people in a rural...
Suppose the national unemployment rate is 3.5%. In a survey of 425 people in a rural Wisconsin county, 22 people are found to be unemployed. County officials apply for state aid based on the claim that the local unemployment rate is higher than the national average. Test this claim at the 0.05 significance level.
The dropout rate in certain school district was 11.9% in 2015. A researcher believes that rate...
The dropout rate in certain school district was 11.9% in 2015. A researcher believes that rate has changed since then. They survey 304 people between the ages of 16 and 24 and found that 35 are high-school dropouts. Test the researcher's claim at the 5% level of significance. State hypothesis Determine level of significance Calculate test statistic and associated p value make a decision state conclusion
we are going to examine one of the main findings of the Framingham study: an association...
we are going to examine one of the main findings of the Framingham study: an association between serum cholesterol (i.e., how much cholesterol is in someone's blood) and whether or not that person develops heart disease. We'll use the following null and alternative hypotheses: Null Hypothesis: In the population, the distribution of cholesterol levels among those who get heart disease is the same as the distribution of cholesterol levels among those who do not. Alternative Hypothesis: The cholesterol levels of...
Suppose grade point averages are normally distributed. We are interested in whether smokers have different gpa’s...
Suppose grade point averages are normally distributed. We are interested in whether smokers have different gpa’s than non-smokers. We know that non-smokers have a mean gpa of 3.1. You take a sample of 16 smokers and find they have a mean gpa of 3.38 with standard deviation of .021. a. Suppose you wanted to test whether the gpas of the two groups different. • write down the null and alternative hypotheses • Construct the test statistic. What is the distribution...
Suppose grade point averages are normally distributed. We are interested in whether smokers have different gpa’s...
Suppose grade point averages are normally distributed. We are interested in whether smokers have different gpa’s than non-smokers. We know that non-smokers have a mean gpa of 3.1. You take a sample of 16 smokers and find they have a mean gpa of 3.38 with standard deviation of .021. a. Suppose you wanted to test whether the gpas of the two groups different. • write down the null and alternative hypotheses • Construct the test statistic. What is the distribution...
found that the fear of losing the good opinion of one’s family and peers kept people...
found that the fear of losing the good opinion of one’s family and peers kept people from driving home drunk. Let’s say we have two independent random samples of people: those who think that their peers would disapprove of them from driving drunk, and those who think that their peers would either not care or approve of their driving drunk. We ask each person in each group to self-report the number of times that he or she has driven drunk...
suppose a random sample of 25 is taken from a population that follows a normal distribution...
suppose a random sample of 25 is taken from a population that follows a normal distribution with unknown mean and a known variance of 144. Provide the null and alternative hypothesis necessary to determine if there is evidence that the mean of the population is greater than 100. Using the sample mean ybar as the test statistic and a rejection region of the form {ybar>k} find the value of k so that alpha=.15 Using the sample mean ybar as the...
A report included the following information on the heights (in.) for non-Hispanic white females. Age Sample...
A report included the following information on the heights (in.) for non-Hispanic white females. Age Sample Size Sample Mean Std. Error Mean 20–39 863 64.7 0.09 60 and older 933 63.2 0.11 (a) Calculate a confidence interval at confidence level approximately 95% for the difference between population mean height for the younger women and that for the older women. (Use μ20–39 − μ60 and older.) Interpret the interval. We cannot draw a conclusion from the given information. We are 95%...
Does where you live affect your insurance rate? The mean auto insurance rate for 29 drivers...
Does where you live affect your insurance rate? The mean auto insurance rate for 29 drivers in a large city is $1,839 per car with a standard deviation of $315 while the mean rate for 23 drivers in a rural area is $1,648 with a standard deviation of $239. At α = 0.05, can we conclude that drivers in the large city pay higher insurance rates than those in the rural area, assuming the population variances are equal? a. Calculate...
6. Personality researchers wondered if there were differences between "dog people" and "cat people" in how...
6. Personality researchers wondered if there were differences between "dog people" and "cat people" in how extraverted they were. To find out if there are any differences, researchers asked people to self-identify as cat or dog people and then complete a personality inventory. The results of the extraversion test (on a scale from 0-100) for the two types of person are presented below. "Dog People" Sample: M = 83, SS = 360, n = 10 "Cat People" Sample: M =...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT