A particular city has an Asian population of 1419 people, out of a total population of 23,609. Conduct a goodness of fit test at the 5% level to determine if the self-reported sub-groups of Asians are evenly distributed. Round expected frequency to two decimal places.
Race | Frequency | Expected Frequency |
---|---|---|
Asian Indian | 134 | |
Chinese | 123 | |
Filipino | 1033 | |
Japanese | 83 | |
Korean | 13 | |
Vietnamese | 11 | |
Other | 22 |
Part (a)
State the null hypothesis.The self-reported sub-groups of Asians are not evenly distributed.The self-reported sub-groups of Asians are evenly distributed.
Part (b)
State the alternative hypothesis.The self-reported sub-groups of Asians are evenly distributed.The self-reported sub-groups of Asians are not evenly distributed.
Part (c)
What are the degrees of freedom?
Part (d)
State the distribution to use for the test.t6
t7
Part (e)
What is the test statistic? (Round your answer to two decimal places.)
Part (f)
What is the p-value? (Round your answer to four decimal places.)H0
is false, then there is a chance equal to the p-value that the value of the test statistic will be equal to or greater than the calculated value.IfH0
is true, then there is a chance equal to the p-value that the value of the test statistic will be equal to or less than the calculated value. IfH0
is true, then there is a chance equal to the p-value that the value of the test statistic will be equal to or greater than the calculated value.IfH0
is false, then there is a chance equal to the p-value that the value of the test statistic will be equal to or less than the calculated value.Part (g)
Sketch a picture of this situation. Label and scale the horizontal axis, and shade the region(s) corresponding to the p-value. (Upload your file below.)
Part (h)
Indicate the correct decision ("reject" or "do not reject" the null hypothesis) and write the appropriate conclusion.(i) Alpha:reject the null hypothesisdo not reject the null hypothesis
Since α > p-value, we reject the null hypothesis.Since α > p-value, we do not reject the null hypothesis. Since α < p-value, we reject the null hypothesis.Since α < p-value, we do not reject the null hypothesis.
There is sufficient evidence to conclude that the self-reported sub-groups of Asians are not evenly distributed.There is not sufficient evidence to conclude that the self-reported sub-groups of Asians are not evenly distributed.
Given that:
Ho : Asians = not evenly distributed
Ha : Asians != not evenly distributed
df = 6 | N = 1419 | ||||
Race | Frequency(O) | Expected Frequency(E) | O-E | (O-E)^2 | (O-E)^2/E |
Asian Indian | 134 | 202.7142857 | -68.7143 | 4721.653061 | 23.29215745 |
Chinese | 123 | 202.7142857 | -79.7143 | 6354.367347 | 31.34642102 |
Filipino | 1033 | 202.7142857 | 830.2857 | 689374.3673 | 3400.719219 |
Japanese | 83 | 202.7142857 | -119.714 | 14331.5102 | 70.69807712 |
Korean | 13 | 202.7142857 | -189.714 | 35991.5102 | 177.5479714 |
Vietnamese | 11 | 202.7142857 | -191.714 | 36754.36735 | 181.3111849 |
Other | 22 | 202.7142857 | -180.714 | 32657.65306 | 161.1018826 |
Test Stastics | 4046.016913 | ||||
p value | 0 |
P value is < 0.0001 |
A particular city has an Asian population of 1419 people, out of a total population of...
According to an independent insurance agent, the following is a breakdown of the amount of life insurance purchased by males in the following age groups. The agent is interested in whether the age of the male and the amount of life insurance purchased are independent events. Conduct a test for independence at the 5% level. Age of Males None $50,000 – $100,000 $100,001 – $150,000 $150,001 – $200,000 $200,000+ 20 – 29 45 15 40 0 5 30 – 39...
The marital status distribution of the U.S. male population, age 15 and older, is as shown below. Marital Status Percent never married 31.3 married 56.1 widowed 2.5 divorced/separated 10.1 Suppose that a random sample of 400 U.S. young adult males, 18 to 24 years old, yielded the following frequency distribution. We are interested in whether this age group of males fits the distribution of the U.S. adult population at the 5% level. Calculate the frequency one would expect when surveying...
Car manufacturers are interested in whether there is a relationship between the size of car an individual drives and the number of people in the driver's family (that is, whether car size and family size are independent). To test this, suppose that 798 car owners were randomly surveyed with the following results. Conduct a test for independence at the 5% level. Family Size Sub & Compact Mid-size Full-size Van & Truck 1 20 35 41 34 2 21 50 69...
According to an article in Newsweek, the natural ratio of girls to boys is 100:105. In China, the birth ratio is 100:114 (46.7% girls). Suppose you don't believe the reported figures of the percent of girls born in China. You conduct a study. In this study, you count the number of girls and boys born in 150 randomly chosen recent births. There are 63 girls and 87 boys born of the 150. Based on your study, do you believe that...
The age distribution of the Canadian population and the age distribution of a random sample of 455 residents in the Indian community of a village are shown below. Age (years) Percent of Canadian Population Observed Number in the Village Under 5 7.2% 43 5 to 14 13.6% 73 15 to 64 67.1% 297 65 and older 12.1% 42 Use a 5% level of significance to test the claim that the age distribution of the general Canadian population fits the age...
The age distribution of the Canadian population and the age distribution of a random sample of 455 residents in the Indian community of a village are shown below. Age (years) Percent of Canadian Population Observed Number in the Village Under 5 7.2% 43 5 to 14 13.6% 73 15 to 64 67.1% 297 65 and older 12.1% 42 Use a 5% level of significance to test the claim that the age distribution of the general Canadian population fits the age...
The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Type of Household Percent of U.S. Households Observed Number of Households in the Community Married with children 26% 101 Married, no children 29% 118 Single parent 9% 28 One person 25% 97 Other (e.g., roommates, siblings) 11% 67 Use a 5% level of significance to test the claim that the distribution of U.S. households fits the...
The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Type of Household Percent of U.S. Households Observed Number of Households in the Community Married with children 26% 100 Married, no children 29% 118 Single parent 9% 30 One person 25% 93 Other (e.g., roommates, siblings) 11% 70 Use a 5% level of significance to test the claim that the distribution of U.S. households fits the...
while her husband spent 2 ½ hours picking out new speakers, a statistician decided to determine whether the percent of men who enjoy shopping for electronic equipment is higher than the percent of women who enjoy shopping for electronic equipment. The population was Saturday afternoon shoppers. Out of 69 men, 25 said they enjoyed the activity. 7 of the 22 women surveyed claimed to enjoy the activity. Interpret the results of the survey. Conduct a hypothesis test at the 5%...
a.) Are the conditions for running this test stratified? - Yes, all observed counts are bigger than 5. - Yes, the mean number of counts is bigger than 10. - Yes, the total number of rolls is bigger than 10. - Yes, all expeced counts are bigger than 10. b.) State the null hypothesis. - The data fit the distribution for a fair six-sided die. - The data do not fit the distribution for fair six-sided die. c.) What are...