The following table shows age distribution and location of a random sample of 166 buffalo in a national park.
Age | Lamar District | Nez Perce District | Firehole District | Row Total |
Calf | 15 | 12 | 14 | 41 |
Yearling | 13 | 11 | 9 | 33 |
Adult | 31 | 28 | 33 | 92 |
Column Total | 59 | 51 | 56 | 166 |
Use a chi-square test to determine if age distribution and location are independent at the 0.05 level of significance.
(a) What is the level of significance?
State the null and alternate hypotheses.
H0: Age distribution and location are not
independent.
H1: Age distribution and location are not
independent.H0: Age distribution and location
are independent.
H1: Age distribution and location are not
independent. H0: Age
distribution and location are not independent.
H1: Age distribution and location are
independent.H0: Age distribution and location
are independent.
H1: Age distribution and location are
independent.
(b) Find the value of the chi-square statistic for the sample.
(Round the expected frequencies to at least three decimal places.
Round the test statistic to three decimal places.)
Are all the expected frequencies greater than 5?
YesNo
What sampling distribution will you use?
binomialchi-square uniformnormalStudent's t
What are the degrees of freedom?
(c) Find or estimate the P-value of the sample test
statistic. (Round your answer to three decimal places.)
p-value > 0.1000.050 < p-value < 0.100 0.025 < p-value < 0.0500.010 < p-value < 0.0250.005 < p-value < 0.010p-value < 0.005
(d) Based on your answers in parts (a) to (c), will you reject or
fail to reject the null hypothesis of independence?
Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis. Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.
(e) Interpret your conclusion in the context of the
application.
At the 5% level of significance, there is sufficient evidence to conclude that age distribution and location are not independent.At the 5% level of significance, there is insufficient evidence to conclude that age distribution and location are not independent.
In: Statistics and Probability
Suppose that 14 children, who were learning to ride two-wheel bikes, were surveyed to determine how long they had to use training wheels. It was revealed that they used them an average of six months with a sample standard deviation of three months. Assume that the underlying population distribution is normal.
1. What distribution should you use for this problem? Why?
2. Construct a 99\% confidence interval for the population mean
length of time using training wheels.
In: Statistics and Probability
The Bureau of Economic Analysis in the U.S. Department of Commerce reported that the mean annual income for a resident of North Carolina is $18,688 (USA Today, August 24, 1995). A researcher for the state of South Carolina wants to see if the mean annual income for a resident of South Carolina is different. A sample of 400 residents of South Carolina shows a sample mean annual income of $16,860 and the population standard deviation is assumed to known, =$14,624. Use a 0.05 level of significance, the researcher wants to test the following hypothesis.
H0: = 18,688 Ha: 18,688
a) Use confidence interval approach to test the hypothesis?
b)What is your conclusion?
C)What are three rejection rules (You have used confidence interval approach in Question 2)?
D)Do three rejection rules lead to the same conclusion? What is your conclusion?
In: Statistics and Probability
Problem 3-15 (Algorithmic) Telephone calls arrive at the rate of 48 per hour at the reservation desk for Regional Airways. (a) Find the probability of receiving 6 calls in a 5-minute interval. If required, round your answer to four decimal places. f(6) = (b) Find the probability of receiving 8 calls in 15 minutes. If required, round your answer to four decimal places. f(8) = (c) Suppose that no calls are currently on hold. If the agent takes 4 minutes to complete processing the current call, how many callers do you expect to be waiting by that time? If required, round your answer to one decimal place. Number of callers = What is the probability that no one will be waiting? If required, round your answer to four decimal places. The probability none will be waiting after 4 minutes is . (d) If no calls are currently being processed, what is the probability that the agent can take 2 minutes for personal time without being interrupted? If required, round your answer to four decimal places. The probability of no interruptions in 2 minutes is .
In: Statistics and Probability
A random sample of 26 seedless watermelons has a mean weight of 32.71 ounces and a standard deviation of 9.48 ounces. Assuming the weights are normally distributed, construct a 90% confidence interval estimate for the population mean weight of all seedless watermelons.
In: Statistics and Probability
A recent study from the Department of Education shows that approximately 11% of students are in private schools. A random sample of 450 students from a wide geographic area indicated that 55 attended private schools.
a) Estimate the true proportion of students attending private schools with 95% confidence.
b) Are the study results by the Department of Education consistent with the sample? Why or why not?
In: Statistics and Probability
You are given the sample mean and the population standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Interpret the results and compare the widths of the confidence intervals.
From a random sample of 48 business days, the mean closing price of a certain stock was $106.06. Assume the population standard deviation is $11.45.
The 90% confidence interval is ( [ ], [ ] ).
In: Statistics and Probability
. USA Today reported that about 47% of the general consumer population in the United States is loyal to the automobile manufacturer of their choice. Suppose that Chevrolet did a study of a random sample of 1006 Chevrolet owners and found that 490 stated that they would buy another Chevrolet. Test the claim by Chevrolet that the population proportion of consumers loyal to its product is different from 47%. Use = 0.05. (a) State the hypotheses for this test. (b) State the p-value for this test. (c) State your conclusion for the test in the context of the problem statement. (d) Give a 95% confidence interval for the proportion of customers loyal to Chevrolet. (e) Is your confidence interval consistent with your conclusion from the hypothesis test? Explain why or why not.
In: Statistics and Probability
. A researcher took a random sample of 600 college students and surveyed them about how many days of class they missed per semester. On average, the sample missed 3.1 days of class. If the sample standard deviation is .45, construct a confidence interval to estimate the true population mean at the 95% confidence level (.05 alpha level). Make sure to interpret your results in a sentence or two! (HINT: Make sure you use the correct formula for this question, keep in mind that you do not know the population standard deviation.)
In: Statistics and Probability
If a couple has six boys and six girls, how many gender
sequences are possible?
(Also, is there a way to do this with a T-83 plus calculator?)
In: Statistics and Probability
Consider the quarterly electricity production for years
1-4:
Year 1 2 3 4
Q1 99 120 139 160
Q2 88 108 127 148
Q3 93 111 131 150
Q4 111 130 152 170
(a) Estimate the trend using a centered moving average.
(b) Using a classical additive decomposition, calculate the
seasonal component.
(c) Explain how you handled the end points. - PLEASE
PROVIDE ANSWER C !
Note: Explain all the steps and computations
In: Statistics and Probability
Find a modified divisor that will give modified quotas to produce the desired number of seats. Use rounding down for the standard quotas.
10 seats
Population
1st Precinct |
135,000 |
2st Precinct |
231,000 |
3st Precinct |
118,000 |
4st Precinct |
316,000 |
TOTAL |
800,000 |
45,000 |
||
65,000 |
||
55,000 |
||
50,000 |
In: Statistics and Probability
A sample of 22 offshore oil-workers took part in a simulated
escape exercise. The sample yielded an average escape time of 388.7
min. and standard deviation of 22.4 min. The 95% confidence
interval for the true average of escape time is:
(379.34 , 398.06)
(378.768 , 398.632)
(380.482 , 396.918)
In: Statistics and Probability
explain how linear regression could be used in business decision making?
In: Statistics and Probability
When do we use “Tukey’s test for Additivity”? Describe how it works.
In: Statistics and Probability