Question

In: Statistics and Probability

In a sample of 100 Oklahoma public schools, it was found that Oklahoma spends 8,097 dollars...

In a sample of 100 Oklahoma public schools, it was found that Oklahoma spends 8,097 dollars per pupil per year, with a standard deviation of 220 dollars. The population average of three surrounding states (Arkansas, Kansas, and Missouri) is 10,039 dollars. (these numbers are accurate – taken from most recent available data for 2016 https://ballotpedia.org/Public_education_in_Oklahoma.). Is per pupil spending for all Oklahoma schools less than the surrounding state population average? If you had to voice your concerns to your elected representative about per pupil spending in Oklahoma public schools, what would you tell them?

Solutions

Expert Solution

Let denotes the average pupil spending for all Oklahoma schools.

Conclusion in problem context :

There is sufficient evidence to support the claim that per pupil spending for all Oklahoma schools is significantly less than the surrounding state population average


Related Solutions

2. A random sample of 100 patients with virus found that it was fatal in 18...
2. A random sample of 100 patients with virus found that it was fatal in 18 people and 182 people eventually made a full recovery. In a random sample of 300 SARS patients, 34 died. a.) Verify your conditions for inference b.) Calculate a 90% confidence interval for the true mean difference in the proportion of virus cases that were fatal and the proportion of SARS cases that were fatal. c.) If you were to create a 95% confidence interval,...
3) A random sample of 100 pumpkins is obtained and the mean circumference is found to...
3) A random sample of 100 pumpkins is obtained and the mean circumference is found to be 40.5 cm. Assuming that the population standard deviation is known to be 1.6 cm, use a 0.05 significance level to test the claim that the mean circumference of all pumpkins is equal to 39.9 cm.
The Nellie Mae organization found that in a random sample of 100 undergraduate students taken in...
The Nellie Mae organization found that in a random sample of 100 undergraduate students taken in 2004, the average credit card balance was $2169. Suppose sample standard deviation of these credit card balances is $1000. Perform a test to test if the average credit card debt exceeds $2000. a) Identify the population    Identify the variable: Identify the parameter b) Calculate T stat Find P-value with DF
A random sample of 100 voters found that 46% were going to vote for a certain...
A random sample of 100 voters found that 46% were going to vote for a certain candidate. Find the 90% confidence interval for the population proportion of voters who will vote for that candidate. A. 38.7% < p < 53.3% B. 37.8% < p < 54.2% C. 41.9% < p < 50.1% D. 39.6% < p < 52.4%
In a large Midwestern university, a random sample of 100 entering freshman in 2009 found that...
In a large Midwestern university, a random sample of 100 entering freshman in 2009 found that 20 finished in the bottom third of their high school class. Admission standards at the university were tightened the next year. In 2011, a random sample of 100 entering freshman found that 10 had finished in the bottom third of their high school class. Let p1 be the proportion of admitted freshman who had finished in the bottom third of their high school class...
In 1995, a random sample of 100 adults that had investments in the stock market found...
In 1995, a random sample of 100 adults that had investments in the stock market found that only 20 said they were investing for the long haul rather than to make quick profits. A simple random sample of 100 adults that had investments in the stock market in 2002 found that 36 were investing for the long haul rather than to make quick profits. Let p1995 and p2002 be the actual proportion of all adults with investments in the stock...
1. A random sample of 100 construction workers found that the average monthly salary was $21,310...
1. A random sample of 100 construction workers found that the average monthly salary was $21,310 USD with a standard deviation of $2405. A random sample of 81 Office workers found that the average salary was $27,041 with a standard deviation of $4651. a.) Calculate a 95% confidence interval for the true mean difference between office workers’ salaries and construction workers’ salaries. b.) Verify your conditions for inference c.) Interpret your confidence interval with a sentence in context. d.) According...
A random sample of 100 postal employees found that the average time these employees had worked...
A random sample of 100 postal employees found that the average time these employees had worked for the postal service was x(bar) = 7 years, with standard deviation s = 2 years. Assume the distribution of the time the population of employees has worked for the postal service is approximately normal, with mean μ. Are these data evidence that μ has changed from the value of 7.5 years of 20 years ago? Carry a test of significance at 1% and...
A random sample found that forty percent of 100 Americans were satisfied with the gun control...
A random sample found that forty percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 99% confidence interval for the true proportion of Americans who were satisfied with the gun control laws in 2017. Fill in the blanks appropriately. A random sample found that forty percent of 100 Americans were satisfied with the gun control laws in 2017. Compute a 99% confidence interval for the true proportion of Americans who were satisfied with the...
Q: The average starting salary of a random sample of 100 high school students was found...
Q: The average starting salary of a random sample of 100 high school students was found to be $31,840. The population standard deviation for all such individuals is known to be $9,840. a. (12) Ten years ago, the average starting salary was $25,000. Does the sample data support the claim that the starting salary for this group has increased? Use alpha = 0.05. b. (6) Describe in general Type I and Type II errors and the Power of the test....
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT