Question

Average cost of a visit to a vet clinic for a dog in 2004 was $60....

Average cost of a visit to a vet clinic for a dog in 2004 was $60. A recent random sample of 55 visits had a mean of $63.50 and a standard deviation of $2.00. Have costs increased? Use  = 0.05.

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Solution :

= 60

= 63.50

s = 2

n = 55

This is the right tailed test .

The null and alternative hypothesis is

H0 :   = 60

Ha : > 60

Test statistic = t

= ( - ) / s / n

= (63.50- 60) /2 / 55

= 12.978

P-value = 0

= 0.05  

0 < 0.05

Reject the null hypothesis .

There is sufficient evidence to suggest that Average cost  increased.

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
The mean body temperature of pugs is supposed to be 102.4 F. A Vet wants to...
The mean body temperature of pugs is supposed to be 102.4 F. A Vet wants to prove that the temperature is actually higher. If A sample of 64 pugs had an average temp of 103 F, with a standard deviation of .2 F, could the vet prove their claim. Use a level of significance of 0.05. What is the null hypothosis, alternative and answer?
A new teenage pregnancy clinic will have a fixed cost of $300,000—start-up costs, hiring advanced practice...
A new teenage pregnancy clinic will have a fixed cost of $300,000—start-up costs, hiring advanced practice nurses and employees, salaries, purchase of equipment, and other miscellaneous items. Also, the teenage pregnancy clinic has variable costs of $200 per patient visit—medical equipment, educational materials, and other miscellaneous items. There will be three levels of patient treatment: Simple = initial assessment and four prenatal visits Moderate = continuous treatment and monthly visits Complex = high-risk treatment Each patient will be charged the...
3. A survey was conducted about the cost for a family of four to visit an...
3. A survey was conducted about the cost for a family of four to visit an amusement park for one day. A sample of 32 families yielded an average cost of $190.28 with a standard deviation of $51.75. Last year, a magazine published that the average cost for a family of four to visit an amusement park was $175. Based on the sample data above, can we conclude that the mean cost is actually higher than this at α=0 3....
20. During a recent year, the average cost of making a movie was 54.8 million dollars....
20. During a recent year, the average cost of making a movie was 54.8 million dollars. This year, a random sample of 15 recent action movies had an average production cost of 62.3 million dollars with a variance of 90.25 million dollars. At the 0.05 level of significance (alpha), can it be concluded that it costs more than average to produce an action movie? NOTE: be sure to take the square root of a given Variance in the word problems...
In 2004, the mean household expenditure on gasoline and motor oil was $1773, according to data...
In 2004, the mean household expenditure on gasoline and motor oil was $1773, according to data obtained from the U.S. Energy Information Administration. An economist wanted to know whether this amount has changed significantly from its 2004 level. In a random sample of 1750 households, he found the mean expenditure (in 2004 dollars) on gasoline and motor oil during the most recent year to be $1803, with a standard deviation of $468. Do these results suggest the amount spend on...
A study showed that in a certain month, the mean time spent per visit to Facebook...
A study showed that in a certain month, the mean time spent per visit to Facebook was 19.5 minutes. Assume the standard deviation of the population is 8 minutes. Suppose that a simple random sample of 100 visits in that month has a sample mean of 21.14 minutes. A social scientist is interested in knowing whether the mean time of Facebook visits has increased. Perform the hypothesis test and compute the P-value. Write down your P-value. Round your answer to...
suppose the population average amount of dog biscuits in a bag is 40 pounds with a...
suppose the population average amount of dog biscuits in a bag is 40 pounds with a standard deviation of 6 pounds. a simple random sample of 64 bags is selected. what is the probability the sample mean weight will be within 1.5 pounds of the population mean?
The business manager of a local health clinic is interested in estimating the difference between the...
The business manager of a local health clinic is interested in estimating the difference between the fees for extended office visits in her center and the fees of a newly opened group practice. She gathered the following information regarding the two offices. Health Clinic Group Practice Sample size 50 visits 45 visits Sample mean $21 $19 Sample Standard deviation $2.75 $3.00 a. Determine the degrees of freedom for the t-distribution. b. Use the p-value approach and test to determine if...
A study showed that in a certain month, the mean time spent per visit to Facebook...
A study showed that in a certain month, the mean time spent per visit to Facebook was 19.5 minutes. Assume the standard deviation of the population is 8 minutes. Suppose that a simple random sample of 100 visits in that month has a sample mean of 21.66 minutes. A social scientist is interested in knowing whether the mean time of Facebook visits has increased. Perform the hypothesis test and compute the P-value. Write down your P-value. You will need it...
During a recent year the average cost of making a movie was $54.8 million. This year,...
During a recent year the average cost of making a movie was $54.8 million. This year, a random sample of 15 recent action movies had an average production cost of $62.3 million with a sample variance of $90.25 million. At an alpha of 0.05 level of significance, can it be concluded that it costs $54.8 million to produce an action movie?
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT