Question

According to my fitbit, in 2018 (n1=49 weeks), I averaged 81,963.6 steps per week with a...

According to my fitbit, in 2018 (n1=49 weeks), I averaged 81,963.6 steps per week with a standard deviation of 11,570.8 steps per week. In 2019 (n2 = 52 weeks), I averaged 82,892.1 steps per week with a standard deviation of 10,560.3 steps per week.

a. conduct a two-sided test of the equality of VARIANCE for these two time periods at the alpha =0.05 level. State null and alternative hypothesis.

b. conduct a two-sided test of the equality of MEANS for these two time periods at the alpha =0.05 level. State null and alternative hypothesis.

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

Homework Answers

Answer #1

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
A random sample of n1 = 52 men and a random sample of n2 = 48...
A random sample of n1 = 52 men and a random sample of n2 = 48 women were chosen to wear a pedometer for a day. The men’s pedometers reported that they took an average of 8,342 steps per day, with a standard deviation of s1 = 371 steps. The women’s pedometers reported that they took an average of 8,539 steps per day, with a standard deviation of s2 = 214 steps. We want to test whether men and women...
Historically, evening long-distance calls from a particular city have averaged 15.2 minutes per call. In a...
Historically, evening long-distance calls from a particular city have averaged 15.2 minutes per call. In a random sample of 35 calls, the sample mean time was 14.3 minutes. Assume the standard deviation is known to be 5 minutes. Using a 0.05 level of significance, is there sufficient evidence to conclude that the average evening long-distance call has decreased? Note: Use the six-steps – clearly labeled (15 pts.) State the null hypothesis and the alternative hypothesis Determine the critical value(s); (draw...
A medical investigation claims that the average number of contaminations per week at the Main Hospital...
A medical investigation claims that the average number of contaminations per week at the Main Hospital is 19.2. A random sample of 10 weeks had a mean number of 20.5. The sample standard deviation is 3.5. Is there enough evidence to reject the investigator’s claim at alpha =.05? Steps to be covered: state the hypothesis and identify the claim find the critical value from the table and mention the acceptance range compute the test value (t test) make the decision...
The mean number of shopping trips per week in a county is 2.2. In a particular...
The mean number of shopping trips per week in a county is 2.2. In a particular neighborhood within the county, a survey based on 50 people reveals that the mean in the neighborhood is 1.85, with a standard deviation of 1.2. Test the null hypothesis that the mean in the neighborhood is no different from that in the county. State the null and alternative hypotheses, find the test statistic, compare it with the critical value, make a decision, and find...
The belief is that the mean number of hours per week of part-time work of high...
The belief is that the mean number of hours per week of part-time work of high school seniors in a city is 10.6 hours. Data from a simple random sample of 25 high school seniors indicated that their mean number of part-time work was 11.4 with a standard deviation of 1.3. Test whether these data cast doubt on the current belief. (use α = 0.05) a. State your null and alternative hypotheses. b.State the rejection region. c.Calculate the test statistic...
Twenty (20) randomly selected students were asked how many hours they spend each week watching television....
Twenty (20) randomly selected students were asked how many hours they spend each week watching television. They reported a sample mean and standard deviation of 26.4 and 8.2 respectively. Assume that the population of times is approximately normal. Test at the 0.05 level the claim that the mean number of hours per week spent by all students is greater than 20. The null and alternative hypothesis to conduct the test is
4. The amount of time in hours that college students spend on facebook per week is...
4. The amount of time in hours that college students spend on facebook per week is normally distributed with a mean of 40. You decide to conduct your own test. In a random sample of 60 college students you find that the mean time per week spent on facebook is 37.8 hrs and the standard deviation is 12.2. Does this mean differ significantly from the national average? Test your hypothesis at the .05 significance level and state your conclusion.
A researcher claims that high-school students exercise an average (mean) of 8 hours per week. (You...
A researcher claims that high-school students exercise an average (mean) of 8 hours per week. (You think that the number is actually higher.) From a sample of 40 students, you find a mean of 9 hours with a sample standard deviation of 1 hour. Conduct a hypothesis test using a 5% significance level. a) What are the null and alternative hypotheses? b) What is the test statistic? c) What is the p-value? d) Do your reject the null hypothesis? Explain...
1. A vending machine pours an average of 8.0 oz of coffee with a standard deviation...
1. A vending machine pours an average of 8.0 oz of coffee with a standard deviation of 0.2 oz if it is functioning properly. An inspector wants to take 16 cups of coffee from the machine to see if the machine is functioning well or not. He wants to have a significance level (alpha ) of 4%. (i) State the null and the alternative hypothesis . Is it a one-sided or two-sided test problem? (ii) Compute the power of test...
In a​ randomized, double-blind​ experiment, 120 babies were randomly divided into a treatment group (n1 =60)...
In a​ randomized, double-blind​ experiment, 120 babies were randomly divided into a treatment group (n1 =60) and a control group ( n2= 60 right). After the​ study, the treatment group had a mean serum retinol concentration of 47.35 micrograms per deciliter (ug/ dL) with a standard deviation of 17.26 ug/dL​, while the control group had a mean serum retinol concentration of 14.72 ug/dL with a standard deviation of 14.21 ug/ dL. Does the treatment group have a higher standard deviation...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT