Question

Many amusement parks base their projections of revenues and profits on the assumption that the average...

Many amusement parks base their projections of revenues and profits on the assumption that the average

young adult visits an amusement park four times per year. To investigate the validity of this assumption, a random sample of

63 young adults is drawn and each is asked to report the number of times they skied the previous year. The sample yields an average of 4.841 and a sample standard deviation of 2, can we infer that the assumption is wrong with 97% confidence?

Hypotheses:

Ho: μ (Click to select)≠>≤=<≥   vs. HA: μ (Click to select)≤<=>≥≠

(Round to two decimals)

Test statistic:

If using the Rejection Region approach:

(Round to two decimals and enter only positive answers)

What value (critical value) marks the rejection region?

If using the P-value approach:

(Round to four decimals)

p-value:

Use the Confidence Interval Method to make a conclusion (Round to two decimals):

Conclusion: Since  (Click to select)falls outsideis inside [, ], (Click to select)Do not reject the null hypothesisReject the alternative hypothesisReject the null hypothesisDo not reject the alternative hypothesis. We have (Click to select)insufficient evidencesufficient evidenceinconclusive evidence to conclude that the (Click to select)population standard deviationpopulation meansample standard deviationsample mean (Click to select)number of times a young adult visits an amusement park is different thannumber of times a young adult visits an amusement park is less thannumber of times a young adult visits an amusement park is more than

0 0
Add a comment Improve this question Transcribed image text
Answer #1

Rvendeite -84 o. , 2520-3-34 2. u- 5468 (.2742)

Add a comment
Know the answer?
Add Answer to:
Many amusement parks base their projections of revenues and profits on the assumption that the average...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions
  • Many Alpine ski centers base their projections of revenues and profits on the assumption that the average Alpine skier...

    Many Alpine ski centers base their projections of revenues and profits on the assumption that the average Alpine skier skis four times per year. To investigate the validity of this assumption, a random sample of 63 skiers is drawn and each is asked to report the number of times he or she skied the previous year. The average number of times from those reported is 4.84. Suppose the population standard deviation is 2, can we infer at the 10% significance...

  • Many ski resorts base their projection of revenues and profits on the assumption that the average...

    Many ski resorts base their projection of revenues and profits on the assumption that the average skier skis 4 times per year. A random sample of 63 skiers is drawn and each skier is asked to indicate the number of times he or she skied the previous year. The dataset Qn9.1 contains the responses for each skier. Assume that the population standard deviation is 2. [ Hint: Use an appropriate Excel template from chapter 9 or formulas] a. Formulate the...

  • Historically, the population of wolves in Canada have been found to have an average adult weight...

    Historically, the population of wolves in Canada have been found to have an average adult weight of 38.9 kilograms. The Canadian Wildlife Service believes that due to climate change, this may have increased. A random sample of 12 wolves is captured and weighed and the following results are obtained. Assume that the weights are normally distributed: 44.8, 40.7, 42.6, 46.8, 48.7, 35.4, 38.9, 39.5, 38.3, 48.9, 48.4, 38.4. At a 0.05 significance level, test the hypothesis that the average weight...

  • A U.S. Web Usage Snapshot indicated a monthly average of 36 Internet visits per user from...

    A U.S. Web Usage Snapshot indicated a monthly average of 36 Internet visits per user from home. A random sample of 24 Internet users yielded a sample mean of 42.1 visits with a standard deviation of 5.3. At the 0.01 level of significance, can it be concluded that this differs from the national average? Step 1: State hypotheses by filling in the symbol (=, <, >, or not equal) and the population mean: Ho: μ Blank H1: μ Blank Step...

  • sample mean = 213.4552 sample Standard deviation = 44.81542 N=50 alpha = .05 SEM = 6.337857477...

    sample mean = 213.4552 sample Standard deviation = 44.81542 N=50 alpha = .05 SEM = 6.337857477 For each of the following hypothesis testing problems, manually calculate the t-statistic, use the 5% level of significance (alpha = 0.05), determine the rejection region, determine the p-value of the t-test, use the 95% confidence interval in part (c) to make a decision about whether or not to reject the null hypothesis. Test the null hypothesis that the true mean is 225 versus the...

  • The average student-loan debt is reported to be $25,235. A student believes that the student-loan debt...

    The average student-loan debt is reported to be $25,235. A student believes that the student-loan debt is higher in her area. She takes a random sample of 100 college students in her area and determines the mean student-loan debt is $27,524 and the standard deviation is $6,000. Is there sufficient evidence to support the student's claim at a 5% significance level? Preliminary: Is it safe to assume that n≤5% of all college students in the local area? Yes No Is...

  • The average student loan debt is reported to be $25,235. A student belives that the student loan debt is higher in...

    The average student loan debt is reported to be $25,235. A student belives that the student loan debt is higher in her area. She takes a random sample of 100 college students in her area and determines the mean to be $27,524 and the standard devition to be $6000. Is there sufficient evidence to support the student' claim at a 5% significance level? a) Determine the null and alternative hypotheses. Ho: d = Ho: Select an answer (Put in the...

  • NEED HELP ASAP - will give thumbs up Multi part question set If you can’t finish...

    NEED HELP ASAP - will give thumbs up Multi part question set If you can’t finish all of it, that’s ok! Would really appreciate any and all help as soon as possible Thank you so much in advance! Thank you!! Migrants through Mexico - 1, NUMERICAL ANSWER There are 7 questions in this set. As you proceed through these questions, use necessary information from previous questions. The Washington Post and the Mexican newspaper Reforma surveyed 1.200 Mexican adults in July...

  • 1.3.3 Question Help * Sample 1 Sample 2 Assume that both populations are normally distributed. a)...

    1.3.3 Question Help * Sample 1 Sample 2 Assume that both populations are normally distributed. a) Test whether μ? μ2 at the α 0.05 level of significance for the given sample data. b) Construct a 95% confidence interval about μ1-2. 16 44.1 12.4 52.5 9.7 EB Click the icon to view the Student t-distribution table a) Perform a hypothesis test. Determine the null and alternative hypotheses Determine the critical value(s). Select the correct choice bElow and fill in the answer...

  • Need a little help. Please help You wish to test the claim that the average IQ...

    Need a little help. Please help You wish to test the claim that the average IQ score is less than 100 at the .10 significance level. You determine the hypotheses are: Ho : μ = 100 H 1 : μ < 100 You take a simple random sample of 41 individuals and find the mean IQ score is 97.2, with a standard deviation of 14.9. Let's consider testing this hypothesis two ways: once with assuming the population standard deviation is...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT