Question

Math 15Assessment # 41. A travel association says that the daily lodging cost for a family...

Math 15Assessment # 41. A travel association says that the daily lodging cost for a family traveling in California is the same as in Florida. The daily lodging cost for 35 families traveling in California is $136 and the standard deviation is $25. The daily lodging cost for 35 families traveling in Florida is $140 and the standard deviation is $30. At α = 0.10 is there enough evidence to reject the travel association’s claim?

Steps a through e

please show work step by step and include formulas. Also please write clearly if your doing it by hand. Thanks in advance.

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 group says that the mean daily lodging cost for a family traveling in Region 1...
A group says that the mean daily lodging cost for a family traveling in Region 1 is the same as in Region 2. The mean daily lodging cost for 32 families traveling in Region 1 is ​$126 with a population standard deviation of ​$2828. The mean daily lodging cost for thirty two families traveling in Region 2 is ​$121 with a population standard deviation of ​$2121. At alpha equals 0.10 is there enough evidence to reject the​group's claim? Complete parts​...
A group says that the mean daily lodging cost for a family traveling in Region 1...
A group says that the mean daily lodging cost for a family traveling in Region 1 is the same as in Region 2. The mean daily lodging cost for 3939 families traveling in Region 1 is ​$141141 with a population standard deviation of ​$2424. The mean daily lodging cost for 3939 families traveling in Region 2 is ​$135135 with a population standard deviation of ​$2525. At alpha equals 0.10 commaα=0.10, is there enough evidence to reject the​ group's claim? Complete...
A tourist agency in Hawaii claims the mean daily cost of meals and lodging for a...
A tourist agency in Hawaii claims the mean daily cost of meals and lodging for a family of 4 traveling in Hawaii is at most $650 with a standard deviation of $40. You work for a consumer protection advocate and want to test this claim. In a random sample of 45 families of 4 traveling in Hawaii, the mean daily cost of meals and lodging is $657. At € α = 0.05, do you have enough evidence to reject the...
A travel association claims that the mean daily food cost for two adults traveling together on...
A travel association claims that the mean daily food cost for two adults traveling together on vacation in San Francisco is $110. A random sample of 14 groups of adults is used to conduct a hypothesis test to show that the mean daily food is greater than 110. Using the mean and sample standard deviation from this sample, the test statistic was calculated to be 1.76. The corresponding p-value of this test is: a. 0.05 < p-value < 0.10 b....
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT