Question

According to the Census Bureau, 3.34 people reside in the typical American household. A sample of 26 households in Arizona re
2 c. Compute the value of the test statistic. (Negative amount should be indicated by a minus sign. Round your answer to 3 de
0 0
Add a comment Improve this question Transcribed image text
Answer #1

Part a)

To Test :-
H0 :- µ <= 3.34
H1 :- µ < 3.34

Part b)

Reject null hypothesis if t < -t(α, n-1)
Critical value t(α, n-1) = t(0.1 , 26-1) = 1.316
t < - 1.316

Part c)

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 2.7 - 3.34 ) / ( 1.17 / √(26) )
t = -2.789

Part d)

Test Criteria :-
Reject null hypothesis if t < -t(α, n-1)
Critical value t(α, n-1) = t(0.1 , 26-1) = 1.316
t < - t(α, n-1) = -2.7892 < - 1.316
Result :- Reject null hypothesis

HO. Mean number of residents less than 3.34 persons

Reject H0, mean number of residents is less than 3.34 persons.

Add a comment
Know the answer?
Add Answer to:
According to the Census Bureau, 3.34 people reside in the typical American household. A sample of...
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
  • According to the Census Bureau, 3.36 people reside in the typical American household. A sample of...

    According to the Census Bureau, 3.36 people reside in the typical American household. A sample of 27 households in Arizona retirement communities showed the mean number of residents per household was 2.70 residents. The standard deviation of this sample was 1.25 residents. At the .10 significance level, is it reasonable to conclude the mean number of residents in the retirement community household is less than 3.36 persons?    a. State the null hypothesis and the alternate hypothesis. (Round your answer...

  • Consider the following hypotheses: He: μ28e The population is normally distributed. A sample produces the following...

    Consider the following hypotheses: He: μ28e The population is normally distributed. A sample produces the following observations: 56 67 62 81 8366 Conduct the test at the 1% level of significance. (You may find lt useful to reference the appropriate table: table or Цеье o. Calculate the value of the test statistic. (Negative value should be Indicated by a minus sign. Round Intermedlate caleulatlons to at least 4 declmal places and final answer to 2 declmal places.) Test statistic b....

  • A cell phone company offers two plans to its subscribers. At the time new subscribers sign...

    A cell phone company offers two plans to its subscribers. At the time new subscribers sign up, they are asked to provide some demographic information. The mean yearly income for a sample of 43 subscribers to Plan A is $58,200 with a standard deviation of $9.000. This distribution is positively skewed: the coefficient of skewness is not larger. For a sample of 42 subscribers to Plan B. the mean income is $58.400 with a standard deviation of $8.700. At the...

  • An automotive dealer advertised in Auto-Trend Magazine that the cost of a hybrid car was less...

    An automotive dealer advertised in Auto-Trend Magazine that the cost of a hybrid car was less than $10,000. Listed below is a total cost in $000 for a sample of 8 hybrid cars. At the .05 significance level, is it reasonable to conclude the mean cost of a hybrid car is less than $10,000? 10.1 9.8 8.8 10.3 9.0 9.7 8.5 8.6 (A) State the null hypothesis and the alternate hypothesis. Use a .05 level of significance. (Enter your answers...

  • The cost of weddings in the United States has skyrocketed in recent years. As a result,...

    The cost of weddings in the United States has skyrocketed in recent years. As a result, many couples are opting to have their weddings in the Caribbean. A Caribbean vacation resort recently advertised in Bride Magazine that the cost of a Caribbean wedding was less than $10,000. Listed below is a total cost in $000 for a sample of 8 Caribbean weddings. At the .005 significance level is it reasonable to conclude the mean wedding cost is less than $10,000...

  • n order to conduct a hypothesis test for the population proportion, you sample 480 observations that...

    n order to conduct a hypothesis test for the population proportion, you sample 480 observations that result in 264 successes. (You may find it useful to reference the appropriate table: z table or t table) Ho: pz 0.60 HA: p0.60 a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic 2.24 a-2. Find the p-value....

  • In order to conduct a hypothesis test for the population proportion, you sample 440 observations that...

    In order to conduct a hypothesis test for the population proportion, you sample 440 observations that result in 220 successes. (You may find it useful to reference the appropriate table: z table or t table Ho: p 0.52; HA: P 0.52 a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) Test statistic a-2. Find the p-value....

  • The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting...

    The number of "destination weddings" has skyrocketed in recent years. For example, many couples are opting to have their weddings in the Caribbean. A Caribbean vacation resort recently advertised in Bride Magazine that the cost of a Caribbean wedding was less than $30,000. Listed below is a total cost in $000 for a sample of 8 Caribbean weddings. At the 0.05 significance level, is it reasonable to conclude the mean wedding cost is less than $30,000 as advertised? 28.7 31.8...

  • The type of household for the U.S. population and for a random sample of 411 households...

    The type of household for the U.S. population and for a random sample of 411 households from a community in Montana are shown below. Observed Number of Households in the Community 90 Type of Household Married with children Married, no children Single parent One person Other (e.g., roommates, siblings) Percent of U.S. Households 26% 29% 9% 25% 11% 126 28 100 67 Use a 5% level of significance to test the claim that the distribution of U.S. households fits the...

  • In order to conduct a hypothesis test for the population proportion, you sample 320 observations that...

    In order to conduct a hypothesis test for the population proportion, you sample 320 observations that result in 128 successes.(You may find it useful to reference the appropriate table: z table or t table HO pz 0.45; HA: p < 0.45. a-1. Calculate the value of the test statistic. (Negative value should be indicated by a minus sign. Round intermediate calculations to at least 4 decimal places and final answer to 2 decimal places.) est statistic a-2. Find the p-value....

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
ADVERTISEMENT