Question

Thi a previous pol, 29% of adults with children under the age of 18 reported that their family ate dinner together seven nigh
Because npo (1-P) = 257.2 > 10 and the sample size is less than 5% of the population size, and the sample can be reasonably a
0 0
Add a comment Improve this question Transcribed image text
Answer #1

p_0=0.29

np_0(1-p_0)=1096*0.29*(1-0.29)=225.67 > 10 and the sample size less than 5% of the population size, and the sample can be assumed resonably random the requirements are satisfied

Add a comment
Know the answer?
Add Answer to:
Thi a previous pol, 29% of adults with children under the age of 18 reported that...
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
  • In a previous poll, 36% of adults with children under the age of 18 reported that...

    In a previous poll, 36% of adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Suppose that, in a more recent poll, 378 of 1108 adults with children under the age of 18 reported that their family ate dinner together seven nights a week. Is there sufficient evidence that the proportion of families with children under the age of 18 who eat dinner together seven nights a week has decreased?...

  • 10.2.17- Assigned Media Question Help In a clinical trial, 22 out of 864 patients taking a...

    10.2.17- Assigned Media Question Help In a clinical trial, 22 out of 864 patients taking a prescription drug daily complained of flulike symptoms. Suppose that it is known that 2.2% of patients taking competing drugs complain of flulike symptoms. Is there sufficient evidence to conclude that more than 2.2% of this drug's users experience flulike symptoms as a side effect at the a = 0.01 level of significance? less than 5% of the population size, and the sample can be...

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.09 hours, with a standard deviation of 2.29 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.31 hours, with a standard deviation of 1.65 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between adults with no...

  • Was the A random sample of 40 adults with no children under the age of 18...

    Was the A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.96 hours, with a standard deviation of 226 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4 23 hours, with a standard deviation of 155 hours. Construct and interpret a 95% confidence interval for the mean difference in leisure time between...

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.55 ​hours, with a standard deviation of 2.28 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.14 ​hours, with a standard deviation of 1.86 hours. Construct and interpret a 90​% confidence interval for the mean difference in leisure time between adults with no...

  • A random sample of 40 adults with no children under the age of 18 years results...

    A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.52 hours, with a standard deviation of 2.36 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.46 hours, with a standard deviation of 1.53 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...

  • A random sample of 40 adults with no children under the age of 18 years resuts...

    A random sample of 40 adults with no children under the age of 18 years resuts in a mean daily leisure time of 5.74 hours, with a standard deviation of 2.32 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.11 hours, with a standard deviation of 1.73 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...

  • 14. A random sample of 40 adults with no children under the age of 18 years results in a mean dai...

    14. A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.23 hours, with a standard deviation of 2.25 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure difference in leisure time between adults with no children and adults with children (1-H2 H represent the adults with children under the age of 18 The e0% (Round...

  • Vy > A random sample of 40 adults with no children under the age of 18...

    Vy > A random sample of 40 adults with no children under the age of 18 years results in a mandaly leisure time of 5.69 hours, with a standard deviation of 2.32 hours. A random sample of 40 adults with children under the age of 18 results in a mean dailyleisure time of 4.25 hours, with a standard deviation of 1.62 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between aduts with no...

  • A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.6...

    A random sample of 40 adults with no children under the age of 18 years results in a mean daily leisure time of 5.69 hours, with a standard deviation of 2.49 hours. A random sample of 40 adults with children under the age of 18 results in a mean daily leisure time of 4.07 hours, with a standard deviation of 1.54 hours. Construct and interpret a 90% confidence interval for the mean difference in leisure time between adults with no...

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