Question

Suppose a poll was taken of 1000 randomly selected voters and the results showed that 53%...

Suppose a poll was taken of 1000 randomly selected voters and the results showed that 53% support Congress’ Jobs Creation Bill. You want to test the hypothesis that the majority of likely voters support Congress’ Bill; in other words to test the hypothesis H0: π ≤ 0.50 versus H1: π > 0.50. If the p-value is .0575, then at 0.05 level of significance, your conclusion is:

Select one:

There is no evidence that likely voters support Congress’ Jobs Creation Bill.

There is evidence that likely voters support Congress’ Jobs Creation Bill.

There is not enough information to make a statement one way or the other.

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

P value =0.0575 > 0.05

If P value ...................then we fail to reject H0

There is no evidence that likely voters support Congress’ Jobs Creation Bill.
Add a comment
Know the answer?
Add Answer to:
Suppose a poll was taken of 1000 randomly selected voters and the results showed that 53%...
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
  • Suppose a poll was taken of 1000 randomly selected voters and the results showed that 53%...

    Suppose a poll was taken of 1000 randomly selected voters and the results showed that 53% support Congress’ Jobs Creation Bill. You want to test the hypothesis that the majority of likely voters support Congress’ Bill; in other words to test the hypothesis H0: π ≤ 0.50 versus H1 : π > 0.50, at .01 level of significance. What is (are) the critical value(s) for the test? Choose the best answer. Select one: ±2.575 2.33 ±2.33 1.96 1.645

  • One month before an​ election, a poll of 660 randomly selected voters showed 57 % planning...

    One month before an​ election, a poll of 660 randomly selected voters showed 57 % planning to vote for a certain candidate. A week later it became known that he had once nbsptried nbspan illegal drug​, and a new poll showed only 54​% of 1080 voters supporting him. Do these results indicate a decrease in voter support for his ​candidacy? ​a) Test an appropriate hypothesis and state your conclusion. ​b) If you concluded there was a​ difference, estimate that difference...

  • 4 Gi According to a recent poll 53 percent of Americans would vote for the incumbent...

    4 Gi According to a recent poll 53 percent of Americans would vote for the incumbent president. If a random sample of 100 people results in 45 percent who would vote for the incumbent, test the claim that the actual percentage is 53 percent. Use a 0.10 significance level. (there are three correct choices in this question) SELECT ALL APPLICABLE CHOICES A) There is enough evidence to reject the Claim B) Ho p2.53 (claim) H1: P<.53 C) There is enough...

  • A poll of 2,142 randomly selected adults showed that 92​% of them own cell phones. The...

    A poll of 2,142 randomly selected adults showed that 92​% of them own cell phones. The technology display below results from a test of the claim that 91​% of adults own cell phones. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.05 significance level to complete parts​ (a) through​ (e). Test of p=0.91 vs p≠0.91 Sample X N Sample p ​95% CI ​Z-Value ​P-Value 1 1970 2,142 0.919701 ​(0.908193​,0.931210​) 1.57 0.117 a. Is the...

  • Hi I need help to find the p-value n a recent poll of 740 randomly selected...

    Hi I need help to find the p-value n a recent poll of 740 randomly selected adults, 586 said that it is morally wrong to not report all income on tax returns. Use a 0.05 significance level to test the claim that 70% of adults say that it is morally wrong to not report all income on tax returns. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, the conclusion about the null hypothesis, and the final conclusion that addresses...

  • a poll of 2094 randomly selected adults showed that 94% of them own cell phones A...

    a poll of 2094 randomly selected adults showed that 94% of them own cell phones A poll of 2,094 randomly selected adults showed that 94% of them own cell phones. The technology display below results from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the binomial distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p*0.92 Samplex N...

  • A local newspaper article reported that at least 50% of the construction jobs in the metropolitan...

    A local newspaper article reported that at least 50% of the construction jobs in the metropolitan New Orleans area are being filled by undocumented foreign workers. Anna Reed believes the actual percentage is much lower than that, and intends to challenge the newspaper’s figure. She took a random sample of 100 construction workers and found that 42 of them are undocumented. 1) What is the margin of error of the survey at the 95% level of confidence? 2) Find a...

  • A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The...

    A poll of 2,084 randomly selected adults showed that 94% of them own cell phones. The technology display below ret from a test of the claim that 92% of adults own cell phones. Use the normal distribution as an approximation to the bin distribution, and assume a 0.01 significance level to complete parts (a) through (e). Test of p = 0.92 vs p+0.92 Z-Value P-value Sample p 95% CI N Sample X 0.000 4.01 (0.930869,0.956847) 1 1967 2,084 0.943858 a....

  • Randomly selected birth records were​ obtained, and categorized as listed in the table to the right....

    Randomly selected birth records were​ obtained, and categorized as listed in the table to the right. Use a 0.01 0.01 significance level to test the reasonable claim that births occur with equal frequency on the different days of the week. How might the apparent lower frequencies on Saturday and Sunday be​ explained? Day nbsp Day Sun Mon Tues Wed Thurs Fri Sat Number of Births Number of Births 48 48 64 64 61 61 60 60 56 56 65 65...

  • ONLY DO NUMBER 3 For this project you will test claims and conjectures using hypothesis testing. ...

    ONLY DO NUMBER 3 For this project you will test claims and conjectures using hypothesis testing. For each hypothesis test, report the following: The null hypothesis, H0 The alternative hypothesis, H1 The test statistic rounded to the nearest hundredth (use T Stats or Proportion Stats in StatCrunch to find test statistics) The P-value for the test (use T Stats or Proportion Stats in StatCrunch to find P-values) The formal decision (Reject H0 or Fail to reject H0, remember that reject...

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