Question

An episode of 60 Minutes had a 15% share of 5000 sample households (750 out of...

An episode of 60 Minutes had a 15% share of 5000 sample households (750 out of 5000). Use an α = 0.01 significance level to test this proportion claim that the total number of households nationwide tuned in was different than (≠) 20%.

Select the correct P-value and appropriate conclusion answer from the list below:

TIP: Conduct the appropriate Hypothesis Test on your TI 83/84 calculator and consider the P-value it provides.

Group of answer choices

My P-value greater than α Alpha, so I Reject Null Hypothesis

My P-value greater than α Alpha, so I Accept Null Hypothesis

My P-value less than α Alpha, so I Accept Null Hypothesis

My P-value less than α Alpha, so I Reject Null Hypothesis

Homework Answers

Answer #1

thank you.

Know the answer?
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
Similar Questions
An episode of 60 Minutes had a 15% share of 5000 sample households (750 out of...
An episode of 60 Minutes had a 15% share of 5000 sample households (750 out of 5000). Use an α = 0.01 significance level to test this proportion claim that the total number of households nationwide tuned in was different than (≠) 20%. Select the correct P-value and appropriate conclusion answer from the list below: TIP: Conduct the appropriate Hypothesis Test on your TI 83/84 calculator and consider the P-value it provides A. My P-value greater than α Alpha, so...
In order to test HO: µ0 = 40 versus H1: µ ≠ 40, a random sample...
In order to test HO: µ0 = 40 versus H1: µ ≠ 40, a random sample of size n = 25 is obtained from a normal population with a known σ = 6. My x-BAR mean is 42.3 from my sample. Using a TI 83/84 calculator, calculate my P-value with the appropriate Hypothesis Test.                                   Use a critical level α = 0.10 and decide to Accept or Reject HO with the valid reason for the decision. Group of answer choices My...
8.3 A recent broadcast of a television show had a 15 ​share, meaning that among 5000...
8.3 A recent broadcast of a television show had a 15 ​share, meaning that among 5000 monitored households with TV sets in​ use, 15% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in​ use, less than 25​% were tuned into the program. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the...
A recent broadcast of a television show had a 15 ?share, meaning that among 5000 monitored...
A recent broadcast of a television show had a 15 ?share, meaning that among 5000 monitored households with TV sets in? use,15?% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in? use, less than 20% were tuned into the program. Identify the null? hypothesis, alternative? hypothesis, test? statistic, conclusion about the null?hypothesis, and final conclusion that addresses the original claim. Use the...
A recent broadcast of a television show had a15 ?share, meaning that among 6000 6000 monitored...
A recent broadcast of a television show had a15 ?share, meaning that among 6000 6000 monitored households with TV sets in? use,15?% of them were tuned to this program. Use a 0.01 significance level to test the claim of an advertiser that among the households with TV sets in? use, less than 25?% were tuned into the program. Identify the null? hypothesis, alternative? hypothesis, test? statistic, P-value, conclusion about the null? hypothesis, and final conclusion that addresses the original claim....
Test the​ hypothesis, using​ (a) the classical approach and then​ (b) the​ P-value approach. Be sure...
Test the​ hypothesis, using​ (a) the classical approach and then​ (b) the​ P-value approach. Be sure to verify the requirements of the test. H0​: p=0.6 versus H1​: p>0.6 n=200​; x=140​, α=0.05 ​(a) Choose the correct result of the hypothesis test for the classic approach below. A.Do not reject the null​ hypothesis, because the test statistic is greater than the critical value. B.Do not reject the null​ hypothesis, because the test statistic is less than the critical value. C.Reject the null​...
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The...
Lightbulbs of a certain type are advertised as having an average lifetime of 750 hours. The price of these bulbs is very favorable, so a potential customer has decided to go ahead with a purchase arrangement unless it can be conclusively demonstrated that the true average lifetime is smaller than what is advertised. A random sample of 44bulbs was selected, the lifetime of each bulb determined, and the appropriate hypotheses were tested using MINITAB, resulting in the accompanying output. Variable...
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. Type of Household Percent of U.S. Households Observed Number of Households in the Community Married with children 26%         92             Married, no children 29%         120             Single parent 9%         29             One person 25%         101             Other (e.g., roommates, siblings) 11%         69   Use a 5% level of significance to test the claim that the distribution of U.S. households fits the...
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. Type of Household Percent of U.S. Households Observed Number of Households in the Community Married with children 26%         103             Married, no children 29%         118             Single parent 9%         34             One person 25%         90             Other (e.g., roommates, siblings) 11%         66             Use a 5% level of significance to test the claim that the distribution of U.S. households fits the...
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. Type of Household Percent of U.S. Households Observed Number of Households in the Community Married with children 26%         96             Married, no children 29%         113             Single parent 9%         32             One person 25%         97             Other (e.g., roommates, siblings) 11%         73             Use a 5% level of significance to test the claim that the distribution of U.S. households fits the...