A survey found that out of a random sample of 200 workers 168 said that they...
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A 2011 NBC News survey found that 76% of a sample of 4522 American teens said...
A 2011 NBC News survey found that 76% of a sample of 4522 American teens said they owned an MP3 player such as an iPod. Assume that the sample was an SRS. Find the 90% confidence interval (+0.0001) for the proportion of all American teens who own an MP3 player. The 90% confidence interval is from to
Out of a random sample of 200 youth, 50 are found to be careless drivers. Out...
Out of a random sample of 200 youth, 50 are found to be careless drivers. Out of a random sample of 400 adults, 64 are found to be careless drivers. a. Determine, at a 1% significance level, if there is good evidence that the proportion of youth who are careless drivers is different than the proportion of adults who are careless drivers b. Form a 99% confidence interval for the difference in proportion of youth and adults who are careless...
ame: A survey conducted by Sallie Mae and Gallup of 1404 respondents found that 323 students...
ame: A survey conducted by Sallie Mae and Gallup of 1404 respondents found that 323 students paid for their education by student loans. Find the 90% confidence of the true proportion of students who paid for their education by student loans. A survey of 1898 people found that 45% of the adults said that dandelions were the toughest weeds to control in their yards. Find the 95% confidence interval of the true proportion who said that dandelions were the toughest...
In a random sample of 200 people, 154 said that they watched educational television. Find the...
In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the population proportion of people who watched educational television. Which formula will you use for this problem? O3 -1.14( ) <-<a+bala ) On= (*12:0) On=p:4 () Oộ - - - - - - Oz-za-(M) <r<+zar (M)
A survey of 355 working parents found 200 who said they spend too little time with...
A survey of 355 working parents found 200 who said they spend too little time with their children because of work commitments. (a) What is the point estimate of the proportion of the population of working parents who feel they spend too little time with their children because of work commitments? (b) At 95% confidence, what is the margin of error? (c) What is the 95% confidence interval estimate of the population proportion of working parents who feel they spend...
A survey was conducted to assess the market share of Ringtone, a telecommunications company. Out of...
A survey was conducted to assess the market share of Ringtone, a telecommunications company. Out of a random sample of 200 people who have mobile phone contracts surveyed, 38 of them have Ringtone contracts. (a) Calculate the sample proportion of people who have Ringtone contracts. (b) Ringtone has established a target market share of at least 15%. Use the sample proportion found in (a), test whether this target has been attained at a 5% significance level. (c) Use the information...
A survey was conducted to assess the market share of Ringtone, a telecommunications company. Out of...
A survey was conducted to assess the market share of Ringtone, a telecommunications company. Out of a random sample of 200 people who have mobile phone contracts surveyed, 38 of them have Ringtone contracts. (a) Calculate the sample proportion of people who have Ringtone contracts. (b) Ringtone has established a target market share of at least 15%. Use the sample proportion found in (a), test whether this target has been attained at a 5% significance level. (c) Use the information...
A survey was conducted to assess the market share of Ringtone, a telecommunications company. Out of...
A survey was conducted to assess the market share of Ringtone, a telecommunications company. Out of a random sample of 200 people who have mobile phone contracts surveyed, 38 of them have Ringtone contracts. (a) Calculate the sample proportion of people who have Ringtone contracts. (b) Ringtone has established a target market share of at least 15%. Use the sample proportion found in (a), test whether this target has been attained at a 5% significance level. (c) Use the information...
6. Trick or Treat In a survey of a random sample of 35 households in the...
6. Trick or Treat In a survey of a random sample of 35 households in the Cherry Creek neighborhood of Denver, it was found that 11 households turned out the lights and pretended not to be home on Halloween. Compute a 90% confidence interval for p, the proportion of all housholds in Cherry Creek that pretend not to be home on Halloween. What assumptions are necessary to calculate the confidence interval of part (a)? Interpretation The national proportion of all...
A random sample of 78 students were interviewed and 59 said they would vote for a...
A random sample of 78 students were interviewed and 59 said they would vote for a democrat in the 2008 election. 1. Let p represent the proportion of all students at this college who will vote for a democrat. Find a point estimate p for p. 2. Find a 90% confidence interval for p. 3. What assumptions are required for the calculations of part (b)? Do you think these assumptions are satisfied? Explain 4. How many more students should be...
A 2011 NBC News survey found that 76% of a sample of 4522 American teens said they owned an MP3 player such as an iPod. Assume that the sample was an SRS. Find the 90% confidence interval (+0.0001) for the proportion of all American teens who own an MP3 player. The 90% confidence interval is from to
ame: A survey conducted by Sallie Mae and Gallup of 1404 respondents found that 323 students paid for their education by student loans. Find the 90% confidence of the true proportion of students who paid for their education by student loans. A survey of 1898 people found that 45% of the adults said that dandelions were the toughest weeds to control in their yards. Find the 95% confidence interval of the true proportion who said that dandelions were the toughest...
In a random sample of 200 people, 154 said that they watched educational television. Find the 90% confidence interval of the population proportion of people who watched educational television. Which formula will you use for this problem? O3 -1.14( ) <-<a+bala ) On= (*12:0) On=p:4 () Oộ - - - - - - Oz-za-(M) <r<+zar (M)
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