Out of 600 people sampled, 132 said they prefer Coke over Pepsi. What is the point...
Out of 600 people sampled, 132 said they prefer Coke over Pepsi.
What is the point estimate for the proportion of ALL people that
prefer Coke over Pepsi?
p^ = 0.22
What is the standard error for the sampling proportions in this
case?
SE=
Give your answers as decimals, to three places.
I can't find the SE for this problem.
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Out of 600 people sampled, 132 said they prefer Coke over Pepsi.
What is the point...
Out of 600 people sampled, 474 had kids. Based on this, construct a 99% confidence interval...
Out of 600 people sampled, 474 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to four places <p<
NOTE VOTE Out of 300 people sampled, 249 preferred Candidate A. Based on this, estimate what...
NOTE VOTE Out of 300 people sampled, 249 preferred Candidate A. Based on this, estimate what proportion of the voting population (TT) prefers Candidate A. Use a 95% confidence level, and give your answers as decimals, to three places. <t< If n=560 and p (p-hat) =0.23, find the margin of error at a 90% confidence level Give your answer to three decimals License Points possible: 1 This is attempt 1 of 1.
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calculate: PART A PART B PART C PART A: Out of 300 people sampled, 204 preferred Candidate A. Based on this, find a 90% confidence level for the true proportion of the voting population (pp) prefers Candidate A. Give your answers as decimals, to three places. < pp < PART B: In a sample of 390 adults, 289 had children. Construct a 95% confidence interval for the true population proportion of adults with children. Give your answers as decimals, to...
"Trydint' bubble gum company claims that 3 out of 10 people prefer their gum to "Eklypse"....
"Trydint' bubble gum company claims that 3 out of 10 people prefer their gum to "Eklypse". The null and alternative hypothesis in symbols would be: OH :: < 0.3 H: > 0.3 OH :P <0.3 H:p > 0.3 OH = 0.3 H:+0.3 OH:p > 0.3 H:p < 0.3 OH: > 0.3 H: <0.3 OH :p=0.3 H:P +0.3 The null hypothesis in words would be: The proportion of people in a sample that prefer Trydint gum is not 0.3 The average...
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Out of 200 people sampled, 94 had children. Based on this, construct a 99% confidence interval for the true population proportion of people with children. Preliminary: Is it safe to assume that n≤5% of all people with children? No Yes Verify nˆp(1−ˆp)≥10. YOU MUST ROUND YOUR ANSWER TO ONE DECIMAL PLACE. nˆp(1−ˆp)= 4.Confidence Interval: What is the 99% confidence interval to estimate the population proportion? YOU MUST ROUND ANSWER TO THREE DECIMALS PLACES. 5. ?????? <p< ????
Out of 600 people sampled, 474 had kids. Based on this, construct a 99% confidence interval for the true population proportion of people with kids. Give your answers as decimals, to four places <p<
NOTE VOTE Out of 300 people sampled, 249 preferred Candidate A. Based on this, estimate what proportion of the voting population (TT) prefers Candidate A. Use a 95% confidence level, and give your answers as decimals, to three places. <t< If n=560 and p (p-hat) =0.23, find the margin of error at a 90% confidence level Give your answer to three decimals License Points possible: 1 This is attempt 1 of 1.
"Trydint' bubble-gum company claims that 3 out of 10 people prefer their gum to "Eklypse". The null and alternative hypothesis in symbols would be: OH: <0.3 H: > 0.3 OH:P < 0.3 H:p > 0.3 OH = 0.3 H:H70.3 Ho:p > 0.3 H:p < 0.3 OH: > 0.3 Hu<0.3 OH,:= 0.3 H:P +0.3 The null hypothesis in words would be: The proportion of people in a sample that prefer Trydint gum is not 0.3 The average of people that prefer...
"Trydint' bubble gum company claims that 3 out of 10 people prefer their gum to "Eklypse". The null and alternative hypothesis in symbols would be: OH :: < 0.3 H: > 0.3 OH :P <0.3 H:p > 0.3 OH = 0.3 H:+0.3 OH:p > 0.3 H:p < 0.3 OH: > 0.3 H: <0.3 OH :p=0.3 H:P +0.3 The null hypothesis in words would be: The proportion of people in a sample that prefer Trydint gum is not 0.3 The average...
In a recent poll, 350 people were asked if they liked skiing, and 55% said they did. Find the margin of error of this poll, at the 90% confidence level. As in the reading, in your calculations: --Use z1.645 for a 90% confidence interval -Use z=2 for a 95% confidence interval -Use z-2.576 for a 99% confidence interval Give your answer rounded to three decimal places. If n=560 and p (p-hat) = 0.7, construct a 99% confidence interval. As in...
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