Question

Survey A asked 1000 people how they liked the new movie Avengers: Endgame and 88\% said they did enjoy it. Survey B also concluded that 82\% of people liked the move but they asked 1600 total moviegoers. Which of the following is true about this comparison?

Confidence intervals
ANSWERED
asked 2021-08-02
Survey A asked 1000 people how they liked the new movie Avengers: Endgame and \(\displaystyle{88}\%\) said they did enjoy it.
Survey B also concluded that \(\displaystyle{82}\%\) of people liked the move but they asked 1600 total moviegoers.
Which of the following is true about this comparison?
1. The margin of error is the same for both.
2. The confidence interval is smaller for Survey B.
3. Increasing the number of people asked does not change the \(\displaystyle{95}\%\) confidence interval.
4. Survey A is more accurate since the percentage is higher.
5. Survey A has more approvals than Survey B.
6. Survey A is better than Survey B since it has a higher percentage.

Answers (1)

2021-08-02

Given data:
Survey A asked 1000 people how they liked the new movie Avengers: Endgame and \(\displaystyle{88}\%\) said they did enjoy it.
Survey B also concluded that \(\displaystyle{82}\%\) of people liked the move but they asked 1600 total moviegoers.
Confidence interval for Survey A
CI(Proportion) \(\displaystyle={\left(\hat{{{p}}}-{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}},\hat{{{p}}}+{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\right)}\)
\(\displaystyle={\left({0.88}-{1.96}\times\sqrt{{{\frac{{{0.88}{\left({1}-{0.88}\right)}}}{{{1000}}}}}},{0.88}+{1.96}\times\sqrt{{{\frac{{{0.88}{\left({1}-{0.88}\right)}}}{{{1000}}}}}}\right)}\)
\(\displaystyle={\left({0.86},{0.9}\right)}\)
Confidence interval for Survey B
CI(Proportion) \(\displaystyle={\left(\hat{{{p}}}-{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}},\hat{{{p}}}+{z}_{{{c}}}\sqrt{{{\frac{{\hat{{{p}}}{\left({1}-\hat{{{p}}}\right)}}}{{{n}}}}}}\right)}\)
\(\displaystyle={\left({0.82}-{1.96}\times\sqrt{{{\frac{{{0.82}{\left({1}-{0.82}\right)}}}{{{1600}}}}}},{0.82}+{1.96}\times\sqrt{{{\frac{{{0.82}{\left({1}-{0.82}\right)}}}{{{1600}}}}}}\right)}\)
\(\displaystyle={\left({0.801},{0.839}\right)}\)
Answer:
2. The confidence interval is smaller for Survey B.

0
 
Best answer

expert advice

Have a similar question?
We can deal with it in 3 hours

Relevant Questions

asked 2021-08-08
Survey A asked 1000 people how they liked the new movie Avengers: Endgame and \(\displaystyle{88}\%\) said they did enjoy it.
Survey B also concluded that \(\displaystyle{82}\%\) of people liked the move but they asked 1600 total moviegoers.
Which of the following is true about this comparison?
The margin of error is the same for both.
The confidence interval is smaller for Survey B.
Increasing the number of people asked does not change the \(\displaystyle{95}\%\) confidence interval.
Survey A is more accurate since the percentage is higher.
Survey A has more approvals than Survey B.
Survey A is better than Survey B since it has a higher percentage.
asked 2021-08-03
A poll asked the question, "What do you think is the most important problem facing this country today?" Twenty-five percent of the respondents answered "crime and violence." The margin of sampling error was plus or minus 3 percentage points. Following the convention that the margin of error is based on a \(\displaystyle{95}\%\) confidence interval, find a \(\displaystyle{95}\%\) confidence interval for the percentage of the population that would respond "crime and violence" to the question asked by the pollsters.
Lower limit \(\displaystyle\%\)
Upper limit \(\displaystyle\%\)
asked 2021-05-14
When σ is unknown and the sample size is \(\displaystyle{n}\geq{30}\), there are tow methods for computing confidence intervals for μμ. Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When \(\displaystyle{n}\geq{30}\), use the sample standard deviation s as an estimate for σσ, and then use the standard normal distribution. This method is based on the fact that for large samples, s is a fairly good approximation for σσ. Also, for large n, the critical values for the Student's t distribution approach those of the standard normal distribution. Consider a random sample of size n = 31, with sample mean x¯=45.2 and sample standard deviation s = 5.3. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's t distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution?
asked 2021-08-01
Of 1000 randomly selected cases of lung cancer 843 resulted in death within 10 years. Construct a \(\displaystyle{95}\%\) two-sided confidence interval on the death rate from lung cancer.
a) Construct a \(\displaystyle{95}\%\) two-sides confidence interval on the death rate from lung cancer. Round your answer 3 decimal places.
\(\displaystyle?\leq{p}\leq?\)
b) Using the point estimate of p obtained from the preliminary sample what sample size is needed to be \(\displaystyle{95}\%\) confident that the error in estimatimating the true value of p is less than 0.00?
c) How large must the sample if we wish to be at least \(\displaystyle{95}\%\) confident that the error in estimating p is less than 0.03 regardless of the value of p?
...