Copy of 5Math 243 Hwk sheet on Ch.5 confidence Intervals
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Feb 20, 2024
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Math 243 Hwk sheet on ch.5
Name: M
Maximum Points: 5 pts
Show all your work and interpret the results for full credit. ●
There are 11 questions
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Please make sure to round the decimals to 4 places. Percentages to 2 decimal places.
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EACH QUESTION WORTH 1 point. This will be converted to 5 pts in grade book
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Use Proper symbols and formulas where needed. Interpret the results.
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Explain the answer in the context of the given question. This means to write a complete sentence after you find the answer. Failure to do so will result in partial credit.
1.
Pg.277 pb.5.7 Refer to Pg.270-272 for formulas. In 2013, the Pew Research Foundation reported that “45% of U.S. adults report that they live with one or more chronic conditions”. However, this value was based on a sample, so it may not be a perfect estimate for the population parameter of interest on its own. The study reported a standard error of about 1.2%, and a normal model may reasonably be used in this setting. Create a 95% confidence interval for the proportion of U.S. adults who live with one or more chronic conditions. Also, interpret the confidence interval in the context of the study.
sample proportion=0.45 standard error=0.012 margin of error=0.02352 we are 95% confident that the proportion of U.S adults who live with one or more chronic conditions is between 0.42648=42.65% and 0.47352=47.35%
Lakshmi Balaji Math 243 Hwk sheet on ch.5 -confidence interval Page 1
2.
Refer to Pg.269-274; also pg.302-306 The Pew research poll asked whether respondents
expected to see a female president in their lifetime. 78% of the 1,835 respondents said “yes”. a)
Construct a 90% confidence interval for the proportion of Americans who expect to see a female president in their lifetime and interpret this interval in the context of the data. **First Identify the unknown population parameter: P +/- z* * sqrt(P * (1 - P)/n)
= 0.78 +/- 1.645 * sqrt(0.78 * 0.22/1835)
= 0.78 +/- 0.0159
= 0.7641, 0.7959 we are 90% confident that the true proportion of americans who expect to see a female president is between
76.41% and 79.58%
b)
How would you expect the width of a 98% confidence interval to compare to the interval you calculated in part (a)? Explain your reasoning. The width of the 98% confidence interval will be wider than the width of 90% confidence interval. c)
Now construct the 98% confidence interval. P +/- z* * sqrt(P * (1 - P)/n)
= 0.78 +/- 2.33 * sqrt(0.78 * 0.22/1835)
= 0.78 +/- 0.0225
= 0.7575, 0.8025 we are 98% confident that the true proportion of americans who expect to see a female president is between
75.75% and 80.25%
d)
In general, what does 98% confidence interval mean to you? Lakshmi Balaji Math 243 Hwk sheet on ch.5 -confidence interval Page 2
The confidence interval tells you how confident you are in your results and in general 98% is very confident. 3.
Refer to Pg.269-274: also pg.302-306 An insurance company checks police records on 582 accidents selected at random and notes that teenagers were at the wheel in 91 of them.
a.
Describe the unknown population parameter that is of interest in the study. Unknown parameter is the proportion of auto accidents that involve teenage drivers.
b.
Construct a 90% confidence interval for the percentage of all auto accidents that involve teenage drivers and interpret it in the context of the data. Use the correct symbols when doing the calculations. n= sample size→ 582 p-hat= sample proportion→ 91/582=0.156 [0.156+ or - 1.64 * sr 0.156(1- 0.156)/582 =[0.156 + or - 0.025] =[0.132, 0.181] we are 90% confident that the true proportion of all auto accidents that involve teenage drivers are between 13.2% and 18.1% c.
Construct a 95% confidence interval for the proportion of all auto accidents that involve teenage drivers and compare it to the 90% confidence interval that you constructed. 0.156+ or - 1.96* sr 0.159(1-0.156)/582 =0.156+ and - 0.030 =0.127, 0.186 As the 95% confidence interval has greater margin of error. It is wider than 90% confidence interval.
d.
A politician urging tighter restrictions on drivers’ licenses issued to teens says, “In one of every five auto accidents, a teenager is behind the wheel.” Does your 95% confidence interval support or contradict this statement? Explain. claim: 1 of every 5, auto accidents involves a teenager p= =0.2 ⅕
Lakshmi Balaji Math 243 Hwk sheet on ch.5 -confidence interval Page 3
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As both the confidence intervals do not include 0.2 (claimed value), our confidence interval contradicts the statement.
e.
Refer to pg.308-309 If we wanted the margin of error to be no larger than 1% at a 95% confidence level, about how many auto accidents would we need to survey? E= margin of error= 1%= 0.01 confidence level= 95%= 0.95 level of significance= 1-0.95= 0.05 critical value of z for two tailed at 0.05 level of significance= z= 1.96 sample proportion=p=0.156 n= (z/E)sq * p * (1-p) = (1.96/0.01)sq * 0.156 * (1-0.156) =5067.4 ~5068
f.
Without doing any calculations, describe what would happen to the confidence interval if we used a large sample.
If we increase the sample size (or use a larger sample size), the margin of error will decrease, hence, the width of confidence interval will decrease.
Lakshmi Balaji Math 243 Hwk sheet on ch.5 -confidence interval Page 4