1. Question: On a busy Sunday morning, a waitress randomly sampled customers about their preference for morning beverages, Specifically, she wanted to find out how many people preferred coffee over tea. The proportion of customers that preferred coffee was 0.42 with a margin of error 0.07. Cons...
1. On a busy Sunday morning, a waitress randomly sampled customers about their preference for morning
beverages, Specifically, she wanted to find out how many people preferred coffee over tea. The proportion
of customers that preferred coffee was 0.42 with a margin of error 0.07.
Construct a confidence interval for the proportion of customers that preferred coffee.
(0.42 - 0.07), (0.42 + 0.07) = (0.35), (0.49)
2. A company sells juice in 1quart bottles. In a quality control test, the company found the mean volume of
juice in a random sample of bottles was X = 31 ounces, with a marginal error of 3 ounces.
Construct a confidence interval for the mean number of ounces of juice bottled by this company.
(31-3), 31+3) = (28), (34)
3. Randomly selected employees at an office were asked to take part in a survey about overtime. The office
manager wanted to find out how many employees worked overtime in the last week. The proportion of employees
that worked overtime was 0.83, with a margin of error of 0.11.
(0.83 – 0.11), (0.83 + 0.11) = (0.72), (0.94)
4. A random sample or garter snakes were measured, and the proportion of snakes that were longer than 20
inches in length recorded. The measurements resulted in a sample proportion of p = 0.25 with a sampling
standard deviation of Op = 0.05.
Write a 68% confidence interval for the true proportion of garter snakes that were over 20 inches in length.
(.25 - .05). (.25 + .05) = (0.20), (.30)
5. The average number of onions needed to make French onion soup from the population of recipes is unknown.
A random sample of recipes yields a sample mean of x = 8.2 onions. Assume the sampling distribution of the
mean has a standard deviation of 2.3 onions.
Use the Empirical Rule to construct a 95% confidence interval for the true population mean number of onions.
Since 95% falls in 2 SD’s the calculation would be (8.2 – 4.6) “4.6 is the margin of error”, (8.2 + 4.6) = (3.6) , (12.8)
6. In a survey, a random sample of adults were asked whether a tomato is a fruit or vegetable. The survey
resulted in a sample proportion of 0.58 with a sampling standard deviation of 0.08 who stated a tomato is a fruit.
Write a 99.7 confidence interval for the true proportion of number of adults who stated the tomato is a fruit.
(0.58 – 3 x 0.08), (0.58 + 3 x 0.08) =(0.58 – .24), (0.58 + .24) = 0.34 + 0.82
7. A college admissions director wishes to estimate the mean number of students currently enrolled. The age of
random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel
to construct a 90% confidence interval for the population mean age. Round your answer to 2 decimal places and
use increasing order.
Use week 6 worksheet to get mean and SD.
Data
25.8 Mean 23.1043
22.2 Sample Standard Deviation 1.3693
Confidence
Level 0.900
n 32
Mean 23.1043
StDev 1.3693
pop stdev no
SE 0.242060
t 1.696
Margin of
Error 0.410534
Lower Limit 22.693766
Upper Limit 23.514834
Lower margin of error = 22.69 and upper limit is 23.51
8. Suppose that the scores of bowlers in a particular league follow a normal distribution such that a standard
deviation of the population is 12. Find the 95% confidence interval of the mean score for all bowlers in this league
using the accompanying data set of 40 random scores. Round your answers to 2 decimal places using ascending
order.
Lower Limit = 90.78 Upper Limit = 98.22
Confidence
Level 0.950
n 40
Mean 94.5000
StDev 12.0000
pop stdev yes
SE 1.897367
, z 1.960
Margin of
Error 3.718839
Lower Limit 90.781161
Upper Limit 98.218839
9. In the survey of 603 adults, 98 said that they regularly lie to people conducting surveys. Create a 99%
confidence interval for the proportion of adults who regularly lie to people conducting surveys. Use excel to
create the confidence interval rounding to 4 decimal places.
Lower Limit = 0.1238 Upper Limit = 0.2012
Confidence Level 0.990
n 603
Number of
Successes 98
Sample Proportion 0.162521
SE 0.015024
z 2.576
Margin of Error 0.038702
Lower Limit 0.123819
Upper Limit 0.201222
10. In a random sampling of 350 attendees at a minor league baseball game, 184 said that they bought food from
the concession stand. Create a 95%confidence interval for the proportion of fans who bought food from the
concession stand. Use excel to create the confidence interval rounding to 4 decimal places.
Lower limit = 0.4734 Upper Limit = 0.5780
Confidence Level 0.950
n 350
Number of
Successes 184
Sample Proportion 0.525714
SE 0.026691
z 1.960
Margin of Error 0.052314
Lower Limit 0.473400
Upper Limit 0.578028
11. Suppose that the weight of tight ends in a football league are normally distributed such that sigma squared =
1,369. A sample of 49 tight ends was randomly selected and the weights are given in the table below. Use Excel to
create a 95% confidence interval for the mean weight of the tight ends in this league. Rounding your answers to 2
decimal places and using ascending order. (Have to get square root of 1369 which is 37). Population sample is
yes .
Lower limit = 241.42 Upper Limit = 262.14
Confidence 0.950
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