Free Printable Finding Confidence Intervals

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Finding Confidence Intervals

Total questions: 25

Worksheet time: 1hrs 8mins

Name

Class

Date

A sample of 30 women finds the mean height to be 62.3 in and the standard deviation to be 1.5 in. Find the margin of error of women's heights with 95% confidence.

1.699

0.56

0.537

19.325

Why do we use a t-distribution instead of a z-distribution for means?

It reflects greater variability present in small samples.

It allows us to decrease the margin of error without increasing our confidence level.

It allows us to use the population standard deviation.

None of these.

Which of the following changes to a study would result in a narrower confidence interval?

increasing the confidence level, increasing the sample size

decreasing the confidence level, decreasing the sample size

increasing the confidence level, decreasing the sample size

decreasing the confidence level, increasing the sample size.

In a crash test of 15 troopers, collision repair costs are found to have a distribution that is approximately normal, with a mean of $1800 and a standard deviation of $950. Construct a 99% confidence interval for the repair cost.

(1070, 2530)

(1273, 2326)

(1799, 1801)

None of these.

Find the t* value for a 99% confidence interval with a sample of 23 people.

1.714

1.711

2.064

None of these.

A research group wishes to estimate the mean amount of time (in hours) that members of a fitness center spend exercising each week. They want to estimate the mean within a margin of error of 0.5 hours with a 95% level of confidence. Previous data suggests that the standard deviation of the population is 2.2. Which of the following is the smallest sample size they could use?

180

A 90% confidence interval for a population mean is determined to be 800 to 900. If the confidence is increased to 95% confidence while the sample statistics and sample size remain the same, the confidence interval

becomes wider

becomes narrower

does not change

A random sample of 100 visitors to a popular theme park spent an average of $142 on the trip with a standard deviation of $47.5. Which of the following would be the 98% confidence interval for the mean money spent by all visitors to this theme park?

($130.77, $153.23)

($132.77, $151.43)

($132.69, $151.31)

($140.88, $143.12)

A quality control specialist at a glass factory must estimate the mean clarity rating for a new batch of glass using a sample of 18 glass sheets from the batch. Past investigations show that clarity ratings are normally distributed. The specialist decides to use a t-distribution rather than a z-distribution because ...

The sample size is too small for a z-distribution

Clarity ratings for the entire bacth are normally distributed.

The t-distribution will create a narrower interval than z.

The standard deviation for the batch is unknown.

10.

A random sample of 100 Woodrow students found that 53 of them were in possession of a pencil. Create a 90% confidence interval for this sample.

(0.45,0.61)

(0.43,0.63)

(0.53,0.63)

11.

The 99% confidence interval for a proportion is (0.54, 0.72). What is p̂?

0.63

0.18

0.09

0.66

12.

Agricultural researchers plant 100 plots with a new variety of corn and measure the mean yield for these plots in bushels per acre. They treat the 100 plots as a simple random sample of possible plots of corn (as researchers often do) and report a 95% confidence interval for the mean corn yield of (128.4, 131.6) bushels per acre. Which of the following is a correct interpretation of the 95% confidence level?

We are 95% confident that the true mean yield for this new variety of corn is captured by the interval (128.4, 131.6) bushels per acre

If many intervals were constructed this way from many independent sets of 100 plots, 95% of the intervals would capture the true mean corn yield.

If many intervals were constructed this way from many independent sets of 100 plots, 95% of the time the true mean corn yield would be in the interval (128.4, 131.6) bushels per acre.

13.

A 90% confidence interval for the average salary of all CEOs in the electronics industry was constructed using the results of a random survey of 45 CEOs. The interval was ($139,048, $154,144). Give a practical interpretation of the interval.

90% of the sampled CEOs have salaries that fell in the interval $139,048 to $154,144

We are 90% confident that the mean salary of all CEOs in the electronics industry falls in the interval $139,048 to $154,144.

90% of all CEOs in the electronics industry have salaries that fall between $139,048 to $154,144

We are 90% confident that the mean salary of the sampled CEOs falls in the interval $139,048 to $154,144.

14.

A random sample of 85 sixth-graders in a large city take a course designed to improve scores on a reading comprehension test. Based on this sample, a 90% confidence interval for the mean improvement in test scores for all sixth-graders in the city taking this course is found to be (12.6, 14.8). Which of the following are the sample mean and margin of error on which this interval is based?

Sample Mean = 13.7; Margin of Error = 1.1

Sample mean = 13.7; Margin of Error = 2.2

Sample mean is unknown; Margin of Error = 2.2

15.

Find the minimum sample size needed to be 4% within the true proportion if you are 95% confident and p-hat = 0.6

961

577

16.

What is the margin of error for the interval (6.8, 12.9)?

9.85

3.05

3.50

6.8

17.

Does the interval (3, 14) support the claim that the mean of a data set is 2.9

Yes

Not enough information.

18.

Which of the following is NOT a parameter?

p̂

19.

The 99% confidence interval for a proportion is (0.54, 0.72). What is p̂?

0.63

0.18

0.09

0.66

20.

A researcher wishes to determine the mean energy consumption of a new light bulb. She takes a random sample of 41 bulbs and determines that the mean consumption is 1.3 watts per hour with a standard deviation of 0.7. When constructing a 97% confidence interval, which would be the most appropriate value of the critical value?

2.072

2.25

2.704

2.807

21.

A survey of 280 homeless persons showed that 63 were veterans. Construct a 90% confidence interval for the proportion of homeless persons who are veterans.

(0.161 to .289)

(0.176 to 0.274)

(0.184 to 0.266)

(0.167 to 0.283)

22.

You take a poll and find that 34% of the 100 people you asked dislike school lunches. Using a 95% confidence level, what is the margin of error?

0.093

0.047

.034

.34

23.

A sample of 45 has mean of 7.1 and Standard Error of 2.3.

Construct a 90% Confidence Interval for the population mean µ.

(3.23, 10.97)

(6.52, 7.68)

(6.54, 7.66)

(3.32, 10.88)

24.

An IQ test was given to a simple random sample of 75 students at a certain college. The sample mean score was 105.2. Scores on this test are known to have a standard deviation of 10. It is desired to construct a 90% confidence interval for the mean IQ scores of the students at the college.

What is the point estimate?

105.2

0.90

25.

Find the Standard Error.

105.2

1.155

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Finding Confidence Intervals

Total questions: 25

Worksheet time: 1hrs 8mins

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