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62. Blink When two lights close together blink alternately, we “see” one light moving back and forth if the time between blinks is short. What is the
longest interval of time between blinks that preserves the illusion of motion? Ask subjects to turn a knob that slows the blinking until they “see” two lights rather than one light moving. A report gives the results in the form “mean plus or minus the standard error of the mean.” Data for 12 subjects are summarized as 251±45 (in milliseconds).
(a) Find the sample standard deviation sx for these measurements. Show your work.
(b) Explain why the interval 251±45 is not a confidence interval.

Short Answer

Expert verified

(a) The standard deviation sxfor the measurements is 155.8846

(b) The provided interval 251±45is not a confidence interval, because t*is unknown.

Step by step solution

01

Part (a) Step 1 : Given information

Data for 12 subjects are summarized as 251±45 (in milliseconds). To Find the sample standard deviation sx for these measurements.

02

Part (a) Step 2 : Explanation

The standard deviation is divided by the square root of the sample size to get the standard error of the mean.

Standard error of mean is E¯=sxn, and multiply with non both sides.

Then, s=E¯×n.

where, Sample size nis 12, Standard error of mean E¯=45, and the population mean x¯is251.
localid="1650092899445" s=E¯×n=45×12=155.8846
Therefore, the standard deviation is localid="1649913384538" 155.8846.

03

Part (b) Step 1 : Given Information

To explain why the provided interval 251±45 is not a confidence interval.

04

Part (b) step 2 : Explanation

The confidence interval is calculated by using the formula=x¯±t*×E¯.

The interval in the problem is x¯±E¯.

And, t*is unknown.

As a result, the provided interval is not a confidence interval.

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Most popular questions from this chapter

Many teens have posted profiles on sites such as Facebook and My Space. A sample survey asked random samples of teens with online profiles if they included false information in their profiles. Of 170younger teens (ages 12to 14) polled, 117said “Yes.” Of 317older teens (ages 15to 17) polled, 152 said “Yes.”6 A 95% confidence interval for the difference in the population proportions (younger teens – older teens) is 0.120 to 0.297. Interpret the confidence interval and the confidence level

In each of the following situations, discuss whether it would be appropriate to construct a one-sample tinterval to estimate the population mean.

(a) We want to estimate the average age at which U.S. presidents have died. So we obtain a list of all U.S. presidents who have died and their ages at death.

(b) How much time do students spend on the Internet? We collect data from the 32members of our AP Statistics class and calculate the mean amount of time that each student spent on the Internet yesterday.

(c) Judy is interested in the reading level of a medical journal. She records the length of a random sample of 100words from a multipage article. The Minitab histogram below displays the data.

One reason for using a tdistribution instead of the standard Normal curve to find critical values when calculating a level C confidence interval for a population mean is that

(a) zcan be used only for large samples.

(b) zrequires that you know the population standard deviation σ.

(c) zrequires that you can regard your data as an SRS from the population.

(d) the standard Normal table doesn't include confidence levels at the bottom.

(e) a zcritical value will lead to a wider interval than a tcritical value.

True or false: The interval from 2.84to 7.55has a 95%chance of containing the actual population standard deviation S. Justify your answer.

56. The SAT again High school students who take the SAT Math exam a second time generally score higher than on their first try. Past data suggest that the score increase has a standard deviation of about 50 points.
How large a sample of high school students would be needed to estimate the mean change in SAT score to within 2 points with 95% confidence? Show your work.

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