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American Alligators. Refer to Exercise 8.78

a. The mean duration for a sample of 612dives was 322seconds. Find a 995confidence interval for based on that data.

b. Compare the 995confidence intervals obtained here and in Exercise 8.78 (b) by drawing a graph similar to Fig. 8.7on page 327

c. Compare the margins of error for the two 99%confidence intervals.

d. What principle is being illustrated?

Short Answer

Expert verified

Part (a) (311.588,332.412)

Part (b)

Part (c) 31.23

Part (d) The sample size is increased and the confidence level is the same, which provides a decreasing margin of error.

Step by step solution

01

Part (a) Step 1: Given information

x=322,n=612 and =100

02

Part (a) Step 2: Concept

The formula used: Margin of error.

03

Part (a) Step 3: Calculation

Using MINITAB, compute a 99%confidence interval estimate of the population mean

Consider x=322,n=612and =100

Procedure for MINITAB:

Step 1: Select Stat >Basic Statistics >1-Sample Zfrom the drop-down menu.

Step 2: In the Summarized Data section, enter 612as the sample size and 322as the mean.

Step 3: In the Standard deviation box, type 100for s

Step 4: Select Options and enter 322as the level of confidence.

Step 5: In the alternative, select not equal.

Step 6: In all dialogue boxes, click OK.

MINITAB output:

One-Sample Z

The assumed standard deviation =100

NMeanSE Mean99% CI
612322.0004.042(311.588, 332.412)

The population mean has a 99%confidence interval estimate of(311.588,332.412) based on MINITAB output.

04

Part (b) Step 1: Explanation

The confidence interval in Exercise 8.78 is shown in the below graph:

The confidence interval in part (a) is shown in the below graph:

05

Part (c) Step 1: Calculation

When (311.588,332.412)is used, get the margin of error for the 99%confidence interval?

The margin of error is,

Margin of error=332.412-311.5882=20.8242=10.412

Thus, the margin of error for the 99%confidence interval is 10.412

The 99%confidence interval for is (306.77,369.23)based on Exercise 8.78

The margin of error is,

Margin of error=369.23-306.772=62.462=31.23

Thus, the margin of error for the confidence interval is 31.23

Comparison:

When compared to the margin of error determined in Exercise 8.78 the margin of error for this exercise is less.

06

Part (d) Step 1: Explanation

The premise is that the sample size is raised while the confidence level remains constant, resulting in a smaller margin of error.

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Most popular questions from this chapter

Find the confidence level and for

a. 90%confidence interval.

b. 94%confidence interval.

A simple randoes sample is taken from a population and yields the following data for a variable of the population:

find a point estimate for the population standard deviation (i.e., the standard deviation of the variable).

One-Sided One-Mean t-Intervals. Presuming that the assumptions for a one-mean t-interval are satisfied, we have the following formulas for (1-)-level confidence bounds for a population mean :

  • Lower confidence bound: x-t-s/
  • Upper confidence bound: x^+ts/n

Interpret the preceding formulas for lower and upper confidence bounds in words.

Find the confidence level and for

a. 85%confidence interval.

b. 95%confidence interval

M\&Ms. In the article "Sweetening Statistics-What M\&M's Can Teach Us" (Minitab Inc., August 2008), M. Paret and E. Martz discussed several statistical analyses that they performed on bags of M\&Ms. The authors took a random sample of 30 small bags of peanut M\&Ms and obtained the following weights, in grams (g).

a. Determine a 95%lower confidence bound for the mean weight of all small bags of peanut M\&Ms. (Note: The sample mean and sample standard deviation of the data are 52.040gand 2.807grespectively.)

b. Interpret your result in pant (a).

c. According to the package, each small bag of peanut M\&Ms should weigh 49.3gComment on this specification in view of your answer to part (b) It provides equal confidence with a greater lower limit.

Part (c) Because the weight of 49.3g is below the 95% lower confidence bound.

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