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Try it

[link] shows a different random sampling of 20 cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cell phones certified for use in the United States. As previously, assume that the population standard deviation is σ = 0.337.

Phone Model SAR Phone Model SAR
Blackberry Pearl 8120 1.48 Nokia E71x 1.53
HTC Evo Design 4G 0.8 Nokia N75 0.68
HTC Freestyle 1.15 Nokia N79 1.4
LG Ally 1.36 Sagem Puma 1.24
LG Fathom 0.77 Samsung Fascinate 0.57
LG Optimus Vu 0.462 Samsung Infuse 4G 0.2
Motorola Cliq XT 1.36 Samsung Nexus S 0.51
Motorola Droid Pro 1.39 Samsung Replenish 0.3
Motorola Droid Razr M 1.3 Sony W518a Walkman 0.73
Nokia 7705 Twist 0.7 ZTE C79 0.869

x ¯ = 0.940

  α 2 = 1 C L 2 = 1 0.93 2 = 0.035

Z 0.035 = 1.812

E B M = ( z 0.035 ) ( σ n ) = ( 1.812 ) ( 0.337 20 ) = 0.1365

x ¯ EBM = 0.940 – 0.1365 = 0.8035

x ¯ + EBM = 0.940 + 0.1365 = 1.0765

We estimate with 93% confidence that the true SAR mean for the population of cell phones in the United States is between 0.8035 and 1.0765 watts per kilogram.

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Notice the difference in the confidence intervals calculated in [link] and the following Try It exercise. These intervals are different for several reasons: they were calculated from different samples, the samples were different sizes, and the intervals were calculated for different levels of confidence. Even though the intervals are different, they do not yield conflicting information. The effects of these kinds of changes are the subject of the next section in this chapter.

Changing the confidence level or sample size

Suppose we change the original problem in [link] by using a 95% confidence level. Find a 95% confidence interval for the true (population) mean statistics exam score.

To find the confidence interval, you need the sample mean, x ¯ , and the EBM .

  • x ¯ = 68
  • EBM = ( z α 2 ) ( σ n )
  • σ = 3; n = 36; The confidence level is 95% ( CL = 0.95).

CL = 0.95 so α = 1 – CL = 1 – 0.95 = 0.05

α 2 = 0.025 z α 2 = z 0.025

The area to the right of z 0.025 is 0.025 and the area to the left of z 0.025 is 1 – 0.025 = 0.975.

z α 2 = z 0.025 = 1.96

when using invnorm(0.975,0,1) on the TI-83, 83+, or 84+ calculators. (This can also be found using appropriate commands on other calculators, using a computer, or using a probability table for the standard normal distribution.)

EBM = (1.96) ( 3 36 ) = 0.98

x ¯ EBM = 68 – 0.98 = 67.02

x ¯ + EBM = 68 + 0.98 = 68.98

Notice that the EBM is larger for a 95% confidence level in the original problem.

Interpretation

We estimate with 95% confidence that the true population mean for all statistics exam scores is between 67.02 and 68.98.

Explanation of 95% confidence level

Ninety-five percent of all confidence intervals constructed in this way contain the true value of the population mean statistics exam score.

Comparing the results

The 90% confidence interval is (67.18, 68.82). The 95% confidence interval is (67.02, 68.98). The 95% confidence interval is wider. If you look at the graphs, because the area 0.95 is larger than the area 0.90, it makes sense that the 95% confidence interval is wider. To be more confident that the confidence interval actually does contain the true value of the population mean for all statistics exam scores, the confidence interval necessarily needs to be wider.

Part (a) shows a normal distribution curve. A central region with area equal to 0.90 is shaded. Each unshaded tail of the curve has area equal to 0.05. Part (b) shows a normal distribution curve. A central region with area equal to 0.95 is shaded. Each unshaded tail of the curve has area equal to 0.025.

    Summary: Effect of Changing the Confidence Level

  • Increasing the confidence level increases the error bound, making the confidence interval wider.
  • Decreasing the confidence level decreases the error bound, making the confidence interval narrower.
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Questions & Answers

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Step 1: Find the mean. To find the mean, add up all the scores, then divide them by the number of scores. ... Step 2: Find each score's deviation from the mean. ... Step 3: Square each deviation from the mean. ... Step 4: Find the sum of squares. ... Step 5: Divide the sum of squares by n – 1 or N.
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The sample of 16 students is taken. The average age in the sample was 22 years with astandard deviation of 6 years. Construct a 95% confidence interval for the age of the population.
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Bhartdarshan' is an internet-based travel agency wherein customer can see videos of the cities they plant to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400 a. what is the probability of getting more than 12,000 hits? b. what is the probability of getting fewer than 9,000 hits?
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Bhartdarshan'is an internet-based travel agency wherein customer can see videos of the cities they plan to visit. The number of hits daily is a normally distributed random variable with a mean of 10,000 and a standard deviation of 2,400. a. What is the probability of getting more than 12,000 hits
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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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