Table 8.2shows a different random sampling of 20cell phone models. Use this data to calculate a 93% confidence interval for the true mean SAR for cellphones certified for use in the United States. As previously, assume that the population standard deviation is σ= 0.337.

Phone ModelSAR Phone ModelSAR
Blackberry pearl81201.48
Nokia E71x1.53
HTC Evo Design 4G0.8
Nokia N750.68
HTC Freestyle1.15
Nokia N791.4
LG Ally1.36
Sagem Puma1.24
LG Fathom0.77
Samsung Fascinate0.57
LG Optimus Vu0.462
Samsung Infuse 4G
0.2
Motorola Cliq XT1.36
Samsung Nexus S0.51
Motorola Droid pro1.39
Samsung Replenish0.3
Motorola Droid Razr M1.3
Sony W518a walkman0.73
Nokia 7705Twist0.7
ZTE C79 0.869

Short Answer

Expert verified

We estimate with 93% confidence that the true population mean is between 0.7996and 1.0724

Step by step solution

01

Given Information

Given in the question, the value as

The population standard deviation is σ= 0.337.

02

Explanation

If x-is the sample mean of a random sample of size n from a normal population with unknown variance σ2, a 100(1-α) % Cl on μis given by

x¯zα2σnμx¯+zα2σn (1)

wherezα2is the upper100α2percentage point of the standard normal distribution.

We know that standard deviation isσ=0.337, a random samplen=20and a sample mean

x¯=1.48+0.8++0.86920=0.936

We need to find a 93% confidence interval estimate for the population mean. Therefore,

α2=10.932=0.035zα2=z0.035=1.81 (2)

03

Explanation

The earlier implication was obtained on a probability table for the standard normal distribution.

From equations (1) and (2) we get

0.9361.810.33720μ0.936+1.810.33720

Therefore,93%Clforμis

0.7996μ1.0724

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