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lara31 [8.8K]
3 years ago
9

A study of 420,076 cell phone users found that 131 of them developed cancer of the brain or nervous system. Prior to this study

of cell phone use, the rate of such
cancer was found to be 0.0224% for those not using cellphones. Complete parts (a) and (b)

a. Use the sample data to construct a 95 i dence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system

%

(Round to three decimal places as needed)
Mathematics
1 answer:
Keith_Richards [23]3 years ago
7 0

Answer:

a) 95% Confidence interval = (0.026%, 0.037%) to 3 d.p

b) The result of the 95% confidence interval does not agree with the previous rate of such cancer because the value 0.0224% does not lie within this confidence interval obtained.

Step-by-step explanation:

Confidence Interval for the population proportion is basically an interval of range of values where the true population proportion can be found with a certain level of confidence.

Mathematically,

Confidence Interval = (Sample proportion) ± (Margin of error)

Sample proportion = (131/420076) = 0.0003118483 = 0.0312%

Margin of Error is the width of the confidence interval about the mean.

It is given mathematically as,

Margin of Error = (Critical value) × (standard Error)

Critical value at 95% confidence interval for sample size of 420,076 is obtained from the z-tables because the sample size is large enough for the properties of the sample to approximate the population properties.

Critical value = 1.960 (from the z-tables)

Standard error = σₓ = √[p(1-p)/n]

n = sample size = 420,076

σₓ = √(0.0003118×0.999688/420076) = 0.0000272421 = 0.0000272

95% Confidence Interval = (Sample proportion) ± [(Critical value) × (standard Error)]

CI = 0.000312 ± (1.96 × 0.0000272)

CI = 0.000312 ± 0.0000534

95% CI = (0.0002586055, 0.0003653945)

95% Confidence interval = (0.000259, 0.000365)

95% Confidence interval = (0.0259%, 0.0365%) = (0.026%, 0.037%) to 3 d.p

b) The result of the 95% confidence interval does not agree with the previous rate of such cancer because the value 0.0224% does not lie within this confidence interval obtained.

Hope this Helps!!!

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