A study of 420,048 cell phone users found that 135 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 0.0333% for those not using cell phones. Complete parts (a) and (b)
a. use the sample data to construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system
?% <p< ?%
round to four decimal places as needed
b. Do cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate of such cancer among those not using cell phones? why or why not?
a. Yes because 0.0333% is not included in the confidence interval
b. No because 0.0333% is not included in the confidence interval
c. Yes because 0.0333% is included in the confidence interval
d. No because 0.0333% is included in the confidence interval
To construct a 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system, we can use the formula for a confidence interval proportion:
CI = p̂ ± z * √( p̂ * (1 - p̂) / n)
Where:
p̂ is the sample proportion (135/420,048)
z is the z-value for a 90% confidence level (1.645, obtained from the standard normal distribution table)
n is the sample size (420,048)
a. Calculating the confidence interval:
p̂ = 135/420,048 = 0.00032142857
CI = 0.00032142857 ± 1.645 * √(0.00032142857 * (1 - 0.00032142857) / 420,048)
Simplifying the formula:
CI = 0.00032142857 ± 1.645 * √(0.00032110869 / 420,048)
CI = 0.00032142857 ± 1.645 * 0.00018439744
CI = 0.00032142857 ± 0.00030307285
CI = (0.00001835572, 0.00062450142)
Rounding to four decimal places, the confidence interval is:
0.0000% < p < 0.0624%
Therefore, the answer to part (a) is:
0.0000% < p < 0.0624%
Now, to answer part (b), we need to compare the rate of cancer in cell phone users (p) to the rate of cancer in non-cell phone users (0.0333%).
Looking at the confidence interval, we see that 0.0333% is not included within the interval. Therefore, we can conclude that cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate among those not using cell phones.
The answer to part (b) is:
a. Yes because 0.0333% is not included in the confidence interval
To construct a confidence interval for the percentage of cell phone users who develop cancer of the brain or nervous system, we can use the sample data provided.
Step 1: Calculate the sample proportion (p̂):
Divide the number of cell phone users who developed cancer (135) by the total number of cell phone users in the sample (420,048):
p̂ = 135 / 420,048 ≈ 0.0003214
Step 2: Calculate the standard error (SE):
The standard error represents the variation in the sample proportion. It is calculated using the following formula:
SE = √(p̂ * (1 - p̂) / n)
where n is the size of the sample (420,048).
SE = √(0.0003214 * (1 - 0.0003214) / 420,048) ≈ 0.0000204
Step 3: Determine the margin of error (ME):
The margin of error represents the range in which the true population proportion is likely to fall. It is calculated by multiplying the standard error by the appropriate critical value (z*), which depends on the desired level of confidence. For a 90% confidence interval, z* is approximately 1.645.
ME = z* * SE = 1.645 * 0.0000204 ≈ 0.0000336
Step 4: Construct the confidence interval:
The confidence interval is calculated by subtracting and adding the margin of error to the sample proportion:
Lower bound (p̂ - ME) = 0.0003214 - 0.0000336 ≈ 0.0002878
Upper bound (p̂ + ME) = 0.0003214 + 0.0000336 ≈ 0.000355
Therefore, the 90% confidence interval estimate of the percentage of cell phone users who develop cancer of the brain or nervous system is approximately 0.0003% to 0.0004%.
For part (b) of the question, we need to consider whether the rate of cancer is different among cell phone users compared to those not using cell phones. Since the confidence interval (0.0003% to 0.0004%) does not contain the rate given for those not using cell phones (0.0333%), we can conclude that cell phone users appear to have a rate of cancer of the brain or nervous system that is different from the rate among those not using cell phones.
Therefore, the correct answer is:
a. Yes because 0.0333% is not included in the confidence interval.
Proportional confidence interval formula:
CI90 = p ± (1.645)[√(pq/n)]
...where p = x/n, q = 1 - p, and n = sample size.
Note: ± 1.645 represents 90% confidence interval.
For p in your problem: 135/420048
For q: 1 - p = q
n = 420048
I let you take it from here to calculate the interval. (Note: convert all fractions to decimals.)
You should be able to answer the second part of this problem once you have calculated the interval.