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White raven [17]
3 years ago
7

397 of 514 randomly selected U.S. adults interviewed said they would not be bothered if the National Security Agency collected r

ecords of personal telephone calls they had made. Is there sufficient evidence to conclude that a majority of U.S. adults feel this way? Test the appropriate hypotheses using a 0.01 significance level. (For z give the answer to two decimal places. For P give the answer to four decimal places.)
z =

P =
Mathematics
1 answer:
Afina-wow [57]3 years ago
8 0

Answer:

z=14.73

p=0.000

for 99% confidence level the null hypothesis is rejected, thus majority of us adults would not bothered if the NSA collected personal records.

Step-by-step explanation:

H_{0}: Half of US adults would not bothered if the NSA collect records of personal telephone calls

H_{A}: Majority of the adults would not bothered if the NSA collect records of personal telephone calls

According to the sample, z-value can be found using the formula:

z=\frac{X-M}{\sqrt{n*p*(1-p)} } where

  • X is the adults, who would not be bothered in the survey (397)
  • M is the mean of the distribution of null hypothesis (257)
  • n is the sample size (514)
  • p is the proportion of the sample, who said they would not bothered ( \frac{397}{514} ≈ 0.77)

Putting these numbers in the formula,

z=14.73  and corresponding p value is p≈0.000

Since p<0.01,  for 99% confidence level we reject the null hypothesis, thus majority of us adults would not bothered if the NSA collected personal records.

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