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

Question

31-10-2019
Mathematics

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

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dogacandu dogacandu · Answer 1 · 2019-10-31T23:17:22+01:00

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:

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

[tex]H_{A}[/tex]: 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=[tex]\frac{X-M}{\sqrt{n*p*(1-p)} }[/tex] 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 ( [tex]\frac{397}{514}[/tex] ≈ 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.