In Exercises 1–4, use these results from a USA Today survey in which 510 people chose to respond to this question that was posted on the USA Today website: “Should Americans replace passwords with biometric security (fingerprints, etc)?” Among the respondents, 53% said “yes.” We want to test the claim that more than half of the population believes that passwords should be replaced with biometric security.

Requirements and Conclusions

a. Are any of the three requirements violated? Can the methods of this section be used to test the claim?

b. It was stated that we can easily remember how to interpret P-values with this: “If the P is low, the null must go.” What does this mean?

c. Another memory trick commonly used is this: “If the P is high, the null will fly.” Given that a hypothesis test never results in a conclusion of proving or supporting a null hypothesis, how is this memory trick misleading?

d. Common significance levels are 0.01 and 0.05. Why would it be unwise to use a significance level with a number like 0.0483?

It is given that out of 510 people who responded to a survey, 53% said “yes” to the question of whether they should replace passwords with biometric security.

The following requirements should be satisfied to test the given claim using the normal approximation method:

• The given sample should be a simple random sample.
• The conditions of the binomial probability distribution should be satisfied.
• The expressions\(np \ge 5\)and\(nq \ge 5\)should be true.

a.

The survey consisted of responses that were voluntarily given by 510 people and were based on their opinions and not randomly selected. Thus, the sample cannot be considered a simple random sample.

The following conditions of the binomial distribution must be met:

• There should be a fixed number of trials. Here, the number of trials is fixed and equal to 510.
• The trials should be independent. Here, the respondents have individually and independently answered the survey question. Thus, the trials can be considered independent.
• Each trial should have two outcomes: success and failure. Here, success can be considered the response ‘yes’, and failure can be considered the response ‘no’.
• The probability of success should be the same for all trials. Here, the probability of a ‘yes’ response is the same for each trial.

The value of n is equal to 510.

The value of p is equal to

\[\begin{array}{c}p = 53\% \\ = \frac{{53}}{{100}}\\ = 0.53\end{array}\].

The following values are obtained to evaluate the third requirement:

\(\begin{array}{c}np = \left( {510} \right)\left( {0.53} \right)\\ = 270.3\\ > 5\end{array}\).

Similarly,

\(\begin{array}{c}nq = \left( {510} \right)\left( {1 - 0.53} \right)\\ = 239.7\\ > 5\end{array}\).

Asethe two expressions are true, the third requirement is fulfilled.

As the first requirement and one of the conditions of the second requirement are not satisfied, the given claim cannot be tested using the normal approximation method.

b.

It is stated that if the P is low, the null must go.

This means that if the p-value is less than the level of significance, the null hypothesis is rejected at the given level of significance.

c.

It is stated that if the P is high, the null will fly.

This means that if the p-value comes out to be greater than the significance level, the null hypothesis is accepted.

In practical usage, the null hypothesis is never accepted.

It is either rejected or failed to reject.

In this case, the trick becomes misleading as it suggests accepting the null hypothesis when in reality, it is never accepted but only failsto reject if the p-value is greater than the significance level.

d.

In simple words, the level of significance denotes the probability value that the result of the claim is due to a random chance.

A level of significance equal to 0.01 or 0.05 suggests that there is a 1% or 5% probability that the result about the claim is by chance and does not exist in reality.

If the p-value is lower than this value, it shows that there is a very low probability that the results have occurred by chance.

A rounded-off value is often accepted as the significance level to provide the researcher with an idea of whether the result/decision is due to chance or if it actually exists so that he/she can use the conclusions for future purposes.

Thus, there is no sense in taking a precise value such as 0.0483 as the significance level because it won’t make much sense in deciding the result of the test, and it won’t be easy for the researcher to compare the significance level with the obtained p-value.