Hypothesis Testing with One Sample

Question

You are conducting a study to see if the accuracy rate for fingerprint identification is significantly more than 0.87. You use a significance level of .

H₀

₁

You obtain a sample of size in which there are 404 successes.

What is the p-value for this sample? (Report answer accurate to four decimal places.) 

p-value =_________________

The p-value is...

A) less than (or equal to) α

B) greater than α

This p-value leads to a decision to...

A) reject the null

B) accept the null

C) fail to reject the null

As such, the final conclusion is that...

A) There is sufficient evidence to warrant rejection of the claim that the accuracy rate for fingerprint identification is more than 0.87.

B) There is not sufficient evidence to warrant rejection of the claim that the accuracy rate for fingerprint identification is more than 0.87.

C) The sample data support the claim that the accuracy rate for fingerprint identification is more than 0.87.



D) There is not sufficient sample evidence to support the claim that the accuracy rate for fingerprint identification is more than 0.87.

Answer 1

Hello Diana,

You wish to carry out a hypothesis test involving a population proportion (designated p) at a significance level α = .05. Here, p is the proportion of correctly identified fingerprints. From the alternative hypothesis H₁, we note that this is a one-tail test (to the right.) The appropriate test statistic is given by

z = (P - p)/√(pq/n),

where

n = sample size

p = population proportion

q = 1 - p

P = sample proportion

[Note that I am using an upper-case p to distinguish the sample proportion from the population proportion.]

To carry out the test, we use the value of the population proportion assumed in the null hypothesis. That is,

p = .87

and thus,

q = 1 - .87 = .13

Note that using these values for p and q (and under the assumption that H_o is true), the distribution of the test statistic is (approximately) equal to the standard normal distribution.

Also, from the given data, n = 459

and

P = "number of successes"/n = 404/459 ≅ .88017

Now we may calculate the observed value of the test statistic

z = (.88017 - .87)/√(.87·.13/459)

z ≅ .6479

z ≅ .65

Next, we determine the p-value. Since the test is a one-tail test to the right, the p-value is the probability that the test statistic Z is greater than the calculated (observed) value of the test statistic. That is

p-value = Prob[Z>.65]

This is the area under the standard normal distribution to the right of .65. From a table of the standard normal distribution, we find that the area to the left of .65 is .7422, so

p-value = 1 - .7422 = .2578

When using the p-value method, H_o is rejected if and only if the p-value is less than the significance level α. In our case, however, the p-value is clearly greater than the significance level of .05. Therefore, we fail to reject the null hypothesis. Since we fail to reject H_o, this means that we do not have statistically significant evidence in support of H₁. Therefore, we can express our final conclusion as:

D) There is not sufficient sample evidence to support the claim that the accuracy rate for fingerprint identification is more than 0.87.

Let me know if you need any further explanation.

William