Question

Let p denote the proportion of students at a particular university that use the fitness center on campus on a regular basis. For a large-sample z test of H0: p = 0.5 versus Ha: p > 0.5, find the P-value associated with each of the given values of the test statistic. (Enter your answers to four decimal places.)
(a) 1.30
(b) 0.77
(c) 1.98
(d) 2.65
(e) −0.16

Answer 1

Answer:

a) P-value = 0.0968

b) P-value = 0.2207

c) P-value = 0.0239

d) P-value = 0.0040

e) P-value = 0.5636

Step-by-step explanation:

As the hypothesis are defined with a ">" sign, instead of an "≠", the test is right-tailed.

For this type of test, the P-value is defined as:

$P-value=P(z>z^*)$

being z* the value for each test statistic.

The probability P is calculated from the standard normal distribution.

Then, we can calculate for each case:

(a) 1.30

$P-value=P(z>1.30) = 0.0968$

(b) 0.77

$P-value=P(z>0.77) = 0.2207$

(c) 1.98

$P-value=P(z>1.98) = 0.0239$

(d) 2.65

$P-value=P(z>2.65) = 0.0040$

(e) −0.16

$P-value=P(z>-0.16) = 0.5636$