Question

Find the p-value based on a standard normal distribution for each of the following standardized test statistics. (a) z = 0.79 for a right tail test for a difference in two proportions Round your answer to two decimal places. p-value = i (b) z = -2.38 for a left tail test for a difference in two means Round your answer to three decimal places. p-value = (c) z = 2.31 for a two-tailed test for a proportion Round your answer to three decimal places. p-value = i

Answer 1

Question (a)

The test statistic is z=0.79, for a right tailed test for a difference in two proportions.

Now, the test is right tailed.

So, the p-value is

p - value P(Z > 0.79

Where, Z is the standard normal variate.

$=1-P(Z\leq 0.79)$

$=1-phi(0.79)$

Where, phi is the distribution function of the standard normal variate.

From the standard normal table, this becomes

==l-0.7852

$=0.2148$

So, the p-value is 0.2148.

Question (b)

The test statistic is z=-2.38 for a left tail test for a difference between two means.

Now, the test is left tailed.

So, the p-value is

$p-value=P(Z<-2.38)$

Where, Z is the standard normal variate.

$=phi(-2.38)$

Where, phi is the distribution function of the standard normal variate.

From the standard normal table, this becomes

$=0.0086$

So, the p-value is 0.0086.

Question (c)

The test statistic is z=2.31 for a two tailed test for proportion.

Now, the test is two tailed.

So, the p-value is

$p-value=P(Z>2.31)+P(Z<-2.31)$

Where, Z is the standard normal variate.

$=1-P(Z<2.31)+P(Z<-2.31)$

$=1-phi(2.31)+phi(-2.31)$

Where, phi is the distribution function of the standard normal variate.

From the standard normal table, this becomes

$=1-0.9896+0.0104$

$=0.0208$

So, the p-value is 0.0208.