Question

In each part, we have given the value obtained for the test statistic, $z$, in a one-mean $z-$test. We have also specified whether the test is two tailed, left tailed, or right tailed. Determine the $P-$value in each case and decide whether, at the $5\%$significance level, the data provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.

a. $z=-1.25$; left-tailed test

b. $z=2.36$; right-tailed test

c. $z=1.83$; two-tailed test

Short answer

For a. the data does not provide sufficient evidence, for b, the data provides sufficient evidence; and for c. the data does not provide sufficient evidence.

Step-by-step solution

Step 1. Given Information

The value obtained for the test statistic, $z$, in one-mean$z-$test.
The tests have been specified to be left-tailed, right-tailed and two-tailed respectively.

Step 2. Solving for a. 

The test static,$z=-1.25$ , which is left-tailed test.
$P-$value $=P(Z\le z)$

$=P(Z\le -1.25)$

$=0.1056$
The $P-$value is $0.1056$, which is greater than$5\%$ level of significance.
That is, $P-$value$>\alpha =0.05$.
Thus, we fail to reject our null hypothesis$H_0$.
Therefore, the data does not provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis or not at the $5\%$significance level.

Step 3. Solving for b.

The test static,$z=2.36$ , which is right-tailed test.
$P-$value $=P(Z\ge z)$

$=P(Z\ge 2.36)$

$=1-P(Z<2.36)$

$=1-0.9909$

role="math" localid="1651239866313" $=0.0091$
The $P-$value is $0.0091$, which is less than$5\%$ level of significance.
That is, $P-$value$<\alpha =0.05$.
Thus, we reject our null hypothesis$H_0$.
Therefore, the data does provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis or not at the$5\%$significance level.

Step 4. Solving for c.

The test static, $z=1.83$, which is two-tailed test.

$P-$value$=2P(Z\ge z)$
role="math" localid="1651240615811" $=2P(Z\ge 1.83)$

$=2[1-P(Z<1.83\left)\right]$

$=2[1-0.9664]$

$=2(0.0336)$

$=0.0672$

The $P-$value is $0.0672$, which is greater than $5\%$level of significance.That is, $P-$valuerole="math" localid="1651240116209" $>\alpha =0.05$.
Thus, we fail to reject our null hypothesis$H_0$.
Therefore, the data does not provide sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis or not at the$5\%$ significance level.