Question

(c) it is desired to test $h_{0}: p = 0.6$ versus $h_{1}: p < 0.6$. compute the test statistic $z$. round the answer to at least two decimal places.
the test statistic is $-0.88$.
part: $3 / 4$
part 4 of 4
(d) compute the $p$-value. round the answer to at least four decimal places.
$p$-value:
do you reject $h_{0}$ at the $0.1$ level?
yes
no

Explanation:

Step1: Find the P - value

Since \(H_1:p < 0.6\), this is a left - tailed test. The test statistic \(z=-0.88\).
We use the standard normal distribution table (or a calculator with a normal distribution function). The P - value for a left - tailed test with \(z\) - score \(z\) is \(P(Z<z)\).
Using a standard normal table or a calculator (e.g., in Excel, we can use the function \(NORM.S.DIST(-0.88,TRUE)\)), we find that \(P(Z < - 0.88)\).

Step2: Calculate the P - value

From the standard normal table, \(P(Z < - 0.88)=0.1894\) (using a more accurate calculation, for example, with a calculator: \(P(Z < - 0.88)\approx0.1894\)).

Step3: Compare the P - value with the significance level

The significance level \(\alpha = 0.1\). Since \(P - value=0.1894>0.1=\alpha\).

Answer:

The \(P\) - value is \(0.1894\). Do you reject \(H_0\) at the \(0.1\) level? No.