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Chapter 8: Q15E (page 333)

Let 碌 denote the true average reaction time to a certain stimulus. For a z test of H0: 碌 =5 versus Ha: 碌 > 5, determine the P-value for each of the following values of the z test statistic.

a.1.42 b. .90 c. 1.96 d. 2.48 e. -.11

Short Answer

Expert verified

a)\(P = 7.78\% \)

b) \(P = 18.41\% \)

c) \(P = 2.50\% \)

d) \(P = 0.66\% \)

e) \(P = 54.38\% \)

Step by step solution

01

Step 1:Null hypothesis.

The null hypothesis, denoted by H0, is the claim that is initially assumed to be true (the 鈥減rior belief鈥� claim). The alternative hypothesis, denoted by Ha, is the assertion that is contradictory to H0.

The null hypothesis will be rejected in favour of the alternative hypothesis only if sample evidence suggests that H0 is false. If the sample does not strongly contradict H0, we will continue to believe in the plausibility of the null hypothesis. The two possible conclusions from a hypothesis-testing analysis are then reject H0 or fail to reject H0.

02

Step 2:Solution for part a).

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = 1.42\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > 1.42)\\ = 1 - P(Z < 1.42)\\ = 1 - 0.9222\\ = 0.0778\end{array}\)

\(P = 7.78\% \)

03

Step 3:Solution for part b).

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = 0.90\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > 0.90)\\ = 1 - P(Z < 0.90)\\ = 1 - 0.8159\\ = 0.1841\end{array}\)

\(P = 18.41\% \)

04

Step 4:Solution for part c).

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = 1.96\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > 1.96)\\ = 1 - P(Z < 1.96)\\ = 1 - 0.9750\\ = 0.0250\end{array}\)

\(P = 2.50\% \)

05

Step 5:Solution for part d).

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = 2.48\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > 2.48)\\ = 1 - P(Z < 2.48)\\ = 1 - 0.9934\\ = 0.0066\end{array}\)

\(P = 0.66\% \)

06

Step 6:Solution for part e).

Given that,

\({H_0}:\mu = 5\)

\({H_a}:\mu > 5\)

\(z = - 0.11\)

Complement rule:

\(P(\overline A ) = 1 - P(A)\)

The P-value is the probability of obtaining a value more extreme or equal to the standardized test statistic z.

Since the alternative hypothesis states , \({H_a}:\mu > 5\), the P-value is the probability to the right of the z-score.

Determine the probability using normal probability table in the appendix, which contain the probability to the left of z-scores(use the complement rule),

\(\begin{array}{l}P = P(Z > - 0.11)\\ = 1 - P(Z < - 0.11)\\ = 1 - 0.4562\\ = 0.5438\end{array}\)

\(P = 54.38\% \)

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Most popular questions from this chapter

The relative conductivity of a semiconductor device is determined by the amount of impurity 鈥渄oped鈥� into the device during its manufacture. A silicon diode to be used for a specific purpose requires an average cut-on voltage of \(.60V\), and if this is not achieved, the amount of impurity must be adjusted. A sample of diodes was selected and the cut-on voltage was determined. The accompanying SAS output resulted from a request to test the appropriate hypotheses.

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