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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

Newly purchased tires of a particular type are supposed to be filled to a pressure of 30 psi. Let 碌 denote the true average pressure. A test is to be carried out to decide whether 碌 differs from the target value. Determine the P-value for each of the following z test statistic values.

a.2.10 b. -1.75 c. -.55 d. 1.41 e. -5.3

The paint used to make lines on roads must reflect enough light to be clearly visible at night. Let \(\mu \) denote the true average reflectometer reading for a new type of paint under consideration. A test of \({H_0}:\mu = 20\) versus \({H_n}:\mu > 20\) will be based on a random sample of size n from a normal population distribution. What conclusion is appropriate in each of the following situations?

\(\begin{array}{l}a.n = 15,t = 3.2,\alpha = .05\\b.n = 9,t = 1.8,\alpha = .01\\c.n = 24,t = - 2\end{array}\)

A regular type of laminate is currently being used by a manufacturer of circuit boards. A special laminate has been developed to reduce warpage. The regular laminate will be used on one sample of specimens and the special

laminate on another sample, and the amount of warpage will then be determined for each specimen. The manufacturer will then switch to the special laminate only if it can be demonstrated that the true average amount of warpage for that laminate is less than for the regular laminate. State the relevant hypotheses, and describe the type I and type II errors in the context of this situation.

The following observations are on stopping distance (ft) of a particular truck at \(20mph\) under specified experimental conditions (鈥淓xperimental Measurement of the Stopping Performance of a Tractor-Semitrailer from Multiple Speeds,鈥� NHTSA, DOT HS 811 488, June 2011):

\(32.1 30.6 31.4 30.4 31.0 31.9\)

The cited report states that under these conditions, the maximum allowable stopping distance is \(30\). A normal probability plot validates the assumption that stopping distance is normally distributed.

a. Does the data suggest that true average stopping distance exceeds this maximum value? Test the appropriate hypotheses using \(\alpha = .01\).

b. Determine the probability of a type II error when a 5 .01, \(\sigma = .65\), and the actual value of \(\mu \) is \(31\). Repeat this for \(\mu = 32\) (use either statistical software or Table A.17).

c. Repeat (b) using \(\sigma = .80\) and compare to the results of (b).

d. What sample size would be necessary to have \(\alpha = .01\) and \(\beta = .10\) when \(\mu = 31\) and \(\sigma = .65\)?

The true average diameter of ball bearings of a certain type is supposed to be .5 in. A one-sample t test will be carried out to see whether this is the case. What conclusion is appropriate in each of the following situations?

a.n= 13, t= 1.6, 伪= .05

b.n= 13, t= -1.6, 伪 =.05

c.n= 25, t= -2.6, 伪 =.01

d.n= 25, t= -3.9

See all solutions

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