/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q19E The melting point of each of 16 ... [FREE SOLUTION] | 91影视

91影视

The melting point of each of 16 samples of a certain brand of hydrogenated vegetable oil was determined, resulting in \(\overline x = 94.32\). Assume that the distribution of the melting point is normal with 蟽 =1.20.

a.Test H0: 碌 =95 versus Ha: 碌鈮 95 using a two -tailed level .01 test.

b.If a level .01 test is used, what is 尾(94), the probability of a type II error when 碌=94?

c.What value of n is necessary to ensure that 尾(94) = .1 when 伪 = .01?

Short Answer

Expert verified

a)The P value\({\rm{0}}{\rm{.0232}}\) is bigger than \(\alpha {\rm{ = 0}}{\rm{.01}}\), do not reject hypothesis \({H_0}\).

b) \(\beta (94) = 0.2266\)

c) \(n = 22\)

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).

The z values are taken from the appendix table A.3 as well as \(\Phi (z)\) values.

Testing \({H_0}:\mu = 95\) versus \({H_a}:\mu \ne 95\) based on a sample of size \(n = 16\), which is from a normal population with standard deviation \(\sigma = 1.2\).

Assumption that the sample is from normal population distribution with the known standard deviation allows using the test statistic,

\(Z = \frac{{\overline X - {\mu _0}}}{{\sigma /\sqrt n }}\)

The value of the statistic, for \(\overline x = 94.32\), the test statistic value is,

\(\begin{array}{l}z = \frac{{\overline x - {\mu _0}}}{{\sigma /\sqrt n }}\\z = \frac{{94.32 - 95}}{{1.2/\sqrt {16} }}\\z = - 2.27\end{array}\)

03

Step 3:Solution for part a): P value.

The P value for the two sides alternative hypothesis is,

\(\begin{array}{l}{\rm{P(Z > - z and Z < z) = P(Z > 2}}{\rm{.27 and Z < - 2}}{\rm{.27)}}\\{\rm{ = 2}}\Phi {\rm{( - 2}}{\rm{.27)}}\\{\rm{ = 2(0}}{\rm{.0116)}}\\{\rm{ = 0}}{\rm{.0232}}\end{array}\)

The P value\({\rm{0}}{\rm{.0232}}\) is bigger than \(\alpha {\rm{ = 0}}{\rm{.01}}\), do not reject hypothesis \({H_0}\).

04

Step 4:Solution for part b).

Using the two-tailed level \(0.01\) test, then

\(\begin{array}{l}{z_{\alpha /2}} = {z_{0.005}}\\ = 2.58\end{array}\)

Type II error probability \(\beta (\mu ')\) for a level \(\alpha \) test, when alternative hypothesis is \({H_a}:\mu \ne {\mu _0}\) is ,

\(\beta (\mu ') = \Phi \left( {{z_{\alpha /2}} + \frac{{{\mu _0} - \mu '}}{{\sigma /\sqrt n }}} \right) - \Phi \left( { - {z_{\alpha /2}} - \frac{{{\mu _0} - \mu '}}{{\sigma /\sqrt n }}} \right)\)

The type II error when \(\mu ' = 94\) is,

\(\begin{array}{l}\beta (94) = \Phi \left( {2.58 + \frac{{95 - 94}}{{1.2/\sqrt {16} }}} \right) - \Phi \left( { - 2.58 - \frac{{95 - 94}}{{1.2/\sqrt {16} }}} \right)\\ = \Phi (5.91) - \Phi (0.75)\\ = 0.9999 - 0.7733\\\beta (94) = 0.2266\end{array}\)

05

Step 5:Solution for part c).

The required sample size \(n\) for which a level \(\alpha \) test produces \(\beta (\mu ') = \beta \) for upper or lower test is,

\(n = {\left( {\frac{{\sigma ({z_\alpha } + {z_\beta })}}{{{\mu _0} - \mu '}}} \right)^2}\)

From the appendix or a software for \({z_{\alpha /2}} = 2.58\) and for \(\beta = 0.01,{z_\beta } = 1.28\)

Hence the necessary sample size is

\(\begin{array}{l}n = {\left( {\frac{{1.2(2.58 + 1.28)}}{{95 - 94}}} \right)^2}\\n = 21.46\end{array}\)

But the integer is needed, round it up to \(n = 22\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

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

Many older homes have electrical systems that use fuses rather than circuit breakers. A manufacturer of 40-amp fuses wants to make sure that the mean amperage at which its fuses burn out is in fact 40. If the mean amperage is lower than 40, customers will complain because the fuses require replacement too often. If the mean amperage is higher than 40, the manufacturer might be liable for damage to an electrical system due to fuse malfunction. To verify the amperage of the fuses, a sample of fuses is to be selected and inspected. If a hypothesis test were to be performed on the resulting data, what null and alternative hypotheses would be of interest to the manufacturer? Describe type I and type II errors in the context of this problem situation.

Each of a group of \(20\) intermediate tennis players is given two rackets, one having nylon strings and the other synthetic gut strings. After several weeks of playing with the two rackets, each player will be asked to state a preference for one of the two types of strings. Let \(p\) denote the proportion of all such players who would prefer gut to nylon, and let \(X\) be the number of players in the sample who prefer gut. Because gut strings are more expensive, consider the null hypothesis that at most \(50\% \) of all such players prefer gut. We simplify this to \({H_0}:p = .5\), planning to reject \({H_0}\) only if sample evidence strongly favors gut strings.

a. Is a significance level of exactly \(.05\) achievable? If not, what is the largest a smaller than \(.05\) that is achievable?

b. If \(60\% \) of all enthusiasts prefer gut, calculate the probability of a type II error using the significance level from part (a). Repeat if 80% of all enthusiasts prefer gut.

c. If \(13\) out of the \(20\) players prefer gut, should \({H_0}\) be rejected using the significance level of (a)?

To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of \(\overline x = 52.7\) and a sample standard deviation of s = 4.8. The conduits were manufactured with the specification that true average penetration be at most 50 mils. They will be used unless it can be demonstrated conclusively that the specification has not been met. What would you conclude?

Pairs of P-values and significance levels, 伪, are given.

For each pair, state whether the observed P-value would lead to rejection of H0 at the given significance level.

a.P颅-value = .084, 伪= .05

b.P颅-value = .003, 伪= .001

c.P-颅value = .498, 伪= .05

d.P-颅value = .084, 伪= .10

e.P-颅value = .039, 伪= .01

f.P-颅value = .218, 伪 = .10

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.