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Chapter 9: Q 44. (page 582)

Watching grass grow The germination rate of seeds is defined as the proportion of seeds that sprout and grow when properly planted and watered. A certain variety of grass seed usually has a germination rate of 0.80. A company wants to see if spraying the seeds with a chemical that is known to increase germination rates in other species will increase the germination rate of this variety of grass. The company researchers spray a random sample of 400grass seeds with the chemical, and 339of the seeds germinate. Do these data provide convincing evidence at the 伪=0.05 significance level that the chemical is

effective for this variety of grass?

Short Answer

Expert verified

The chemical are effective of grass of germination process.

Step by step solution

01

Given Information

It is given that n=400

x=339

p=0.05

02

Simplification

Null hypothesis is: H0:p=0.80

Alternate hypothesis: Ha:p鈮�0.80

Sample proportion is p^=xn

p^=339400=0.8475

Standard Deviation is: 蟽p^=p(1-p)n

蟽p^=0.8(1-0.8)400=0.02

Zscore is Z=p^-p蟽p^

Z=0.8475-0.800.02=2.375

The probability value is P(x<Z)=0.99123

P(x>Z)=1-0.99123=0.0087745

P=0.0087745<0.05, null hypothesis is rejected.H0:p=0.80

Chemical is effective for variety of grass.

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Most popular questions from this chapter

Making conclusions A student performs a test of H0:渭=12versus Ha:渭鈮�12

at the 伪=0.05significance level and gets a P-value of 0.01. The

student writes: 鈥淏ecause the P-value is small, we reject H0. The data prove that Hais true.鈥� Explain what is wrong with this conclusion.

An opinion poll asks a random sample of adults whether they favor banning ownership of handguns by private citizens. A commentator believes that more than half of all adults favor such a ban. The null and alternative hypotheses you would use to test this claim are

邪.H0:p鈭�=0.5;Ha:p鈭�>0.5H0:p^=0.5;Ha:p^>0.5.

b. H0:p=0.5;Ha:p>0.5H0:p=0.5;Ha:p>0.5.

c. H0:p=0.5;Ha:p<0.5H0:p=0.5;Ha:p<0.5.

d. H0:p=0.5;Ha:p鈮�0.H0:p=0.5;Ha:p鈮�0.5.

e. H0:p>0.5;Ha:p=0.5H0:p>0.5;Ha:p=0.5.

Walking to school A recent report claimed that 13%of students typically walk to school. DeAnna thinks that the proportion is higher than 0.13at her large elementary school. She surveys a random sample of 100students and finds that 17typically walk to school. DeAnna would like to carry out a test at the 伪=0.05significance level of H0:p=0.13versus Ha:p>0.13, where p= the true proportion of all students at her elementary school who typically walk to school. Check if the conditions for performing the significance test are met.

Don't argue Refer to Exercise 2. Yvonne finds that 96 of the 150 students (64%) say they rarely or never argue with friends. A significance test yields a P-value of0.0291 Interpret the P-value.

According to the Bureau of Labor Statistics, the average age of American workers is 41.9years. The manager of a large technology company believes that the company鈥檚 employees tend to be younger, on average. So she takes a random sample of 12 employees and records their ages.

Here are the data:

27 38 32 24 30 47 42 38 27 43 37 33

a. State appropriate hypotheses for testing the manager鈥檚 belief. Be sure to define the parameter of interest.

b. State the conditions for performing a test of the hypotheses in (a), and determine whether each condition is met.

c. The P-value of the test is0.003. Interpret this value. What conclusion would you make?
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