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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 α=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:p0.80

Sample proportion is p^=xn

p^=339400=0.8475

Standard Deviation is: σp^=p(1-p)n

σp^=0.8(1-0.8)400=0.02

Zscore is Z=p^-pσp^

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:μ=12versus Ha:μ12

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

student writes: “Because the P-value is small, we reject H0. The data prove that Hais true.” Explain what is wrong with this conclusion.

You are testing H0:μ=10against Ha:μ<10based on an SRS of20

observations from a Normal population. The t statistic is t=2.25

The P-value

a. falls between 0.01 and 0.02.

b. falls between 0.02 and 0.04.

c. falls between 0.04 and 0.05.

d. falls between 0.05 and 0.25.

e. is greater than 0.25.

Stating hypotheses

a. A change is made that should improve student satisfaction with the parking situation at a local high school. Before the change, 37%of students approve of the parking that's provided. The null hypothesis H0:p>0.37H0:p>0.37is tested against the alternative Ha: p=0.37Ha:p=0.37

b. A researcher suspects that the mean birth weights of babies whose mothers did not see a doctor before delivery is less than 3000 grams. The researcher states the hypotheses as

H0:x-=3000grams-5H0:x¯=3000grams

Ha: x-<3000grams Ha:x¯<3000grams

explain what's wrong with the stated hypotheses. Then give correct hypotheses.

A company that manufactures classroom chairs for high school students

claims that the mean breaking strength of the chairs is 300 pounds. One of the chairs collapsed beneath a 220-pound student last week. You suspect that the manufacturer is exaggerating the breaking strength of the chairs, so you would like to perform a test of H0:μ=300Ha:μ<300where μ is the true mean breaking strength of this company’s classroom chairs.

a. The power of the test to detect that μ=294 based on a random sample of 30

chairs and a significance level of α=0.05 is 0.71. Interpret this value.

b. Find the probability of a Type I error and the probability of a Type II error for the test in part (a).

c. Describe two ways to increase the power of the test in part (a).

Battery life A tablet computer manufacturer claims that its batteries last an average of 10.5 hours when playing videos. The quality-control department randomly selects 20 tablets from each day’s production and tests the fully charged batteries by playing a video repeatedly until the battery dies. The quality control department will discard the batteries from that day’s production run if they find convincing evidence that the mean battery life is less than 10.5 hours. Here are a dot plot and summary statistics of the data from one day:

a. State appropriate hypotheses for the quality-control department to test. Be sure to define your parameter.

b. Check if the conditions for performing the test in part (a) are met.

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