Question

Information on a large packet of seeds indicates that the germination rate is p0=92%. Some researchers...

Information on a large packet of seeds indicates that the germination rate is p0=92%. Some researchers claim that the proportion of germination should be higher than that.

(a) (1 point) For testing this claim, what is the appropriate null hypothesis and alternative hypothesis?

(b) (2 points) Now 300 seeds are chosen randomly with 290 germinated to test this claim. Compute the test statistic and get the rejection region at the significance level α = 0.05.

(c) (2 points) Report the p value.

(d) (1 point) Given the significance level α = 0.05, make a conclusion about your test. Please interpret the conclusion.

(e) (1 point) Based on the conclusion you made in last step, what possible error could you make? Type I or Type II?

(f) (3 points) If the true germination rate p ∗ = 94%, what’s the power of the test?

Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

Dear your satisfaction is more important, I provided answers with detailed explanation. If you are satisfied with my explanations please THUMBS UP.. Thanks inadvance for your support dear....

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Information on a packet of seeds claims that the germination rate is 92%. What’s the probability...
Information on a packet of seeds claims that the germination rate is 92%. What’s the probability that more than 95% of the 200 seeds in the packet will germinate?
Suppose that I buy a packet of seeds that carries the claim that the germination rate...
Suppose that I buy a packet of seeds that carries the claim that the germination rate for seeds of this type is 93%. Assume that my packet of seeds contains a random sample of seeds, so that the germination of one seed is independent of the germination of any other seed. If there are 120 seeds in this packet, what is the approximate probability that I get a germination rate of less than 0.9? That is, what is P(p-hat <...
Researchers studying infant head circumferences (in centimeters) wish to test if the mean head circumference differs...
Researchers studying infant head circumferences (in centimeters) wish to test if the mean head circumference differs from the historical mean of 34.5 cm. The researchers obtain a sample of 15 infants with a mean head circumference of 34.86 cm and a standard deviation of 0.58 cm. At the α = 0.05 significance level, test the claim HO: µ = 34.5 HA: µ ≠ 34.5 a) Is this an upper tail, lower tail, or two-tail test? (5 points) b) Are we...
Researchers are testing commercial air filtering systems made by the Winston Industrial Supply Company and the...
Researchers are testing commercial air filtering systems made by the Winston Industrial Supply Company and the Barrington Filter company. Random samples are tested for each company, and the filtering efficiency is scored on a standard scale, with the results shown. (Higher scores correspond to better filtering.) Winston: n = 18, sample mean = 85.7, and sample standard deviation = 5.8. Barrington: n = 24, sample mean = 80.6, and sample standard deviation = 6.1. (a) At the 0.05 significance level,...
In a two-tailed hypothesis test of the mean using a 0.05 level of significance, researchers calculated...
In a two-tailed hypothesis test of the mean using a 0.05 level of significance, researchers calculated a p-value of 0.03. What conclusion can be drawn? The alternative hypothesis should be rejected because the p-value is so small. The null hypothesis is true because the p-value is less than the level of significance. The alternative hypothesis is 3% likely to be true. The null hypothesis should be rejected because the p-value is less than the level of significance. 1.The alternative hypothesis...
A poll of 2,133 randomly selected adults showed that 94​% of them own cell phones. The...
A poll of 2,133 randomly selected adults showed that 94​% of them own cell phones. The technology display below results from a test of the claim that 92​% of adults own cell phones. Use the normal distribution as an approximation to the binomial​ distribution, and assume a 0.01 significance level to complete parts​ (a) through​ (e). Test of pequals 0.92vs pnot equals 0.92 Sample X N Sample p ​95% CI ​Z-Value ​P-Value 1 1996 2 comma 133 0.935771 ​(0.922098​,0.949444 ​)...
In a recent​ poll, 801 adults were asked to identify their favorite seat when they​ fly,...
In a recent​ poll, 801 adults were asked to identify their favorite seat when they​ fly, and 476 of them chose a window seat. Use a 0.05 significance level to test the claim that the majority of adults prefer window seats when they fly. Identify the null​ hypothesis, alternative​ hypothesis, test​ statistic, P-value, conclusion about the null​ hypothesis, and final conclusion that addresses the original claim. Use the​ P-value method and the normal distribution as an approximation to the binomial...
5. A consumer watchdog group tests the gas mileage for a specific model of car using...
5. A consumer watchdog group tests the gas mileage for a specific model of car using a (random) sample of 14 cars.          This sample has a mean gas mileage of 27.6 mpg and a standard deviation of 3.2 mpg. (a) Find a 99% confidence interval for the mean gas mileage of all cars of this model.                                  [8 pts] (b) Complete each step below to perform a (traditional) hypothesis test to test the claim in bold made by...
9. The data in the following table summarizes information I collected from my 32 algebra students....
9. The data in the following table summarizes information I collected from my 32 algebra students. Test the claim that passing the course and doing the homework are independent. Did all of their homework Passed the Course      Yes                 No Yes 16                    3 No 5                     8 USE A 0.05 SIGNIFICANCE LEVEL TO TEST THE CLAIM: H0: The row and column variables are independent. (generic null hypothesis) H1: The row and column variables are dependent. (generic alternative hypothesis) Test Statistic: X2=_________ Q34(ROUND...
Use the given information to find the P-value. Also use a 0.05 significance level and state...
Use the given information to find the P-value. Also use a 0.05 significance level and state the conclusion about the null hypothesis (reject the null hypothesis or fail to reject the null hypothesis). (a) The test statistic in a right-tailed test is z = 2.00 (b) The test statistic of z = -1.75 is obtained when testing the claim that p = 1/3
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT