The time needed for all college students to complete a certain paper-and-pencil maze follows a normal distribution with a population mean of 30 seconds and a population standard deviation of 3 seconds. You wish to see if the mean time μ (where μ stands for population mean) is changed by vigorous exercise, so you have a group of nine college students exercise vigorously for 30 minutes and then complete the maze. It takes them an sample average of 31.2 seconds to complete the maze. Use this information to test the hypotheses:

H0: μ = 30,

Ha: ≠30

at the significance level of 0.01.

You conclude:


a) There is not enough evidence to support the claim.

b) There is enough evidence to support the claim.

Answers

Answer 1
Answer:

Answer:

The correct option is a

Step-by-step explanation:

From the question we are told that

   The population mean is  \mu = 30\ seconds

    The standard deviation is  \sigma = 3 \ seconds

     The sample size is  n  = 9

    The null hypothesis is H_o: \mu  = 30

     The alternative hypothesis is  H_a  :  \mu \ne 30

    The level of significance is  \alpha = 0.01

    The sample mean is  \= x = 31.2

Generally the test statistics is mathematically represented as

      z = ( \= x - \mu )/( (\sigma)/(√(n) ) )

=>   z = ( 31.2 - 30)/( (3)/(√(9) ) )

=>   z = 1.2

From the z table  the area under the normal curve to the right corresponding to  1.2  is    

         P(Z >  1.2) =  0.11507

Generally the p-value is mathematically represented as

         p-value = 2 * P(Z >  1.2)

=>       p-value = 2 *0.11507

=>       p-value = 0.230

From the value obtained w can see that

         p-value  > \alpha hence

The decision rule is  

Fail to reject the null hypothesis

The conclusion is there is not enough evidence to support the claim

Answer 2
Answer:

Final answer:

To test the hypotheses, use a one-sample t-test to compare the sample mean to the population mean. Calculate the t-value and compare it to the critical t-value at the significance level of 0.01. If the calculated t-value falls within the critical region, reject the null hypothesis. Otherwise, fail to reject the null hypothesis.

Explanation:

To test the hypotheses, we can use a one-sample t-test since we know the population standard deviation and the sample size is small. The null hypothesis (H0) is that the population mean (μ) is equal to 30 seconds, while the alternative hypothesis (Ha) is that the population mean is not equal to 30 seconds.

Using the given information, we can calculate the t-value by using the formula:

t = (sample mean - population mean) / (population standard deviation / sqrt(sample size))

Once we have the t-value, we can compare it to the critical t-value at the significance level of 0.01. If the calculated t-value falls within the critical region, we reject the null hypothesis. If not, we fail to reject the null hypothesis and conclude that there is not enough evidence to support the claim.

Learn more about Hypothesis testing here:

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Answer:

x=2

Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Answers

Answer:

k -= 1

Step-by-step explanation:

For the first picture :
The answer is k=2

For the second picture:
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Answers

Answer:

The expression which is equivalent to

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Todd and Garrett began arguing about who did better on their tests, but they couldn't decide who did better given that they took different tests. Todd took a test in Math and earned a 74.6, and Garrett took a test in English and earned a 68.8. Use the fact that all the students' test grades in the Math class had a mean of 70.6 and a standard deviation of 11.9, and all the students' test grades in English had a mean of 63.7 and a standard deviation of 8.6 to answer the following questions. Required:
a. Calculate the z-score for Todd's test grade.
b. Calculate the z-score for lan's test grade.
c. Which person did relatively better?

Answers

Part A

For Todd's class, we have this given info:

  • mu = 70.6 = population mean of math scores
  • sigma = 11.9 = population standard deviation of math scores

Compute the z score for x = 74.6

z = (x-mu)/sigma

z = (74.6 - 70.6)/(11.9)

z = 0.34 approximately

Side note: Convention usually has z scores rounded to two decimal places. If your teacher instructs otherwise, then go with those instructions of course.

Answer:  0.34

======================================================

Part B

For Garret's English class, we have:

  • mu = 63.7
  • sigma = 8.6

Compute the z score for x = 68.8

z = (x-mu)/sigma

z = (68.8 - 63.7)/(8.6)

z = 0.59 approximately

Answer:  0.59

======================================================

Part C

Garret has the higher z score, which means that Garret did relatively better to his classmates compared to Todd's performance (in relation to his classmates). The z score is the distance, in units of standard deviation, the score is from the mean. Positive z values are above the mean, while negative z values are below the mean.

Answer: Garret did relatively better