# 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.

The correct option is a

Step-by-step explanation:

From the question we are told that

The population mean is

The standard deviation is

The sample size is  n  = 9

The null hypothesis is

The alternative hypothesis is

The level of significance is

The sample mean is

Generally the test statistics is mathematically represented as

=>

=>

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

Generally the p-value is mathematically represented as

=>

=>

From the value obtained w can see that

hence

The decision rule is

Fail to reject the null hypothesis

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

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.

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## Related Questions

Having trouble finding x

x=2

Step-by-step explanation:

All of the angles equal 360. So if side z is equal to 100 then 360-200 = 160. The side on the left also is equal to the right side. 160-140 =20. 20/10 is 2. X is equal to 2.

which of the following statistic measures the most frequently occuring value in a set of data A.) range B.) mean C.) mode D.) median ​

This is mode defined

Step-by-step explanation:

Helllllpppppp pleasxee huryyyyy

Bottom left

Step-by-step explanation:

Function: each x only has one y, and this is the only graph that fits the description

The answer is the bottom left one!

Pls help me pls pls il need help i will mark brainliest

k -= 1

Step-by-step explanation:

For the first picture :

For the second picture:
The answer is graph 2 because it goes through the origin(0,0) and it creates a straight line

For the third picture:
B. Rebecca is incorrect because 7 cups of flour will make 14 batches of cookies

Which expression is equivalent to b^m x b^n? A. b^m+n B. b^m-n C. b^mx n D. b^m/n

The expression which is equivalent to

Option A

The formula for multiplying exponents are such below.

(b^m)^n = b^mn
b^m/b^n=b^(m-n)
b^m x b^n=b^(m+n)

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?

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.

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

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