A researcher used the technique with 9 students and observed that they had a mean of 10.8 hours with a standard deviation of 1.5. A level of significance of 0.05 will be used to determine if the technique performs differently than the traditional method. Assume the population distribution is approximately normal. Find the value of the test statistic. Round your answer to three decimal places.

The degrees of freedom are given by:

And the p value would be given by:

Since the p value is higher than the the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from the traditional methods.

Step-by-step explanation:

Assuming this first part of the problem obtained from the web: "Using traditional methods, it takes 11.0 hours to receive a basic driving license. A new license training method using Computer Aided Instruction (CAI) has been proposed"

Information given

represent the mean height for the sample  

represent the sample standard deviation

sample size  

represent the value that we want to test

represent the significance level

t would represent the statistic  

represent the p value

Hypothesis to test

We want to check if the true mean for this case is equal to 11 or not, the system of hypothesis would be:  

Null hypothesis:  

Alternative hypothesis:  

The statistic would be given by:

 (1)  

Replacing the info given we got:

The degrees of freedom are given by:

And the p value would be given by:

Since the p value is higher than the the significance level of 0.05 we have enough evidence to FAIL to reject the null hypothesis and we can conclude that the true mean is not significantly different from the traditional methods.

the multiples of 15 are 15, 30,45, 60, 75, 90, 105, 120, 135, 150, 165, 180, 195,   210, 225, 240, 255, 270, 285, 300, 315, 330, 345, 360, 375,   390, 405, 420, 435, 450, 465, 480, 495, 510, 525, 540, 555, 570, 585, 600, 615, 630, 645, 660, 675, 690, 705, 720, and 735.

the factors of 15 are 1, 3, 5, and 15.

hope that , fam!

-une, your fren : )

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation: