1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
horrorfan [7]
3 years ago
6

he Nassau Bahamas Community College (NBC) claims that students who take statistics spend on average 19 hours a week working. The

population standard deviation is not known. Statistics students in MAT 102 believe students work more than 19 hours because they want to make more money.
Mathematics
1 answer:
Virty [35]3 years ago
3 0

Using the t-distribution, it is found that since the <u>test statistic is greater than the critical value</u> for the right-tailed test, it is found that there is enough evidence to conclude that the students' claim that they work more than 19 hours is correct.

At the null hypothesis, it is <u>tested if they spend on average 19 hours a week working</u>, that is:

H_0: \mu = 19

At the alternative hypothesis, it is <u>tested if they spend more than 19 hours a week working</u>, that is:

H_1: \mu > 19

We have the <u>standard deviation for the sample</u>, hence, the <em>t-distribution</em> is used.

The test statistic is given by:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

The parameters are:

  • \overline{x} is the sample mean.
  • \mu is the value tested at the null hypothesis.
  • s is the standard deviation of the sample.
  • n is the sample size.

Searching the problem on the internet, it is found that the values of the <em>parameters </em>are:

\overline{x} = 24.2, \mu = 19, s = 12.59, n = 25

Hence, the value of the <em>test statistic</em> is:

t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{24.2 - 19}{\frac{12.59}{\sqrt{25}}}

t = 2.07

The critical value for a <u>right-tailed test</u>, as we are testing if the mean is greater than a value, with a <u>significance level of 0.05</u> and 25 - 1 = <u>24 df</u> is of t^{\ast} = 1.71

Since the <u>test statistic is greater than the critical value</u> for the right-tailed test, it is found that there is enough evidence to conclude that the students' claim that they work more than 19 hours is correct.

To learn more about the t-distribution, you can take a look at brainly.com/question/13873630

You might be interested in
Decrease £2155.45 by 18.5% Give your answer rounded to 2 DP ​
RSB [31]

Answer:

£1756.70

Step-by-step explanation:

2155.45 * .185 = 398.75825

21.55.45 - 398.75825 = 1756.69175

Hope this helps

3 0
3 years ago
Read 3 more answers
The figure shown has a parallelogram on top and a rectangle below it:
qwelly [4]

Answer:

954.07 square inches

Step-by-step explanation:

You multiply all the mesurements together.

4 0
3 years ago
Read 2 more answers
2 lb 6oz = ________lb
krok68 [10]
2.6 lb

Hope this helps!

5 0
4 years ago
Can someone pls give me the answer to this I need it ASAP
Lady bird [3.3K]
I think it’s A. Remember m stands for the slope of a line
7 0
3 years ago
Helpppppp meeeeeeeer
elixir [45]
1/5 is your answer, sorry if it is wrong but I think it os right :)
5 0
3 years ago
Other questions:
  • Select the statement that correctly describes the solution to this system of equations.
    11·1 answer
  • 5.65 times 3.4 estimate
    14·2 answers
  • Can somone please help me break this down
    10·1 answer
  • Rewrite with only sin x and cos x. <br><br> cos 3x
    8·1 answer
  • Evaluate x2 dv, e where e is the solid that lies within the cylinder x2 + y2 = 4, above the plane z = 0, and below the cone z2 =
    14·1 answer
  • Can anyone help me with number 17?
    8·1 answer
  • Find the mean absolute deviation of the set of data. 6, 6, 8, 10, 10
    8·1 answer
  • Can someone help me with my last homework math related problem thank you.
    8·2 answers
  • I need help with this asap
    14·1 answer
  • Last Sunday 1,575 people visited the amusement park.
    13·2 answers
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!