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
Lina20 [59]
3 years ago
6

Data were collected over a 10-year timespan from a sample of penguins that were randomly given either metal or electronic tags.

One variable examined is the length of foraging trips. Longer foraging trips can jeopardize both breeding success and survival of chicks waiting for food. Mean length of 344 foraging trips was 12.70 days for metal-tagged penguins. Mean length of 512 foraging trips was 11.60 days for electronic-tagged penguins. An estimate of the standard error for this difference of means is SE=0.283
Mathematics
1 answer:
taurus [48]3 years ago
7 0

Question:

Data were collected over a 10-year timespan from a sample of 100 penguins that were randomly given either metal or electronic tags. One variable examined is the length of foraging trips. Longer foraging trips can jeopardize both breeding success and survival of chicks waiting for food. Mean length of 344 foraging trips was 12.70 days for metal-tagged penguins. Mean length of 512 foraging trips was 11.60 days for electronic-tagged penguins. An estimate of the standard error for this difference of means is SE=0.283.

a) We will address the question of whether foraging trips are longer on average among metal-tagged penguins than among electronic-tagged penguins. State hypotheses in terms of two means.

b. Calculate the sample statistic xM - xE

c. Calculate a t-test statistic, using the given estimate of SE.

d. What are the correct (conservative) degrees of freedom for a t-distribution for this test?

e. Use t-distribution methods to find the p-value and draw a rough curve with appropriate shaded region. You may give either a precise p-value from software or bounds on a p-value from Table A.

f. State the conclusion of the test in context, using nontechnical language.

Answer:

See explanation below

Step-by-step explanation:

a)  The null and alternative hypotheses are:

H0 : uM - uE = 0

H1 : uM - uE > 0

b) Calculating the sample statistic,

xM-xE, we have:

Given xM = 12.70 days

xE = 11.60 days

xM - xE = 12.70 - 11.60 = 1.10 days

Therefore sample statistic = 1.10

c) Calculating the test statistic, using the given estimate of SE, we have:

Given standard error, SE = 0.283

\frac{xM - xE}{SE} = \frac{12.70 - 11.60}{0.283} = 3.89

Therefore, t test = 3.89

d) The correct degrees of freedom.

We have:

Min(n_M-1, n_E-1 ) = Min(344-1, 512-1) = Min(343, 511)

df = 343

e) p-value:

Pvalue = P(Z > 3.89) = 0.00000602

Since p value is less than significance level of 0.05, we reject null hypothesis H0.

f) Conclusion:

There is enough evidence to conclude that foraging trips are longer on average among metal-tagged penguins than among electronic-tagged penguins.

You might be interested in
How much dose 23k of gold cost today???
Dvinal [7]
 $825.61  - It is 37.07 per gram
5 0
3 years ago
Read 2 more answers
Graph the system and name three ordered pair solutions. y < 4 and x < 1
miskamm [114]
The 2nd part:
( -2,-1)
(0,3)
( -4,2)
6 0
3 years ago
Sasaasaasasaassaaaassjsjsjhdhhsjwia
yKpoI14uk [10]

Answer:

vdksirbdhkbwgxi2ndhk djzm ez my dude

4 0
3 years ago
Joseph is working two summer jobs, making $12 per hour babysitting and making $6 per hour walking dogs. In a given week, he can
marin [14]

Answer:

There is no possible values for the number of whole hours walking dog

Step-by-step explanation:

Let x is the number of hours working as a babysitting (x ≥ 0)

Let y is the number of hours working as a walking dogs person (y ≥0)

We have a system of inequalities:

  • 12x + 6y ≥ 90
  • x + y ≤ 9

If Joseph worked 2 hours babysitting:

2+ y ≤ 9 <=> y ≤ 7 (y > 0) and

12x + 6y ≥ 90 <=> y ≥  11

so there is no possible values for the number of whole hours walking dogs:

Hope it will find you well.

5 0
4 years ago
find the probability of having 2,3, or 4 successes in five trials of a binomial experiment in which the probability of success i
Nataly [62]
Let p be the probability of success, p = 0.4.

Pr[2 success] = nC2 * p^2 * (1-p)^3 = 0.346
Pr[3 success] = nC3 * p^3 * (1-p)^2 = 0.230
Pr[4 success] = nC4 * p^4 * (1-p)^1 = 0.077

So, the probability of getting 2, 3, or 4 success, would be 0.356 + 0.230 + 0.077 = 0.663, alternatively, 66.3%.
8 0
4 years ago
Other questions:
  • what does 10/6 equal I just wanna make sure I got the right answer in my calculator because my calculator acts worried sometimes
    11·1 answer
  • How many multiples of 3 are there between 10 and 226?
    7·2 answers
  • A simple random sample of 30 filtered​ 100-mm cigarettes is obtained from a normally distributed​ population, and the tar conten
    11·1 answer
  • Rita spent 50 hours making a set of ceramic tiles.her client paid her $800.how much did Rita make per hour​
    8·2 answers
  • What is 1 fourth+1 fourth +1 fourth+ 1 fourth
    15·2 answers
  • Hello, i am taking math assignments and i was wondering could anyone help me. These problems are new and confusing to me so if y
    13·1 answer
  • F(x)= 4x+r? - 1x?<br> Standard form
    8·1 answer
  • Write 37 and 2/3% as a fraction in simplest form.<br> URGENT
    10·1 answer
  • NO LINKS OR ANSWERING WHAT YOU DON'T KNOW?
    13·2 answers
  • A manager at a shopping mall counts 244 males and n females as they enter the mall. The manager realizes that 25% of all males a
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!