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olga55 [171]
3 years ago
13

Ln (1/x)+ln (2x^3)=ln3

Mathematics
1 answer:
Lapatulllka [165]3 years ago
3 0
\ln \frac{1}{x} +\ln2x^3=\ln3
\ln \frac{2x^3}{x} = \ln3
\ln2x^2=\ln3
2x^2=3
x=\sqrt{ \frac{3}{2} }
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Data were collected over a 10-year timespan from a sample of penguins that were randomly given either metal or electronic tags.
taurus [48]

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.

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Can someone help me with this
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Answer:

we need to see the table above

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