Find the unit rate for the following. REMEMBER, UNIT RATE IS HOW MUCH IS ONE OF SOMETHING.

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.

7 0

3 years ago

The Wilson family spend $600 per month on groceries for their family.if this is 15% of their monthly budget, then what is their

Alex_Xolod [135]

Answer:

90

Step-by-step explanation:

8 0

3 years ago

marathon runner runs26.2 miles at the pace of 6.2 minute per mile round off answer to the nearest minute how long will it take t

Scorpion4ik [409]

There are several methods of rounding, but in this case the 'best' one is probably to round up. Rounding down suggests that Kathy actually ran faster.

<span>26.2 (mile) * 6.2 (minutes/mile) = 162.44 rounded up => 163 minutes.</span>

3 0

3 years ago

An Austrian study was completed to determine if untrained sea lions and sea lionesses could follow various experimenter-given cu

Sholpan [36]

Using the z-distribution, it is found that since the test statistic is greater than the critical value for the right-tailed test, this result shows that Zwerg can correctly follow this type of direction by an experimenter more than 50% of the time.

<h3>What are the hypothesis?</h3>

At the null hypothesis, it is tested if Zwerg cannot correctly follow this type of direction by an experimenter more than 50% of the time, that is:

$H_0: p \leq 0.5$

At the alternative hypothesis, it is tested if Zwerg can correctly follow this type of direction by an experimenter more than 50% of the time, that is:

$H_1: p > 0.5$

<h3>Test statistic</h3>

The <em>test statistic</em> is given by:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

In which:

$\overline{p}$ is the sample proportion.

p is the proportion tested at the null hypothesis.

n is the sample size.

For this problem, the parameters are:

$n = 48, \overline{p} = \frac{37}{48} = 0.7708, p = 0.5$

The value of the <em>test statistic</em> is:

$z = \frac{\overline{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$

$z = \frac{0.7708 - 0.5}{\sqrt{\frac{0.5(0.5)}{48}}}$

$z = 3.75$

Considering a <u>right-tailed test</u>, as we are testing if the proportion is greater than a value, with a <u>significance level of 0.05</u>, the critical value for the z-distribution is $z^{\ast} = 1.645$ .

Since the test statistic is greater than the critical value for the right-tailed test, this result shows that Zwerg can correctly follow this type of direction by an experimenter more than 50% of the time.

To learn more about the z-distribution, you can take a look at brainly.com/question/16313918

3 0

3 years ago

Help!<br> Ratios and Porportions

SOVA2 [1]

We learned that ratios are value comparisons, and proportions are equal ratios. Ratios can be written with colons or as fractions. So, to compare the number of girls to boys in a litter of puppies, we can write 2:4 or 2/4 to say that there are two girls to four boys.

7 0

3 years ago