Algal blooms can have negative effects on an ecosystem by dominating its phytoplankton communities. Gonyostomum semen is a nuisance alga infesting many parts of northern Europe. Could the overall biomass of G. semen be controlled by grazing zooplankton species? A research team examined the relationship between the net growth rate of G. semen and the number of Daphnia magna grazers introduced in test tubes. Net growth rate was computed by comparing the initial and final abundance of G. semen in the experiment, with a negative value indicative of a decrease in abundance. Here are the findings:Number of D. magna grazers 1 2 3 4 5 6Net growth rate of G. semen −1.9 −2.5 −2.2 −3.9 −4.1 −4.3Using the above information, software tells us that the least-squares slope is b = −0.5286 with standard error SEb = 0.1059.(a) What is the t statistic for testing H0: β = 0? (Round your answer to two decimal places.)t = (b) How many degrees of freedom does t have?df = Use this table to approximate the P-value of t against the one-sided alternative Ha: β > 0 (as the grazers would be expected to reduce algal growth).0.001 < P < 0.00250.0025 < P < 0.005 0.005 < P < 0.0100.010 < P < 0.0200.020 < P < 0.025(c) What do you conclude? (Use α = 0.05.)There is strong evidence that the population slope β is negative. There is not enough evidence to conclude that the population slope β is negative.

Step-by-step explanation:

Hello!

To study if the biomass of Gonyostomum semen can be controlled by grazing zooplankton species the research team examined the relationship between:

Y: Net grow rate of G. semen in a test tube.

X: Number of D. magna grazers introduced in test tubes.

Since they are testing the relationship between two variables, i.e. how one variable affects the other, the researchers made a linear regression analysis and estimated the slope of the regression:

The point estimate of the slope: b= -0.5286

Standard deviation of the slope: Sb= 0.1059

n= 6 test tubes.

a)

For the hypotheses:

H₀: β = 0

H₁: β ≠ 0

The formula for the t-statistic is:

b)

The statistic has n-2 degrees of freedom.

So in this case you are working with a Student's t with 6-2= 4 degrees of freedom.

c)

Considering a one-sided alternative hypothesis H₁: β > 0, the calculated value of the statistic under the null hypothesis doesn't change, what change is the direction and size of the rejection region. In item a, the hypotheses had a two-tailed rejection region, for this item the rejection region is one-tailed to the right. This means that you'll reject at hight values of the statistic. The p-value is also one-tailed and you'll find it in the right tail of the distribution (That's why it's positive):

P(t₄≥4.99)= 1 - P(t₄<4.99)

Looking at the t-table. For this distribution, the value of 4.99 is between the t values that accumulate P(t₄<4.604)= 0.995 and P(t₄<7.173)= 0.999 of probability, so it will be between:

1 - 0.995= 0.005

and

1 - 0.999= 0.001

Using the given table and considering the value is closer to 4.064 than 7.173, the p-value will be approximate:

0.00025 < P < 0.005

The p-value is less than the significance level, α: 0.05, so the decision is to reject the null hypothesis. You can conclude that the population slope is not negative (β > 0)

⇒There is not enough evidence to conclude that the population slope β is negative.

I hope you have a SUPER day!

answer: the answer is 35%

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step-by-step explanation: he ehe heh step by the thy