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2 years ago

PLZ HELLLP 50 POINTS, WILL GIVE BRAINLY PLZZZZZZZZZZ

Mathematics

2 answers:

ki77a [65]2 years ago

7 0

Answer:

Transgenic Organism

My reasoning:

A transgenic organism is an organism which has been modified with genetic material from another species. The biggest use of transgenic organisms is for agriculture growth, the development of genetically modified crops.

They also research with transgenic mice. Mice can be modified with human DNA for the purpose of testing medical treatments and seeing how they might behave in a human.

Researchers are concerned though that interbreeding between transgenic and wild organisms could have unforeseen consequences and could reign havoc on our ecosystem.

~That's All Folks~

-Siascon

Ludmilka [50]2 years ago

4 0

transgenic organisms is the answer

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A corporation randomly selects 150 salespeople and finds that 66% who have never taken a self- improvement course would like to

Citrus2011 [14]

Answer:

We conclude that there is no difference in the two population proportions using α = 0.05.

Step-by-step explanation:

We are given that a corporation randomly selects 150 salespeople and finds that 66% who have never taken a self- improvement course would like to do so.

The firm did a similar study 10 years ago and found that 70% of a random sample of 160 salespeople wanted a self-improvement course.

Let $p_1 (\pi_1)$ = true proportion of workers who would like to attend a self-improvement course in the recent study.

$p_2 (\pi_2)$ = true proportion of workers who would like to attend a self-improvement course in the past study.

SO, Null Hypothesis, $H_0$ : $p_1=p_2$ {means that there is no difference in the two population proportions}

Alternate Hypothesis, $H_A$ : $p_1\neq p_2$ {means that there is a difference in the two population proportions}

The test statistics that would be used here Two-sample z test for proportions;

T.S. = $\frac{(\hat p_1-\hat p_2)-(p_1-p_2)}{\sqrt{\frac{\hat p_1(1-\hat p_1)}{n_1}+\frac{\hat p_2(1-\hat p_2)}{n_2} } }$ ~ N(0,1)

where, $\hat p_1$ = sample proportion of salespeople who would like to attend a self-improvement course in recent study = 66%

$\hat p_2$ = sample proportion of salespeople who would like to attend a self-improvement course in past study = 70%

$n_1$ = sample of salespeople in recent study = 150

$n_2$ = sample of salespeople in past study = 160

So, the test statistics = $\frac{(0.66-0.70)-(0)}{\sqrt{\frac{0.66(1-0.66)}{150}+\frac{0.70(1-0.70)}{160} } }$

= -0.755

The value of z test statistics is -0.755.

Now, at 0.05 significance level the z table gives critical values of -1.96 and 1.96 for two-tailed test.

Since our test statistic lies within the range of critical values of z, so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.

Therefore, we conclude that there is no difference in the two population proportions.

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3 years ago