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Alexeev081 [22]
2 years ago
15

A corn field has an area of 48,000 square. Ft. The width of the field is 300 feet.

Mathematics
1 answer:
iren2701 [21]2 years ago
8 0

Step-by-step explanation:

A) width × length = area

300 ft × l = 48000 sq ft

B) l = 48000 ÷ 300

l = 160 ft

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We conclude that the vaccine appears to be​ effective.

Step-by-step explanation:

We are given that a doctor released the results of clinical trials for a vaccine to prevent a particular disease.

The subjects in group 1​ (the experimental​ group) were given the​ vaccine, while the subjects in group 2​ (the control​ group) were given a placebo. Of the 200 comma 000 children in the experimental​ group, 38 developed the disease. Of the 200 comma 000 children in the control​ group, 81 developed the disease.

Let<em> </em>p_1<em> = proportion of subjects in the experimental​ group who developed the disease.</em>

<em />p_2<em> = proportion of subjects in the control​ group who developed the disease.</em>

So, Null Hypothesis, H_0 : p_1\geq p_2       {means that the vaccine does not appears to be​ effective}

Alternate Hypothesis, H_A : p_1       {means that the vaccine appears to be​ effective}

The test statistics that would be used here <u>Two-sample z proportion</u> <u>statistics</u>;

                       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 children in the experimental​ group who developed the disease = \frac{38}{200,000} =  0.00019

\hat p_2 = sample proportion of children in the control​ group who developed the disease = \frac{81}{200,000} =  0.00041

n_1 = sample of children in the experimental​ group = 200,000

n_2 = sample of children in the control​ group = 200,000

So, <u><em>test statistics</em></u>  =  \frac{(0.00019-0.00041)-(0)}{\sqrt{\frac{0.00019(1-0.00019)}{200,000}+\frac{0.00041(1-0.00041)}{200,000}  } }

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The value of z test statistics is -4.02.

Now, at 0.01 significance level the z table gives critical values of -2.33 for left-tailed test.

Since our test statistics is less than the critical value of z as -4.02 < -2.33, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the vaccine appears to be​ effective.

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