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Anna [14]
3 years ago
12

Compare the y-intercepts and the rates of changes of the following items.

Mathematics
1 answer:
professor190 [17]3 years ago
3 0

Answer:

i need points

Step-by-step explanation:

yesur

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The St. Croix junior hockey
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Answer:

14

Step-by-step explanation:

56 divided by 4 is 14

7 0
3 years ago
Many elementary school students in a school district currently have ear infections. A random sample of children in two different
Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school n_1 = 42

sample size for the second school n_2  = 34

so 16 out of 42 i.e x_1 = 16 and 18 out of 34 i.e x_2 = 18 have ear infection.

the proportion of students with ear infection Is as follows:

\hat p_1 = \dfrac{16}{42} = 0.38095

\hat p_2 = \dfrac{18}{34}  =  0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}

\bar p = \dfrac{16 +18}{42 +34}

\bar p = \dfrac{34}{76}

\bar p = 0.447368

\bar p + \bar  q = 1 \\ \\ \bar q = 1 -\bar  p \\  \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632

The pooled standard error can be computed by using the formula:

S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} +  \dfrac{\bar p \bar p}{n_2} }

S.E = \sqrt{ \dfrac{  0.447368 *  0.552632}{ 42} +  \dfrac{ 0.447368 *  0.447368}{34} }

S.E = \sqrt{ \dfrac{  0.2472298726}{ 42} +  \dfrac{ 0.2001381274}{34} }

S.E = \sqrt{ 0.01177284105}

S.E = 0.1085

The test statistics is ;

z = \dfrac{\hat p_1 - \hat p_2}{S.E}

z = \dfrac{0.38095- 0.5294}{0.1085}

z = \dfrac{-0.14845}{0.1085}

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

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