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Jet001 [13]
3 years ago
5

Suppose that 20% of students at high school A and 18% of students at high school B participate on a school athletic team. Indepe

ndent random samples of 30 students from each school are selected and asked if they participate on a school athletic team. Let ˆ p A represent the sample proportion of students at school A who participate on a school athletic team and ˆ p B represent the sample proportion of students at school B who participate on a school athletic team.
Is the 10% condition met?
_____ because ___ is less than 10% of all students at high school A and 30 is ___ 10% of all students at high school B.
The standard deviation of ˆ p A − ˆ p B is approximately ______. The difference (high school A – high school B) in the ________ that participate on a school athletic team _______ varies by about ____ from the true difference in proportions of _____.
Mathematics
1 answer:
ANEK [815]3 years ago
6 0

Answer:

B is more of a reasonable answer

Step-by-step explanation:

i  got it correct on the test

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A 12 inch pizza is cut into 8 equal slices. What is the length of the crust of 1 piece of pizza? Use 3.14 for pie
andrew-mc [135]
1.5 iches is the answer
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Solve x2 - 8x - 9 = 0. Rewrite the equation so that it is of the form x2 + bx = c.
barxatty [35]

Answer:

x=9,-1  and x^2+(-8)x=9

Explanation:

we have been given with the quadratic equation x^2-8x-9

we compare the given quadratic equation with general quadratic equation

general quadratic is ax^2+bx+c=0

from given quadratic equation a=1,b= -8,c= -9

substituting these values in the formula for discriminant D=b^{2}-4ac

D=(-8)^2-4(1)(-9)=100

Now, to find the value of x

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Now, substituting the values we will get

x=\frac{-(-8)\pm\sqrt{100}} {2}= \frac{8\pm10}{2}=9,-1

And rewritting the given equation by  shifting 9 to right hand side of the given equation and taking minus inside the bracket so as to convert it in the form of

x^2+bx=c

4 0
3 years ago
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Assume that you have a sample of n 1 equals 6​, with the sample mean Upper X overbar 1 equals 50​, and a sample standard deviati
tigry1 [53]

Answer:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

df=6+5-2=9

p_v =P(t_{9}>2.656) =0.0131

Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

Step-by-step explanation:

Data given

n_1 =6 represent the sample size for group 1

n_2 =5 represent the sample size for group 2

\bar X_1 =50 represent the sample mean for the group 1

\bar X_2 =38 represent the sample mean for the group 2

s_1=7 represent the sample standard deviation for group 1

s_2=8 represent the sample standard deviation for group 2

System of hypothesis

The system of hypothesis on this case are:

Null hypothesis: \mu_1 \leq \mu_2

Alternative hypothesis: \mu_1 > \mu_2

We are assuming that the population variances for each group are the same

\sigma^2_1 =\sigma^2_2 =\sigma^2

The statistic for this case is given by:

t=\frac{(\bar X_1 -\bar X_2)-(\mu_{1}-\mu_2)}{S_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}

The pooled variance is:

S^2_p =\frac{(n_1-1)S^2_1 +(n_2 -1)S^2_2}{n_1 +n_2 -2}

We can find the pooled variance:

S^2_p =\frac{(6-1)(7)^2 +(5 -1)(8)^2}{6 +5 -2}=55.67

And the pooled deviation is:

S_p=7.46

The statistic is given by:

t=\frac{(50 -38)-(0)}{7.46\sqrt{\frac{1}{6}+\frac{1}{5}}}=2.656

The degrees of freedom are given by:

df=6+5-2=9

The p value is given by:

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Since the p value is higher than the significance level given of 0.01 we don't have enough evidence to conclude that the true mean for group 1 is significantly higher thn the true mean for the group 2.

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Correct option is the second option .


This is wrong DONT use it
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