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puteri [66]
3 years ago
6

The enrollment at Temple Junior High School has a change of ─ 60 over a 3 year period. What is the average change per year?

Mathematics
1 answer:
GaryK [48]3 years ago
4 0

Answer:

- 20

Step-by-step explanation:

Total Change in enrollment = - 60

Period of time at which change occurred = 3 years

Average change per year :

Total change in enrollment / period at which change occurred

= - 60 / 3

= - 20

Average change of - 20 enrollments per year

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Teen obesity:
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Answer:

95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

Step-by-step explanation:

We are given that 15% of a random sample of 300 U.S. public high school students were obese.

Firstly, the pivotal quantity for 95% confidence interval for the population proportion is given by;

                        P.Q. = \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }  ~ N(0,1)

where, \hat p = sample % of U.S. public high school students who were obese = 15%

           n = sample of U.S. public high school students = 300

           p = population percentage of all U.S. public high school students

<em>Here for constructing 95% confidence interval we have used One-sample z proportion statistics.</em>

<u></u>

<u>So, 95% confidence interval for the population proportion, p is ;</u>

P(-1.96 < N(0,1) < 1.96) = 0.95  {As the critical value of z at 2.5% level

                                             of significance are -1.96 & 1.96}  

P(-1.96 < \frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } } < 1.96) = 0.95

P( -1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < {\hat p-p} < 1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

P( \hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } < p < \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } ) = 0.95

<u>95% confidence interval for p</u> = [\hat p-1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } } , \hat p+1.96 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }]

  = [ 0.15-1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } } , 0.15+1.96 \times {\sqrt{\frac{0.15(1-0.15)}{300} } } ]

  = [0.110 , 0.190]

Therefore, 95% confidence interval for the percentage of all U.S. public high school students who are obese is [0.110 , 0.190].

6 0
3 years ago
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