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kirza4 [7]
3 years ago
10

(CO 4) In a sample of 8 high school students, they spent an average of 24.8 hours each week doing sports with a sample standard

deviation of 3.2 hours. Find the 95% confidence interval, assuming the times are normally distributed.
(21.60, 28.00)

(22.12, 27.48)

(22.66, 26.94)

(24.10, 25.50)
Mathematics
1 answer:
AveGali [126]3 years ago
6 0

Answer:  

(22.12, 27.48)

Step-by-step explanation:

Given : Significance level : \alpha: 1-0.95=0.05

Sample size : n= 8 , which is a small sample (n<30), so we use t-test.

Critical values using t-distribution: t_{n-1,\alpha/2}=t_{7,0.025}=2.365

Sample mean : \overline{x}=24.8\text{ hours}

Standard deviation : \sigma=3.2\text{ hours}

The confidence interval for population means is given by :-

\overline{x}\pm t_{n-1,\alpha/2}\dfrac{\sigma}{\sqrt{n}}

i.e. 24.8\pm(2.365)\dfrac{3.2}{\sqrt{8}}

24.8\pm2.67569206001\\\\\approx24.8\pm2.68\\\\=(24.8-2.68, 24.8+2.68)=(22.12, 27.48)

Hence, the 95% confidence interval, assuming the times are normally distributed.=  (22.12, 27.48)

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Mkey [24]

Answer:

  • <u>120 pens and 200 pencils.</u>

<u></u>

Explanation:

You can set a system of two equations.

<u>1. Variables</u>

<u />

  • x: number of pens
  • y: number of pencils

<u>2. Cost</u>

  • <em>each pen costs</em> $1, then x pens costs: x
  • <em>each pencil costs</em> $0.5, then y pencil costs: 0.5y

  • Then, the total cost is: x + 0.5y

  • The cost of the whole purchase was $ 220, then the first equation is:

          x + 0.5y = 220 ↔ equation (1)

<u>3. </u><em><u>There were 80 more pencils than pens</u></em>

Then:

  pencils  =      80   +   pens

       ↓                               ↓

       y        =      80   +      x        ↔ equation (2)

<u>4. Solve the system</u>

i) Substitute the equation (2) into the equation (1):

  • x + 0.5(80 + x) = 220

ii) Solve

  • x + 40 + 0.5x = 220
  • 1.5x = 180
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  • x = 120 pens

iii) Substitute x = 120 into the equation (2)

  • y = 80 + 120
  • y = 200 pencils

Solution: 120 pens and 200 pencils ← answer

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