Question

1S or, be Math 120 - Exam #3 Review Part 2 Name yr са pe pe 9. A population has standard deviation o = 21.3. a. How large a sample must be drawn so that a 99% confidence interval for u will have a margin of error equal to 2.5? 2 a ho H ol at H lo an n=10 (1-P)/2012 nr 211-8) = 5.5264 SOOS VOTO :21.3 2 2.5 -р nil b. If the required confidence level were 95%, would

Answer 1

Question-9

Here the values are,

$\sigma=21.3,\alpha=0.01, Margin\ of \ error (M.E)=2.5,z_{\alpha/2}=z_{0.005}=2.576$

we know that,

Margin of error error = 70/2 * n

Or, M.E= 2a/2 x n

$Or,n=\left ( \frac {z_{\alpha/2}\times \sigma}{M.E} \right )^2$

Or, n = (2.576 x 21.3 2.5

$Or,n=481.693634$

$Or,n\simeq 482$

The sample size should be 482 to get 99% confidence interval for population mean with margin of error 2.5

Let the regression equation of time based on week is,

$i.e, \hat{y}=b_{0}+b_{1}x$

Week (x) Time (y) ху x^2 3 4 5 6 7 8 43 39 35 33 32 30 30 28 26 25 321 43 78 105 132 160 180 210 1 4 9 16 25 36 49 224 64 9 81 10 55 234 250 1616 100 385 Total

Here the values are given by,

$\therefore \sum x=55,\sum y=321,\sum xy=1616,\sum x^2=385,N=10$

The coefficients of regression equation is given by,

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$\therefore b_{0}=\frac{321\times 385-55\times 1616}{10\times 385-55^2}$

$\therefore b_{0}=42.0666667$

$\therefore b_{0}\simeq 42.0667$[ round to four decimal place]

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$\therefore b_{1}=\frac{10\times 1616-55\times 321}{10\times 385-55^2}$

$\therefore b_{1}=-1.812121212$

$\therefore b_{1}\simeq -1.8121$ [ round to four decimal place]

The regression equation is given by,

$\therefore \hat{y}=42.0667-1.8121x$