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erma4kov [3.2K]2 years ago

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the professor is decorating his home and wants to buy some plants. he is interested in three types of plants costing $7, $10, an

Luda [366]

X quantity of first plant which cost $7
y quantity of second plant which cost $10
z quantity of third plant which cost $13

{x+y+z=15
{7x+10y+13z=150
There are 8 solutions with spending $150
Five of them are:
1x$7 + 13x$10 + 1x$13
2x$7 + 11x$10 +2x$13
3x$7 + 9x$10 +3x$13
5x$7 + 5x$10 + 5x$13
6x$7 + 3x$10 + 6x$13

3 0

2 years ago

Construct a confidence interval for p1-p2 at the given level of confidence.

Georgia [21]

Using the z-distribution, the 90% confidence interval for the difference of proportions is given by: (-0.0534, 0.0434).

<h3>What is the mean and the standard error for the distribution of differences?</h3>

For each sample, the mean and the standard error are given as follows:

$p_1 = \frac{30}{235} = 0.1277, s_1 = \sqrt{\frac{0.1277(0.8723)}{235}} = 0.0218$ .

$p_2 = \frac{39}{294} = 0.1327, s_1 = \sqrt{\frac{0.1327(0.8673)}{294}} = 0.0198$ .

For the distribution of differences, they are given by:

$\overline{p} = p_1 - p_2 = 0.1277 - 0.1327 = -0.005$ .

$s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0218^2 + 0.0198^2} = 0.0294$

<h3>What is the confidence interval?</h3>

The interval is given by:

$\overline{p} \pm zs$

In this problem, we have a 90% confidence level, hence $\alpha = 0.9$ , z is the value of Z that has a p-value of $\frac{1+0.9}{2} = 0.95$ , so the critical value is z = 1.645.