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mamaluj [8]
3 years ago
12

What is the Factor −14 out of −12x−54y.

Mathematics
1 answer:
SVETLANKA909090 [29]3 years ago
8 0

2 is your answer duuuuuuuuuuuude

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Sarina has some ribbon. After she used 24.7 cm of it, she had 50 cm left. How much ribbon did Sarita start with?
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A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the
lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

Step-by-step explanation:

(a)

Let <em>X</em> = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

<em>X</em> = 610

<em>n</em> = 1000

Confidence level = 90%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{610}{1000}=0.61

The critical value of <em>z</em> for a 90% confidence level is:

z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

<u>Interpretation</u>:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let <em>X</em> = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

<em>X</em> = 440

<em>n</em> = 1000

Confidence level = 95%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{440}{1000}=0.44

The critical value of <em>z</em> for a 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend  is (0.42, 0.46).

<u>Interpretation</u>:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

  1. The confidence level
  2. Sample size
  3. Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of <em>z</em> in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

4 0
3 years ago
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