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Genrish500 [490]
2 years ago
6

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

ir significant others or friends, and 440 said that they are more likely to complain to their dog than to a friend. Suppose that it is reasonable to consider this sample as representative of the population of dog owners.(a) Construct a 90% confidence interval for the proportion of dog owners who take more pictures of their dog than other cant others or friends (Use a table or technology. Round your answers to three decimal places.)(________,_________)Interpret the interval,1. We are 90% confident that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends fals directly in the middle of this interval 2. 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 directly in the middle of this interval We are 90% confident that the mean number of dog owners who take more pictures of their dog than of their significant others or friends fails within this interval 3. 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 this interval 4. We are 90% confident that the true proportion of dog owners who take more pictures of the dog than of their ugnificant others or friends fails within this interval(b) Construct a 95% confidence interval for the proportion of dog owners who are more likely to complain to their dog than to a friend. (Use a table or technology. Round your answers to three decimal places.)(_______,_______)Interpret the interval,1. 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 this interval 2. We are 95% confident that the true proportion of dog owners who are more likely to complain to their dog than to a friend within this interval 3. we are 95% confident that the mean number of dog owners who are more to complain to the dog than to a friend directly within this interva4. There is 95% a chance that the true proportion of o wners who are more to come to the dog than to a friends directly into this interval 5. we are 95%content at the true proportion of owners who are more to come to the dog than to a friends directly into this interval(c) Give two reasons why the confidence First, the confidence level in part(b) is wider than the part(a).First, the confidece level in part(b) is________ the confidence level in part(a) is, so the critical value of part(b) is______the critical value of part(a), Second the _______ in part(b) is ______than in part(a).
Mathematics
1 answer:
lbvjy [14]2 years ago
4 0

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).

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