Complete the table for the given rule. Rule: y is 0.75 greater than x x y 0 3 9

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

Step-by-step explanation:

(a)

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

Given:

X = 610

n = 1000

Confidence level = 90%

The (1 - α)% 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 z 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).

Interpretation:

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 X = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

X = 440

n = 1000

Confidence level = 95%

The (1 - α)% 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 z 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).

Interpretation:

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:

The confidence level
Sample size
Standard deviation.

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

Because of this the critical value of z 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 more than the confidence level in part(a) is, so the critical value of part(b) is more than the critical value of part(a), Second the margin of error in part(b) is more than than in part(a).

4 0

1 year ago

Is 9760-5,220 more or less then 4,000

givi [52]

It's more than 4,000

7 0

2 years ago

Read 2 more answers

Tickets were sold at four different gates of a high school football stadium. The graph below shows the percent of the total tick

MaRussiya [10]

you have to add 90 +90 3 times then theres your answer :)

6 0

2 years ago

At a display booth at an amusement park, every visitor gets a gift bag. Some

kotegsom [21]

The bag will often contain all three items every 70 bags ⇒ answer C

Step-by-step explanation:

At a display booth at an amusement park, every visitor gets a gift bag

according to these information:

1. Hat Every 2nd visitor

2. T-shirt Every 7th visitor

3. Backpack Every 10th visitor

We need to know how often a bag will contain all three items

To solve this problem you must find a number divisible by 2, 7, and 10

To find that number start with the multiples of the largest one and

check them with the other two numbers

∵ The largest number is 10

∵ The multiples of 10 are 10 , 20 , 30 , 40 , 50 , 60 , 70 , 80 , .......

- Look for the first multiple of 10 is divisible by 7

∵ 70 is the first multiple of 10 divisible by 7

∵ 70 is divisible by 2 because its even number

∴ 70 is the smallest number divisible by 2, 7 and 10

The bag will often contain all three items every 70 bags

Learn more:

You can learn more about numbers in brainly.com/question/537998

#LearnwithBrainly

4 0

1 year ago

The Naturally Made Bath and Body store pays $550 a month for rent and utilities. The average cost for its products to be manufac

Oksi-84 [34.3K]

The break-even point occurs when the cost function equals the revenue function: 550 + 3.00x = 5.50x
Naturally Made needs to sell 220 products to break even
The cost or revenue when it sells the break-even number of products will be $1,210

4 0

2 years ago

Read 2 more answers