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nalin [4]
2 years ago
14

A bank wants to get new customers for their credit card. They try two different approaches in their marketing campaign. The firs

t promises a "cash back" reward, and the second promises low interest rates. A sample of 500 people is mailed the first brochure; of these, 125 get the credit card. A separate sample of 500 people is mailed the second brochure; 150 get the credit card. Are the two campaigns equally attractive to customers? Find the test statistic.
a. z = -1.89
b. z = 2.80
c. z = 0.25
Mathematics
1 answer:
AleksandrR [38]2 years ago
7 0

Answer:

the answer is z = 2.80*****

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What are the roots of f(x) = x2 – 48?
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Answer:

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sqrt = Square Root

sqrt(48) = 6.928

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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
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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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The national mean annual salary for a school administrator is $90,00 a year (The Cincinnati Enquirer, April 7, 2012). A school o
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Answer:31) The Coca-Cola Company reported that the mean per capita annual sales of its beverages in the United Sates was 423 eight-ounce servings. Suppose you are curious whether the consumption of Coca-Cola beverages is higher in Atlanta. A sample of 36 individuals from the Atlanta area showed a sample mean annual consumption of 460.4 eight ounce servings with a standard deviation of s=101.9 ounces. Using a=.05, do the sample results support the conclusion that mean annual consumption of Coca-Cola beverage products is higher in Atlanta

Step-by-step explanation:29) The national mean annual salary for a school administrator is $90,000 a year. (The Cincinnati Enquirer, April 7, 2012) A school official took a sample of 25 school administrators in the state of Ohio to learn about salaries in that state to see if they differed from the national average.

a) Formulate hypotheses that can be used to determine whether the population mean annaual administrator salary in Ohio differs from the nation mean of $90,000.

b) The sample data for 25 Ohio administrators is contained in the file named Administrator. What is the p-value for you hypothesis test in part (a)?

c) A a=.05 can your null hypothesis be rejected? What is your conclusion?

d) Repeat the preceding hypothesis test using the critical value approach.

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Write an absolute value equation to satisfy the given solution set shown on a number line.
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An example can be:

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To construct this, we first find the midpoint M of our set, in this case is 4.

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