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3 years ago

12

What addition statement does this show?

1 answer:

kotykmax [81]3 years ago

7 0

B. -3.4 + (-5.7)= -9.1

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<span>Tell if you need logic behind some other answers too. Specify the questions.</span>

8 0

3 years ago

Read 2 more answers

A marketing research assistant for a veterinary hospital surveyed a random sample of 457 pet owners. Respondents were asked to i

Alekssandra [29.7K]

Answer:

The probability content for the confidence interval (3.49, 3.69) is 96%.

Step-by-step explanation:

The sample selected to determine the mean number of times the pet owners visited their veterinarian each year is of size, <em>n</em> = 475.

The sample selected is quite large, i.e. <em>n</em> > 30.

According to the Central Limit Theorem if we have a population with mean <em>μ</em> and standard deviation <em>σ</em> and we take appropriately huge random samples (<em>n</em> ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.

Then, the mean of the distribution of sample mean is given by,

$\mu_{\bar x}=\mu\approx\bar x$ ; for <em>n</em> → ∞.

And the standard deviation of the distribution of sample means is given by,

$\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}\approx \frac{s}{\sqrt{n}}$ ; for <em>n</em> → ∞.

So, the random variable <em>X</em>, defined as the number of visits to the veterinarian each year, follows a Normal distribution.

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

$CI=\bar x\pm z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

The information provided is:

$\bar x=3.59\\s=1.045\\$

The confidence interval is (3.49, 3.69).

The margin of error of the confidence interval for the population mean is:

$MOE=\frac{UL-LL}{2}= z_{\alpha/2}\times \frac{\sigma}{\sqrt{n}}$

Compute the MOE as follows:

$MOE=\frac{UL-LL}{2}=\frac{3.69-3.49}{2}=0.10$

Compute the critical value of <em>z</em> as follows:

$MOE=z_{\alpha/2}\times \frac{s}{\sqrt{n}}\\0.10=z_{\alpha/2}\times \frac{1.045}{\sqrt{475}}\\z_{\alpha/2}=2.0856\\z_{\alpha/2}\approx2.09$

Compute the value of $P(-z_{\alpha/2}$ as follows:

$P(-z_{\alpha/2}$

$=P(Z$

Thus, the probability content for the confidence interval (3.49, 3.69) is 96%.

7 0

4 years ago