Mathematics, 14.05.2021 05:00 andrealfar1010

A gaming company would like to test the hypothesis that the average age of someone who uses gaming console A is different from the average age of someone who uses gaming console B. A random sample of 36 gaming console A users had an average age of 34.2 years while a random sample of 30 gaming console B users had an average age of 32.7 years. Assume that the population standard deviation for the age of gaming console A and gaming console B users is 3.9 and 4.0 years, respectively. This company would like to set α = 0.10. a) What is the critical value for this hypothesis test?
a) 2.33
b) 1.28
c) 1.645
d) 1.96
b) What is the 90% confidence interval for the difference in population means?
a) (-0.11, 3.11)
b) (-1.70, 4.70)
c) (0.65, 2.35)
d) (1.18, 1.82)

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Answer from: smagallanes

Question a:

c) 1.645

Question b:

a) (-0.11, 3.11)

Step-by-step explanation:

Before answering the question, we need to understand the central limit theorem and subtraction between normal variables:

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean $\mu = p$ and standard deviation $s = \sqrt{\frac{p(1-p)}{n}}$

Subtraction between normal variables:

When two normal variables are subtracted, the mean is the difference of the means, while the standard deviation is the square root of the sum of the variances.

A random sample of 36 gaming console A users had an average age of 34.2 years, with a standard deviation of 3.9 years.

This means that $\mu_A = 34.2, s_A = \frac{3.9}{\sqrt{36}} = 0.65$

A random sample of 30 gaming console B users had an average age of 32.7 years, with a standard deviation of 4 years.

This means that $\mu_B = 32.7, s_B = \frac{4}{\sqrt{30}} = 0.73$

Distribution of the difference in population means:

$\mu = \mu_A - \mu_B = 34.2 - 32.7 = 1.5$

$s = \sqrt{s_A^2+s_B^2} = \sqrt{0.65^2+0.73^2} = 0.9774$

a) What is the critical value for this hypothesis test?

We test if the means are different, which means that we have a two-tailed test.

We have the standard deviations for the population, which means that we have a Z test.

Since it is a two-tailed Z-test, the critical value is Z with a p-value of 1 - (0.1/2) = 1 - 0.05 = 0.95, so, looking at the z-table, Z = 1.645, which is option C.

b) What is the 90% confidence interval for the difference in population means?

We have that to find our $\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = \frac{1 - 0.9}{2} = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of $1 - \alpha$ .

That is z with a pvalue of 1 - 0.05 = 0.95 , so Z = 1.645.

Now, find the margin of error M as such

M = zs

M = 1.645*0.9774 = 1.61

The lower end of the interval is the sample mean subtracted by M. So it is 1.5 - 1.61 = -0.11

The upper end of the interval is the sample mean added to M. So it is 1.5 + 1.61 = 3.11

The correct answer is given by option A.

Answer from: Quest

-14

step-by-step explanation:

2x2 – 2(x+4)

=2x2-2（5+4）

=4-2×9

=4-18

=-14

Answer from: Quest

Answer is either b or c

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