Mathematics, 22.07.2021 16:40 dfbernalloboguerrero

A contributor for the local newspaper is writing an article for the weekly fitness section. To prepare for the story, she conducts a study to compare the exercise habits of people who exercise in the morning to the exercise habits of people who work out in the afternoon or evening. She selects three different health centers from which to draw her samples. The 57 people she sampled who work out in the morning have a mean of 5.2 hours of exercise each week. The 70 people surveyed who exercise in the afternoon or evening have a mean of 4.5 hours of exercise each week. Assume that the weekly exercise times have a population standard deviation of 0.6 hours for people who exercise in the morning and 0.4 hours for people who exercise in the afternoon or evening. Let Population 1 be people who exercise in the morning and Population 2 be people who exercise in the afternoon or evening. Step 1 of 2: Construct a 95% confidence interval for the true difference between the mean amounts of time spent exercising each week by people who work out in the morning and those who work out in the afternoon or evening at the three health centers. Round the endpoints of the interval to one decimal place, if necessary.

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The 95% confidence interval for the true difference between the mean amounts of time spent exercising each week by people who work out in the morning and those who work out in the afternoon or evening at the three health centers is (0.5, 0.9).

Step-by-step explanation:

Before building the confidence intervals, we need to understand the central limit theorem and subtraction of normal variables.

Central Limit Theorem

The Central Limit Theorem establishes 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.

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.

In the morning:

Sample of 57, mean of 5.2, standard deviation of 0.6, so:

$\mu_1 = 5.2$

$s_1 = \frac{0.6}{\sqrt{57}} = 0.0795$

In the afternoon/evening:

Sample of 70, mean of 4.5, standard deviation of 0.4, so:

$\mu_2 = 4.5$

$s_2 = \frac{0.2}{\sqrt{70}} = 0.0239$

Distribution of the difference:

$\mu = \mu_1 - \mu_2 = 5.2 - 4.5 = 0.7$

$s = \sqrt{s_1^2+s_2^2} = \sqrt{0.0795^2 + 0.0239^2} = 0.083$

Confidence interval:

The confidence interval is:

$\mu \pm zs$

In which

z is the z-score that has a p-value of $1 - \frac{\alpha}{2}$ .

95% confidence level

So $\alpha = 0.05$ , z is the value of Z that has a p-value of $1 - \frac{0.05}{2} = 0.975$ , so Z = 1.96 .

The lower bound of the interval is:

$\mu - zs = 0.7 - 1.96*0.083 = 0.5$

The upper bound of the interval is:

$\mu + zs = 0.7 + 1.96*0.083 = 0.9$

Answer from: Quest

First step is to get rid of the square root. remember that a square root is the same as raising to the (1/2) power. see below: $\sqrt{\frac{ {72x}^{16} }{ {50x}^{{{36} }}}} = \frac{ { \sqrt{72}x}^{ {16}^{ (\frac{1}{2}) } } }{ { \sqrt{50} x}^ {{36}^{ (\frac{1}{2}) } }}$ multiply exponents per exponent rules to get: $\frac{ { \sqrt{72}x}^{8} }{ { \sqrt{50}x}^{18} }$ remember you subtract exponents from the denominator and add exponents in the numerator so: $\sqrt{ \frac{72}{ 50}} \times ( {x}^{(8 - 18) } ) = \sqrt{ \frac{36}{25} } \times ( {x}^{ - 10} )$ simplify further: $\frac{6}{5} {x}^{ - 10} = \frac{6}{5 {x}^{10} }$ so a is the correct choice. if you don't understand this google exponent rules and it will you understand. look at the image results.
Plz hurrywhat is the simplified form of ? assume x ≠ 0.

Plz hurrywhat is the simplified form of  ? assume  x  ≠ 0.

Answer from: Quest

First, think of what you know. you know that the number t the end of the equation will need to be subtracted because it is the money he lost from buying the lawn mower. and your first two weeks have negative profit, so that tells you that c and d are wrong. now you can plug any of the weeks into a and b to see which one works. i'll start with plugging week 1 into a. $y = 60(1) - 90 \\ y = 60 - 90 \\ y = - 30$ this isn't right because it says your answer for week one should be -90.so now we need to try the same thing with b. $y = 60(1) - 150 \\ y = 60 - 150 \\ y = - 90$ this one got you the right answer, (sometimes if you only check with one of the weeks multiple will work so you need to check with more of the weeks, but it worked out fine here)this means that b is the correct answer.let me know if you need more clarification. i hope this : )

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