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2 years ago

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According to a recent study, US adults spent on average 29 hours per month using the PC internet. The time spent on the PC inter

net follows a normal distribution with mean 29 hours per month and standard deviation 12 hours per month.What percent of US adults spend above 29 hours per month on PC internet? g

1 answer:

Andrei [34K]2 years ago

5 0

Answer: The percent of US adults spend above 29 hours per month on PC internet = 50%

Step-by-step explanation:

Given: , US adults spent on average 29 hours per month using the PC internet.

As the me spent on the PC internet follows a normal distribution.

And there is 50% of the population lies above the mean and below the mean.

Then, the percent of US adults spend above 29 hours per month on PC internet = 50%

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$\mathbb{A~N~S~W~E~R~:}$

$= r$ · $2$

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$\mathfrak{apply~the~fraction~rule: }$ $a$ $\frac{b}{c} =\frac{a~. ~b }{c}$

$\mathfrak{apply~rule: }$ $a$ · $1 = a$

$r$ · $1$ · $4 = r$ · $4$

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So, your answer will be is :

$= r$ · $2$

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What are the distinct factors of 75

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3 years ago

Simplify (9x-1)^-1/2 - (x+2)(9x-1)^-1/2

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Step-by-step explanation:

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The following data lists the ages of a random selection of actresses when they won an award in the category of Best Actress, al

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Answer:

a) $p_v =P(t_{(9)}$

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best Actors at 1% of significance.

b) The 99% confidence interval would be given by (-21.469;2.069)

c) We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two

Step-by-step explanation:

Part a

Let put some notation

x=actor's age , y = actress's age

x: 58 41 36 36 34 33 48 37 37 43

y: 26 27 34 26 35 29 23 42 30 34

The system of hypothesis for this case are:

Null hypothesis: $\mu_y- \mu_x \geq 0$

Alternative hypothesis: $\mu_y -\mu_x$

The first step is calculate the difference $d_i=y_i-x_i$ and we obtain this:

d: -32, -14, -2, -10, 1, -4, -25, 5, -7, -9

The second step is calculate the mean difference

$\bar d= \frac{\sum_{i=1}^n d_i}{n}= -9.7$

The third step would be calculate the standard deviation for the differences, and we got:

$s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =11.451$

The 4 step is calculate the statistic given by :

$t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{-9.7 -0}{\frac{11.451}{\sqrt{10}}}=-2.679$

The next step is calculate the degrees of freedom given by:

$df=n-1=10-1=9$

Now we can calculate the p value, since we have a left tailed test the p value is given by:

$p_v =P(t_{(9)}$

The p value is higher than the significance level given 0.01, so then we can conclude that we FAIL to reject the null hypothesis. And we can say that the true difference for Best Actresses is not significantly lower than the mean for Best Actors at 1% of significance.

Part b

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".

The margin of error is the range of values below and above the sample statistic in a confidence interval.

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The confidence interval for the mean is given by the following formula:

$\bar d \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

Since the Confidence is 0.99 or 99%, the value of $\alpha=0.01$ and $\alpha/2 =0.005$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.005,9)".And we see that $t_{\alpha/2}=3.25$

Now we have everything in order to replace into formula (1):

$-9.7-3.25\frac{11.451}{\sqrt{10}}=-21.469$

$-9.7+3.25\frac{11.451}{\sqrt{10}}=2.069$

So on this case the 99% confidence interval would be given by (-21.469;2.069)

Part c

We got the same conclusion as part a, sicne the confidence interval contains the value 0, we FAIL to reject the null hypothesis that the difference between the two means is 0.

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2 years ago