Mathematics, 25.03.2020 00:10 Jroc23

39: Suppose that the amount of time teenagers spend on the internet is normally distributed with a standard deviation of 1.5 hours. A sample of 100 teenagers is selected at random, and the sample mean computed as 6.5 hours. a. Determine the 95% confidence interval estimate of the population mean. b. Interpret what the confidence interval estimate tells you.

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Answer from: nickespinosa11

Following are the solution to the given question:

In the given scenario, we assume that the true population will be $\mu$ . At $5\%$ level of significance, it cuts off the value that $\pm 1.96$ .

$\to -1.96 \leq \frac{X-\mu}{sd_x} \times \sqrt{n} \leq 1.96$

$\to X = 6.5 \ \ and\ \ sd_x = 1.5 \\\\\to -1.96 \leq \frac{6.5 - \mu}{1.5} \times \sqrt{100} \leq 1.96\\\\ \to -1.96 \leq \frac{6.5 - \mu}{1.5} \times 10 \leq 1.96\\\\ \to -.294 \leq 6.5 - \mu \leq 0.294 \\\\ \to -6.794 \leq - \mu \leq -6.206$

A population parameter is estimated with a 95% confidence interval. It suggests that if 100 similar intervals are established, 95% of the genuine population mean will fall inside this interval.

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Answer from: cheyenne481

a) The 95% confidence interval estimate of the population mean is between 6.206 hours and 6.794 hours.

b) The confidence interval means that we are 95% sure that the true population mean for the amount of time teenagers spend on the internet is between 6.206 hours and 6.794 hours.

Step-by-step explanation:

a)Confidence interval:

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

$\alpha = \frac{1-0.95}{2} = 0.025$

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

So it is z with a pvalue of 1-0.025 = 0.975 , so z = 1.96

Now, find M as such

$M = z*\frac{\sigma}{\sqrt{n}}$

In which $\sigma$ is the standard deviation of the population and n is the size of the sample.

$M = 1.96*\frac{1.5}{\sqrt{100}} = 0.294$

The lower end of the interval is the sample mean subtracted by M. So it is 6.5 - 0.294 = 6.206 hours.

The upper end of the interval is the sample mean added to M. So it is 6.5 + 0.294 = 6.794 hours.

The 95% confidence interval estimate of the population mean is between 6.206 hours and 6.794 hours.

b)Interpretation:

The confidence interval means that we are 95% sure that the true population mean for the amount of time teenagers spend on the internet is between 6.206 hours and 6.794 hours.

Answer from: Quest

Hold on lemme do the work

Answer from: Quest

1^3 + 2^3 + 3^3 +4^3 = (1+2+3+4)^2

step-by-step explanation:

i am only able to answer the first part. i am not sure about the second part. sorry

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39: Suppose that the amount of time teenagers spend on the internet is normally distributed with a s...

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