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b) If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.

Step-by-step explanation:

Previous concepts

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 data is:

661 595 548 730 791 678 672 491 492 583 762 624 769 729 734 706

Part a

Compute the sample mean and sample standard deviation.

In order to calculate the mean and the sample deviation we need to have on mind the following formulas:

$\bar X= \sum_{i=1}^n \frac{x_i}{n}$

$s=\sqrt{\frac{\sum_{i=1}^n (x_i-\bar X)}{n-1}}$

=AVERAGE(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

On this case the average is $\bar X= 660.313$

=STDEV.S(661, 595, 548, 730, 791, 678, 672, 491 ,492, 583, 762 ,624 ,769, 729, 734, 706)

The sample standard deviation obtained was s=95.898

Find the critical value t* Use the formula for a CI to find upper and lower endpoints

In order to find the critical value we need to take in count that our sample size n =16<30 and on this case we don't know about the population standard deviation, so on this case we need to use the t distribution. Since our interval is at 99% of confidence, our significance level would be given by $\alpha=1-0.99=0.01$ and $\alpha/2 =0.005$ . The degrees of freedom are given by:

$df=n-1=16-1=15$

We can find the critical values in excel using the following formulas:

"=T.INV(0.005,15)" for $t_{\alpha/2}=-2.95$

"=T.INV(1-0.005,15)" for $t_{1-\alpha/2}=2.95$

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

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$

And if we find the limits we got:

$660.313- 2.95\frac{95.898}{\sqrt{16}}=589.588$

[tex]660.313+ 2.95\frac{95.898}{\sqrt{16}}=731.038/tex]

So the 99% confidence interval would be given by (589.588;731.038)

Part b

If we see the confidence interval the 620 is included on the interval we don't have enough evidence to reject the rating is 620. But since we need a score that at least 620 and our lower limit is 589.588 we cant conclude that all the clients would have a score that at least of 620.

4 0

3 years ago

In four years, 40% of a radioactive element decays. Find its half-life. Round to one decimal place.

White raven [17]

To find the answer, use $k = \frac{2.303 }{ 4} (log\frac {100 }{ 40})$ , and then use the equation $t=\frac{.693}{k}$ To find it's half life.

5 0

3 years ago

Read 2 more answers

Landen Company uses a single plantwide overhead rate of $100 per direct labor hour (DLH). Landen has two products as follows: ba

KATRIN_1 [288]

Answer:

$5,500

Step-by-step explanation:

Since Landen Company uses a single overhead rate of $100 per direct labor hour, the total amount allocated to the deluxe and basic chairs is given by the sum of the DLH used up for both products multiplied by the overhead rate:

$A =(20+35)*\$100\\A=\$5,500$

The total amount allocated to these products is $5,500.

8 0

3 years ago