Solve for c. a(b − c)=d Attachment added

The authors of a paper describe an experiment to evaluate the effect of using a cell phone on reaction time. Subjects were asked

Serggg [28]

Answer:

a) The 99% confidence interval would be given by (24.409;24.979)

b) n=464

Step-by-step explanation:

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".

Part a

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

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

In order to calculate the critical value $t_{\alpha/2}$ we need to find first the degrees of freedom, given by:

$df=n-1=47-1=46$

Since the Confidence is 0.95 or 95%, the value of $\alpha=0.05$ and $\alpha/2 =0.025$ , and we can use excel, a calculator or a table to find the critical value. The excel command would be: "=-T.INV(0.025,46)".And we see that $t_{\alpha/2}=2.01$

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

$525-2.01\frac{75}{\sqrt{47}}=503.01$

$525+2.01\frac{75}{\sqrt{47}}=546.99$

So on this case the 95% confidence interval would be given by (503.01;546.99)

Part b

The margin of error is given by this formula:

$ME=t_{\alpha/2}\frac{s}{\sqrt{n}}$ (1)

And on this case we have that ME =7 msec, we are interested in order to find the value of n, if we solve n from equation (1) we got:

$n=(\frac{t_{\alpha/2} s}{ME})^2$ (2)

The critical value for 95% of confidence interval is provided, $t_{\alpha/2}=2.01$ from part a, replacing into formula (2) we got:

$n=(\frac{2.01(75)}{7})^2 =463.79 \approx 464$

So the answer for this case would be n=464 rounded up to the nearest integer

8 0

3 years ago

The life times of light bulbs produced by a particular manufacturer have a mean of 1,200 hours and a standard deviation of 400 h

Nezavi [6.7K]

Answer:

a. 1200hours

b. σ² = 400² =160000hour

c. standard error = 133.33

d. for 9 lightbulb it is likely that 9*0.146 = 1.31 bulbs will have lives of fewer than 1,050 hours?

Step-by-step explanation:

mean of 1,200 hours

σ, a standard deviation of 400 hours.

Sample size, N = nine bulbs,

a) mean of the sample mean lifetime: is given as 1,200 hours

b) variance of the sample mean is the square of the σ, a standard deviation of 400 hours.

σ² = 400² =160000hours

c) What is the standard error of the sample mean?

The standard deviation of a sampling distribution of mean values is called the standard error of the means,

standard error of the means, σx = σ √N

The formula for the standard error of the means is true for all values infinite number of sample, N.

σx = σ √N

=400 √9 = 400/3 =133.3333

d) the probability that, on average, those nine lightbulbs have lives of fewer than 1,050 hours

The area under part of a normal probability curve is directly proportional to probability and the value is calculated as

z = ( x₁−x) /σ

where z = propability of normal curve

x₁ = variate mean = 1050hours

x = mean of 1200hours

σ = standard deviation = 400

applying the formula,

z= (1050-1200)/400

z = 150/400 =0.375

Using a table of partial areas beneath the standardized normal curve (see Table of normal curve, a z-value of 0.375 corresponds to an area of 0.1460 between the mean value.

Thus the probability of a lightbulbs having lives of fewer than 1,050 hours is 0.1460.

for 9 lightbulb it is likely that 9*0.146 i.e. 1.31 bulbs will have lives of fewer than 1,050 hours?

8 0

3 years ago

93^2 -83^2 SMART CALCULATED

12345 [234]

Answer:

Step-by-step explanation:

-1760 is the answer 83 times 83 minus 93 times 93 is -1760

8 0

2 years ago

Students at the college are allowed to work on campus

Ahat [919]

Answer: theres a website called weegy did you try that?

5 0

3 years ago

What are the factors of 5x2 + 39x − 8?

prisoha [69]

Answer:

C. (5x - 1)(x + 8)

Step-by-step explanation:

In a factoring problem by hand, the common steps to solve would be the "AM" method, or (Add-Multiply). Simply stated, finding what adds to the x-coeficient and what multiplies to the constant. In this case, what adds to 39(x) and what multiplies to 8. Just reading that, we can see that there aren't any numbers that can do that, so we have to improvise. Set up the problem equal to something that it can factor into:

Multiply the 5 and -8 = -40: this is what we will factor
The factors with add to 39 are 40 and -1.
Set up the problem:

$5x^{2}+39x-8 = 5x^{2}+40x-x-8$

Now, we can fo the math:

$5x^{2} -8 = 5x^{2} -8$

8(5x - 1) + x(5x - 1)

Factor (5x - 1) from this:

(5x - 1)(x + 8)

And we have our answer.

4 0

2 years ago