Mathematics, 10.12.2019 23:31 Anliz0179

Asoftware company allows their programmers to work from home an average of 20 hours per week. a department of 15 programmers work a mean of 22 hours per week from home with a standard deviation of 5.6 hours. at is their sufficient evidence to conclude a difference? give a 95% confidence interval for the true mean number of hours working from home

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Answer from: Theresab2021

The 95% confidence interval would be given by (18.906;25.094)

Since the confidence interval contains the value 20, we don't have enough evidence to conclude a difference.

Step-by-step explanation:

1) 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".

$\bar X =22$ represent the sample mean for the sample

$\mu$ population mean (variable of interest)

s=5.6 represent the sample standard deviation

n=15 represent the sample size

2) Solution to the problem

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=15-1=14

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(0.025,14)".And we see that $t_{\alpha/2}=2.14$

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

$22-2.14\frac{5.6}{\sqrt{15}}=18.906$

$22+2.14\frac{5.6}{\sqrt{15}}=25.094$

So on this case the 95% confidence interval would be given by (18.906;25.094)

Since the confidence interval contains the value 20, we don't have enough evidence to conclude a difference.

Answer from: Quest

The correct answer is .44

Answer from: Quest

answer:

total cost for tiles and paints is $924.

step-by-step explanation:

we have been given that a community hall is in the shape of a cuboid. the hall is 40m long 15m high and 3m wide.

the paint will be required for 4 walls and ceiling.

let us find area of walls and ceiling.

$\text{area of walls and ceiling}=(2*40*15)+(2*3*15)+(40*3)$

$\text{area of walls and ceiling}=1200+90+120$

$\text{area of walls and ceiling}=1410$

therefore, the area of walls and ceiling is 1410 square meters.

given: cost for 10 litre of paint is $10 and 10 litre paint covers 25 square meter. therefore,

$\text{ the total painting cost}=10*(\frac{1410}{25})$

$\text{ the total painting cost}=10*56.4=564$

therefore, the total painting cost is $564.

tiles will be required for floor. let us find the area of floor.

$\text{area of floor} = 40*3\text{ square meters}$

$\text{area of floor} =120\text{ square meters}$

given: 1m squared floor tiles costs $3. so,

$\text{total cost for tiles} = 3*120 = 360$

therefore, the total tiles cost is $360.

now let us find combined total cost of tiles and paint.

$\text{combined total cost}= 564+360 = 924$

therefore, the combined total cost of tiles and paint is $924.

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Asoftware company allows their programmers to work from home an average of 20 hours per week. a depa...