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slega [8]
3 years ago
14

The accounting department analyzes the variance of the weekly unit costs reported by two production departments. A sample of 16

cost reports for each of the two departments shows cost variances of 2.5 and 5.5, respectively. Is this sample sufficient to conclude that the two production departments differ in terms of unit cost variance? Use = .10. State the null and alternative hypotheses.
Mathematics
1 answer:
dusya [7]3 years ago
5 0

Answer:

F=\frac{s^2_2}{s^2_1}=\frac{5.5}{2.5}=2.2

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have n_2 -1 =16-1=15 and for the denominator we have n_1 -1 =16-1=15 and the F statistic have 15 degrees of freedom for the numerator and 15 for the denominator. And the P value is given by:

p_v =2*P(F_{15,15}>2.2)=0.138

For this case the p value is highert than the significance level so we haev enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviations are not significantly different

Step-by-step explanation:

Information given

n_1 = 16 represent the sampe size 1

n_2 =16 represent the sample 2

s^2_1 = 2.5 represent the sample deviation for 1

s^2_2 = 5.5 represent the sample variance for 2

\alpha=0.10 represent the significance level provided

The statistic is given by:

F=\frac{s^2_2}{s^2_1}

Hypothesis to test

We want to test if the variations in terms of the variance are equal, so the system of hypothesis are:

H0: \sigma^2_1 = \sigma^2_2

H1: \sigma^2_1 \neq \sigma^2_2

The statistic is given by:

F=\frac{s^2_2}{s^2_1}=\frac{5.5}{2.5}=2.2

Now we can calculate the p value but first we need to calculate the degrees of freedom for the statistic. For the numerator we have n_2 -1 =16-1=15 and for the denominator we have n_1 -1 =16-1=15 and the F statistic have 15 degrees of freedom for the numerator and 15 for the denominator. And the P value is given by:

p_v =2*P(F_{15,15}>2.2)=0.138

For this case the p value is highert than the significance level so we haev enough evidence to FAIL to reject the null hypothesis and we can conclude that the true deviations are not significantly different

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e-lub [12.9K]

Answer:

60 minutes

Step-by-step explanation:

We solve this question using Lowest Common Factor method

Find and list multiples of each number of minutes( 12 and 10 minutes) until the first common multiple is found. This is the lowest common multiple.

Multiples of 10:

10, 20, 30, 40, 50, 60, 70, 80

Multiples of 12:

12, 24, 36, 48, 60, 72, 84

Therefore,

LCM(10, 12) = 60

The number if minutes until they both drip again is 60 minutes

7 0
3 years ago
Solve for -2<-3-3x<0
gregori [183]

-3 from all ends to get 1<-3x<3 then divide by -3 on all ends to get -1/3>x>-1 you flip the sign because of the negative,


8 0
3 years ago
A company with a large fleet of cars wants to study the gasoline usage. They check the gasoline usage for 50 company trips chose
Nookie1986 [14]

Answer:

The 95% confidence interval is given by (25.71536 ;28.32464)

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

Step-by-step explanation:

Notation and definitions  

n=50 represent the sample size  

\bar X= 27.2 represent the sample mean  

s=5.83 represent the sample standard deviation  

m represent the margin of error  

Confidence =88% or 0.88

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".  

Calculate the critical value tc  

In order to find the critical value is important to mention that 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 88% of confidence, our significance level would be given by \alpha=1-0.88=0.12 and \alpha/2 =0.06. The degrees of freedom are given by:  

df=n-1=50-1=49  

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

qt(0.06,49)

[1] -1.582366

qt(1-0.06,49)

[1] 1.582366

The critical value tc=\pm 1.582366  

Calculate the margin of error (m)  

The margin of error for the sample mean is given by this formula:  

m=t_c \frac{s}{\sqrt{n}}  

m=1.582366 \frac{5.83}{\sqrt{50}}=14.613  

With R we can do this:

m=1.582366*(5.83/sqrt(50))

m

[1] 1.304639

Calculate the confidence interval  

The interval for the mean is given by this formula:  

\bar X \pm t_{c} \frac{s}{\sqrt{n}}  

And calculating the limits we got:  

27.02 - 1.582366 \frac{5.83}{\sqrt{50}}=25.715  

27.02 + 1.582366 \frac{5.83}{\sqrt{50}}=28.325

Using R the code is:

lower=27.02-m;lower

[1] 25.71536

upper=27.02+m;upper

[1] 28.32464

The 95% confidence interval is given by (25.71536 ;28.32464)  

And if we need to round we can use the following excel code:

round(lower,2)

[1] 25.72

round(upper,2)

[1] 28.32

And the interval would be (25.72; 28.32)  

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Answer:

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Step-by-step explanation:

The volume formula

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Given

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Find the volume by substituting the values into formula

  • V = 8.15*4 = 32.6 cm³
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