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kaheart [24]
3 years ago
7

5. We want to compare two different groups of students, students taking Composition 1 in a tradition lecture format and students

taking Composition 1 in a distance learning format. We know that the mean score on the research paper is 85 for both groups. What additional information would be provided by knowing the standard deviation?
Mathematics
1 answer:
RUDIKE [14]3 years ago
3 0

Answer:

By knowing the standard deviation, one gets the idea of how the value is scattered or dispersed about the mean.

Step-by-step explanation:

Let us first define standard deviation.

As it is known that the standard deviation is a measure of dispersion which express the spread of observation in terms of the average of deviations of observations from some central values.

Measure of dispersion gives us an idea about homogeneity or heterogeneity of the distribution.

Standard deviation is supposed almost an ideal measure of dispersion except the general nature of extracting the square root.  

Thus for the given question, if we want to compare the two different groups of students whose mean score is 85. Here the standard deviation for both the groups interprets an idea about how the individual score for each group scattered or varied about the mean score i.e. 85.

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The price of a wallet is $31.95 and the sales tax is 7.5%.
Musya8 [376]

Answer:

34.35

Step-by-step explanation:

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Which statements are true? Check all that apply
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Read the paragraph, then go back and see which of the points match up and which don’t
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Find the point estimate for the true difference between the given population means. Round your answer to three decimal places.
uranmaximum [27]

Answer:

Point Estimate for different between population means = - 0.99

Step-by-step explanation:

We are given data of two samples and we have to find the best point estimate of the true difference between two population means. Remember that in absence of data about population the best estimator is the sample data. So, we will find the means of both sample data and find the difference of that means. This difference between the means of sample data will be the best point estimate for the true difference between the population means.

Formula to calculate the mean is:

Mean=\frac{\text{Sum of Values}}{\text{Number of Values}}

Mean of Sample 1:

Mean=\frac{1414}{11}=128.55

Mean of Sample 2:

Mean=\frac{1684}{13}=129.54

Therefore the best point estimate for difference between two population means would be = Mean of Sample 1 - Mean of Sample 2

= 128.55 - 129.54

= - 0.99

8 0
3 years ago
The success of an airline depends heavily on its ability to provide a pleasant customer experience. One dimension of customer se
tankabanditka [31]

Answer:

A)H0: μ1 − μ2 = 0

Ha: μ1 − μ2 ≠ 0

(b)  Means

Company A___50.6___ min.

Company B___52.75___ min.

c)The t-value is -0.30107.

The p-value is 0 .764815.

C) Do not reject H0. There is no statistical evidence that one airline does better than the other in terms of their population mean delay time.

Step-by-step explanation:

A)H0: μ1 − μ2 = 0

i.e there is no difference between the means of delayed flight for two different airlines

Ha: μ1 − μ2 ≠ 0

i.e there is a difference between the means of delayed flight for two different airlines

(b)

Company A___50.6___ min.

Company B___52.75___ min.

Mean of Company A = x`1= ∑x/n =34+ 59+ 43+ 30+ 3+ 32+ 42+ 85+ 30+ 48+ 110+ 50+ 10+ 26+ 70+ 52+ 83+ 78+ 27+ 70+ 27+ 90+ 38+ 52+ 76/25

= 1265/25= 50.6

Mean of Company B = x`2= ∑x/n =

=46+ 63+ 43+ 33+ 65+ 104+ 45+ 27+ 39+ 84+ 75+ 44+ 34+ 51+ 63+ 42+ 34+ 34+ 65+ 64/20

= 1055/20= 52.75

<u><em>Difference Scores Calculations</em></u>

Company A

Sample size for Company A= n1= 25

Degrees of freedom for company A= df1 = n1 - 1 = 25 - 1 = 24

Mean for Company A= x`1=  50.6

Total Squared Difference (x-x`1) for Company A= SS1: 16938

s21 = SS1/(n1 - 1) = 16938/(25-1) = 705.75

Company B

Sample size for Company B= n1= 20

Degrees of freedom for company B= df2 = n2 - 1 = 20 - 1 = 19

Mean for Company B= x`2=  52.75

Total Squared Difference (x-x`2) for Company B= SS2=7427.75

s22 = SS2/(n2 - 1) = 7427.75/(20-1) = 390.93

<u><em>T-value Calculation</em></u>

<u><em>Pooled Variance= Sp²</em></u>

Sp² = ((df1/(df1 + df2)) * s21) + ((df2/(df2 + df2)) * s22)

Sp²= ((24/43) * 705.75) + ((19/43) * 390.93) = 566.65

s2x`1 = s2p/n1 = 566.65/25 = 22.67

s2x`2 = s2p/n2 = 566.65/20 = 28.33

t = (x`1 - x`2)/√(s2x`1 + s2x`2) = -2.15/√51 = -0.3

The t-value is -0.30107.

The total degrees of freedom is = n1+n2- 2= 25+20-2=43

The critical region for two tailed test at significance level ∝ =0.05 is

t(0.025) (43) = t > ±2.017

Since the calculated value of t=  -0.30107.  does not fall in the critical region t > ±2.017, null hypothesis is not rejected that is there is no difference between the means of delayed flight for two different airlines.

The p-value is 0 .764815. The result is not significant at p < 0.05.

7 0
3 years ago
Six times a number is three thousand six hundread what is the number
Alecsey [184]
Divide 3600 by 6 and its 600
7 0
3 years ago
Read 2 more answers
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