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dangina [55]
2 years ago
5

Oishi and Schimmack (2010) report that people who move from home to home frequently as children tend to have lower than average

levels of well-being as adults. To further examine this relationship, a psychologist obtains a sample of young adults who each experienced 5 or more different homes before they were 16 years old. These participants were given a standardized well-being questionnaire for which the general population has an average score of μ . The well-being scores for this sample are as follows: 38, 37, 41, 35, 42, 40, 33, 33, 36, 38, 32, 39. On the basis of this sample, is well-being for frequent movers significantly different from well-being in the general population? Use a two-tailed test with . Compute the estimated Cohen’s d to measure the size of the difference. Write a sentence showing how the outcome of the hypothesis test and the measure of effect size would appear in a research report.
Mathematics
1 answer:
user100 [1]2 years ago
5 0

Answer:

The well-being for frequent movers is significantly different from well-being in the general population. ( Alternate Hypothesis accepted )

cohen's d = -0.91 , ( Large Effect )

Step-by-step explanation:

Given:-

- A sample of size n = 12

- The population mean u_p = 40

- The sample was taken as:

                     38, 37, 41, 35, 42, 40, 33, 33, 36, 38, 32, 39

Find:-

On the basis of this sample, is well-being for frequent movers significantly different from well-being in the general population? Use a two-tailed test with α = 0.05.

Solution:-

- State the hypothesis for sample mean u_s is same as population mean u_p.

                    Null Hypothesis: u_s = 40

                    Alternate Hypothesis: u_s ≠ 40

- The rejection criteria for the Null hypothesis can be modeled by T-value ( n < 30 ) with significance level α = 0.05.

                    DOF = n - 1 = 12 - 1 = 11

                    Significance level α = 0.05

                    t_α/2 = t_0.025 = +/- 2.201

- For the statistic value we have to compute sample mean u_s given by:

             u_s = Σ xi / n

             u_s = (38 + 37 + 41 + 35 + 42 + 40 + 33 + 33 + 36 + 38 + 32 + 39) / 12

             u_s = 37

- For the statistic value we need population standard deviation S_p given by:

            S_p = S_s / √n

Where, S_s : Sample standard deviation.

            S_s^2 = Σ (xi - u_s)^2 / (n-1)

            =[ 2*(38-37)^2 +  (37-37)^2 + (41-37)^2 + (35-37)^2 + (42-37)^2 + (40-37)^2 + 2*(33-37)^2 + (36-37)^2 + (32-37)^2 + (39-37)^2 ] / ( 11 )

            S_s^2 = [ 2 + 0 + 16 + 4 + 25 + 9 + 32 + 1 + 25 + 4 ] / 11

            S_s^2 = 10.73

            S_s = 3.28

The population standard deviation ( S_p ) is:

            S_p = 3.28 / √12

            S_p = 0.95

- The T-statistics value is computed as follows:

            t = ( u_s - u_p ) / S_p

            t = ( 37 - 40 ) / 0.95 = -3.16

- Compare the T-statistics (t) with rejection criteria (t_α/2).

            -3.16 < -2.201

            t < t_α/2 ...... Reject Null Hypothesis.

- The well-being for frequent movers is significantly different from well-being in the general population. ( Alternate Hypothesis accepted )

- The cohen's d is calculated as follows:

         cohen's d = ( u_s - u_p ) / S_s

         cohen's d = ( 37 - 40 ) / 3.28 = -0.91 ,     ( Large Effect )    

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We dropped perpendiculars from C and D to intersect AB at Q and P respectively.

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Since, Sin(75^0)=\frac{CQ}{8}

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In a similar manner we can find BQ as:

Cos(75^0)=\frac{BQ}{8}

BQ\approx2.07 ft

All these values can be found in the diagram attached.

Thus, because of the inherent symmetry of the isosceles trapezoid, PQ can be found as:

PQ=22-(AP+QB)=22-(2.07+2.07)=17.86

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AC=\sqrt{(AQ)^2+(QC)^2}=\sqrt{(AP+PQ)^2+(QC)^2}=\sqrt{(2.07+17.86)^2+(7.73)^2}\approx21.38 feet.

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Second Question

For this question we can directly apply the formula for the area of a triangle using sines which is as:

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Area=\frac{1}{2}\times 218.5\times 224.5\times sin(58.2^0)\approx20845 m^2

Therefore, Option D is the correct option.

Third Question

For this question we will apply the Sine Rule to the \Delta ABC given to us.

Thus, from the triangle we will have:

\frac{AB}{Sin(\angle C)}=\frac{BC}{Sin(\angle A)}

\frac{c}{Sin(\angle C)}=\frac{a}{Sin(\angle A)}

\frac{17}{Sin(25^0)}=\frac{a}{Sin(45^0)}

This gives a to be:

a\approx28.44

Which is not close to any of the given options.

Fourth Question

Please find the second attachment for a better understanding of the solution provided her.

As can be clearly seen from the attached diagram, we can apply the Cosine Rule here to find the return distance of the plane which is CA.

AC=\sqrt{(AB)^2+(BC)^2-2(AB)(BC)\times Cos(\angle B)}

\therefore AC=\sqrt{(172.20)^2+(111.64)^2-2(172.20)(111.64)\times Cos(177.29^0)}\approx283.8 miles.

Thus, Option D is the answer.





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2 years ago
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