4. Confidence Intervals : The dataset for this Project represents a sample of da

Question

4. Confidence Intervals: The dataset for this Project represents a sample of data from a larger population.

Use this dataset sample to calculate the 95% confidence interval for the true population mean time that all students spend studying per week. If a student spends 15 hours per week studying, is this significantly different from the population mean? Explain. You may choose to use Excel or you may do this by hand.

In your own words, explain why and how sample statistics are used to estimate population parameters. Using the dataset, create an example to support your explanation

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Gender (1:female, 2:male) Age GPA SAT Score (0 - 1600) Year In School (1: freshman, 2:sophmore, 3:junior, 4:senior, 5: grad student) Class Section       (1: section1, 2:secion2, 3:section3, 4:section4) Hours spent on study per week (Range: 0 - 20) Final Score     (0 - 100) Project Score (0 - 100) Political Orientation                                     ( from -3 to +3 ) NOTE: -3 is fully liberal, +3 is fully conservative, and 0 is fully moderate. 518027 2 25 2.59 850 3 2 6.5 75 69 -1 591694 2 31 3.49 1100 4 1 11.6 89 81 0 561363 2 29 3.78 1322 5 1 7.4 75 86 2 535607 2 26 3.91 1400 5 1 13.3 90 80 3 524861 2 25 3.47 1249 3 1 8.9 89 94 3 573174 2 30 3.36 1265 5 1 12.3 87 89 1 581530 2 24 3.14 1189 3 2 10.9 94 97 -2 531316 2 25 2.80 870 3 2 8.1 81 91 1 534953 2 31 2.93 980 1 1 1.6 45 51 0 564779 2 23 3.49 1345 5 3 13.5 89 80 0 581076 2 28 3.32 1267 3 1 6.0 85 87 -2 542612 2 22 3.72 1458 5 3 8.7 87 79 -2 554740 2 28 3.08 1000 4 3 11.3 96 97 0 514150 2 24 3.90 1590 4 2 18.4 97 90 -3 510145 2 28 3.33 1345 5 1 9.0 89 92 1 511050 2 24 1.78 650 1 1 2.4 53 56 1 591316 2 21 3.22 1250 3 3 11.9 83 73 0 567689 1 27 3.36 1061 3 4 5.4 60 67 -1 558905 1 26 3.47 1102 4 2 6.7 71 79 2 525398 1 27 2.89 945 1 2 6.3 67 71 -1 570432 1 24 3.34 1080 1 1 8.3 82 87 1 593296 1 18 2.96 997 2 1 11.4 93 100 1 560608 1 28 3.10 1054 3 4 3.5 56 60 0 553457 1 29 3.38 1305 3 3 6.9 69 76 -1 575282 1 30 3.09 1040 5 1 11.0 78 89 -1 538785 1 24 3.59 1200 1 1 10.6 97 99 -1 511569 1 34 3.93 1506 1 2 17.2 99 90 -2 541697 1 29 3.02 1020 5 2 4.6 46 52 -1 586836 1 29 3.45 1256 5 3 13.7 86 79 -2 554301 1 31 3.49 1200 2 4 14.2 92 92 3 515986 1 31 3.54 1300 5 4 12.8 91 87 -3 597586 1 18 3.49 1300 3 2 12.4 88 75 1 583975 1 25 3.82 1432 4 4 9.0 98 90 2 522767 1 26 1.86 500 1 4 6.0 61

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Explanation / Answer

Here the variable of interest is time that all students spend studying per week.

We have to find confidence interval for population mean and hypothesis testing.

Here we have to test the hypothesis that,

H0 : mu = 15 Vs H1 : mu not= 15

where mu is population mean time that all students spend studying per week.

Assume alpha = level of significance = 5% = 0.05

Here sample data is given so we use one sample t-test.

The test statistic is,

t = (Xbar - mu) / (sd/sqrt(n))

where Xbar is sample mean

mu is population mean

sd is sample standard deviation.

n is sample size.

Now we have to find p-value.

p-value we can find using EXCEL.

syntax :

=TDIST(x, deg_freedom, tails)

where x is absolute value of test statistic and

deg_freedom = n-1

tails = 2

Decision rule : If P-value < alpha then reject H0 at 5% level of significance otherwise fail to reject H0.

The 95% confidence interval for population mean is,

Xbar - E < mu < Xbar + E

where E is margin of e%rror.

E = (tc* sd) / sqrt(n)

where tc is critical value for t-distribution.

c = confidence level = 95%

tc we can find using EXCEL.

syntax :

=TINV(probability, deg_freedom)

where probability = 1 - c

deg_freedom = n-1

We can do one sample t-test in MINITAB.

steps :

ENTER data into MINITAB sheet --> Stat --> Basic statistics --> 1-Sample t --> Samples in columns : select data column --> Perform hypothesis test --> Hypothesized mean : 15 --> Options --> Confidence level :95.0 --> Alternative : not equal --> ok --> ok

————— 6/17/2017 7:38:45 PM ————————————————————

Welcome to Minitab, press F1 for help.

One-Sample T: Hours

Test of mu = 15 vs not = 15

Variable N Mean StDev SE Mean 95% CI T P
Hours 34 9.46471 3.95236 0.67782 (8.08566, 10.84375) -8.17 0.000

Test statistic = -8.17

P-value = 0.000

P-value < alpha

Reject H0 at 5% level of significance.

Conclusion : There is sufficient evidence to say that population mean time that all students spend studying per week is different from 15 hours per weeks.

95% confidence interval for population mean mean time that all students spend studying per week is (8.086, 10.844).

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