Use SPSS for this exam, be sure to include all the following information to get

Question

Use SPSS for this exam, be sure to include all the following information to get fill credits: (1) null and alternative hypotheses (15%) (2) name of the test (15%, explain why you chose it) (3) the test-statistic and its degree of freedom or ANOVA table (if applicable) (30%) (4) p-value 15%, and (5) scientific conclusion (25%, using alpha = 0.05 significance). Assume that the following data is randomly sampled from a normal population of D. melanogaster, body mass (mg) is measured for each fly: 0.627. 0.642, 0.661, 0.620, 0.592, and 0.5449. (a) Test whether the flies were drawn from a fly population of mean body mass 0.500 mg. (b) Another fly sample is randomly collected and measured (in mg): 0.653, 0.684. 0.609, 0.712, and 0.638. Test whether these two samples were selected from the populations with equal means? The following table lists total hot cocoa sales for five stores during four times of the year. Apply an analysis of variance to this data, report your complete ANOVA table. Test the null hypothesis that hot cocoa sales are the same for all four months surveyed. Assume that the March is a control group. Use an appropriate test to compare the control mean hot cocoa consumption with the other means. Fit a simple linear model of Y on one independent variable X_2only, report (1) the Y intercept (b) and regression coefficient; (2) test H_0: beta = 0. Fit a linear model of Y on ALL Xs, report (1) the Y intercept and partial regression coefficients; (2) test if there is a significant multiple regression relationship b Y and Xs; (3) test H_0: beta_3 = 0 (for X_s).

Explanation / Answer

Answer to question# 1)

Part a)

The hypothesis would be :

Null hypothesis: Mean body mass is : 0.500

Alternate hypothesis : Mean body mass is different from 0.500

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The test that applies is one sample mean T test

this is because the population Standard deviation is not provided in the question

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Enter the data in data view, after creating the variable bodymass1

Click on analyze tab , compare means , one sample T test

In the one sample T test window:

enter the variable bodymass1

enter the mean = 0.500

click ok to get the output

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We get:

Test Statistic t = 1.444

df = 7

P value = 0.19189

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Conclusion: Since the P value 0.19189 is greater than alpha 0.05 , we fail to reject the null

Thus we conclude that Mean body mass is 0.500 mg

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Answer to part b)

Now in variable view create the second variable as Samplenumber

Now in data view enter the new sample body mass in column 1 only and enter the sample number 1 for first seven observations and 2 for next seven set of observations

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Now click on analyze > compare mean > Independent sample T test

enter Testing variable as bodymass1 and grouping variable as samplenumber

The click on "Define groups"

You can define the groups as : 1 = sample 1 and 2 = sample 2

The click continue and ok

click ok to get the output

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We get Test Statistic T = 0.822668

df = 6

P value = 0.4422

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Conclusion: Since the P value 0.4422 > significance level 0.05 , we fail to reject the null

Thus we conclude that the two samples have same mean.

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