Question

Problem 1) Consider the example we used in class on daily activity and obesity. Briefly, 10 lean and 10 obese volunteers were recruited to wear a sensor that monitored their every move for 10 days. The time that each subject spent walking/standing, sitting, and lying down were recorded. For more details see example 21.2 in the text. In addition to the variable we looked at in class, the data were analyzed to see if there was a difference in the average time the subjects spent lying down. The data are in the Excel file, tab lying down.

a) For one experimental unit, what is the response variable? Categorical or quantitative?

b) Verify the two conditions for inference: 2 independent random samples? is n1 ≥ 30 or are the data approximately normally distributed or have no extreme asymmetry or outliers? Is n2 ≥ 30 or are the data approximately normally distributed or have no extreme asymmetry or outliers?

c) Fill in the table with the summary statistics using Excel or StatKey (NOT by hand) for each group. Designate which is Group 1 and Group 2

Group LyingDownTime

lean 555.5

lean 450.65

lean 537.362

lean 489.269

lean 514.081

lean 506.5

lean 467.7

lean 567.006

lean 531.431

lean 396.962

obese 521.044

obese 514.931

obese 563.3

obese 532.208

obese 504.931

obese 448.856

obese 460.55

obese 509.981

obese 448.706

obese 412.919

Conduct a 4-step significance test to determine if there is evidence of a difference in the average time the subjects spent lying down. Be sure to state the conclusion in the context of the problem, not statistics jargon. e) Calculate a 95% confidence interval. (Hint: it should look very odd.) Remember that with a quantitative response variable, a 95% confidence interval leads to a two-tailed significance test with α = 0.05 as our cutoff. If we reject Ho, then we would not expect µ0 to be inside the confidence interval. Look at your confidence interval. Does it contain the value for µ0? Look at the conclusion of the significance test. Are they consistent?

Answer 1

RESPONSE VARIABLE IS LYING DOWN TIME

VARIABLE IS QUANTITATIVE

SAMPLES ARE INDEPENDENT

DATA ARE APPROXIMATELY NORMALLY DISTRIBUTED.

OUTLIERS ARE NONE.

NULL HYPOTHESIS H0: $\mu(L)-\mu(O)=0$

ALTERNATIVE HYPOTHESIS Ha: $\mu(L)-\mu(O)\ne0$

alpha=0.05

Firstly we will check are variances are equal or not

NULL HYPOTHESIS H0: $\sigma^21=\sigma^2$ 2

So by using excel data analysis tool

we have

From this table as we can see P value =0.3735 >0.05 level of significance therefore FAIL TO REJECT NULL HYPOTHESIS H0.

THEREFORE VARIAINCES ARE EQUAL.

Now we can test

From above table we can see that P value = 0.6593 >0.05 level of significance therefore NOT SIGNIFICANT

Decision: FAIL TO REJECT NULL HYPOTHESIS HO.

CONCLUSION: WE DONT HAVE SUFFICIENT EVIDENCE TO SHOW THAT THERE IS DIFFERENCE IN MEAN LYING TIME OF LEAN AND OBESE PEOPLE.

e) Pooled Variance
s^2p = ((df1)(s^21) + (df2)(s^22)) / (df1 + df2) = 43909.11 / 18 = 2439.39

Standard Error
s(M1 - M2) = √((s^2p/n1) + (s^2p/n2)) = √((2439.39/10) + (2439.39/10)) = 22.09

Confidence Interval
μ1 - μ2 = (M1 - M2) ± ts(M1 - M2) = 9.91 ± (2.1 * 22.09) = 9.91 ± 46.4051

μ1 - μ2 = (M1 - M2) = 9.91, 95% CI [-36.4951, 56.3151].

You can be 95% confident that the difference between your two population means (μ1 - μ2) lies between -36.4951 and 56.3151.

SINCE CONFIDENCE INTERVAL DOES CONTAIN HYPOTHESIZED VALUE "0" THEREFORE NOT SIGNIFICANT. FAIL TO REJECT H0 .HENCE WE CAN CONCLUDE THAT WE DONT HAVE SUFFICIENT EVIDENCE TO SHOW THAT THERE IS DIFFERENCE IN MEAN LYING TIME OF LEAN AND OBESE PEOPLE.