Statistics and Compute Mean Sum

Ch.7 # 14Evolutionary theories often emphasize that humans have adapted to their physical environment. One such theory hypothesizes that people should spontaneously follow a 24-hour cycle of sleeping and waking—even if they are not exposed to the usual pattern of sunlight.

To test this notion, eight paid volunteers were placed (individually)b in a room in which there was no light from the outside and no clocks or other indications of time. They could turn the lights no and off as they wished. After a month in the room, each individual tended to develop a steady cycle. Their cycle at the end of the study was as follows: 25, 27, 25, 23, 24, 25, 26, and 25.

Using the 5% level of significance, what should we conclude about the theory that 24 hours is the natural cycle? (That is, does the average cycle length under these conditions differ significantly from the 24 hours?) (a) Use the steps of hypothesis testing. (b) Sketch the distributions involved. (c) Explain your answer to someone who has never taken a course in statistics. Solution:

(a) Use the steps of hypothesis testing.Size of sample, n = 8Degree of freedom = n-1 = 8-1 = 7Sum of sample = i=1∑n=8xi = (25+27+25+23+24+25+26+25) = 200 | Time (in Hours)| Sum| Mean(xm)| (xi- xm)2| Standard Deviation| 1| 25| 200| Mean = Sum/n= 200/8= 25| 0| σ = √( i=1∑n=8(xi-xm)2/(n-1))= √(10/7)= √1.4285= 1.195| 2| 27| | | 4| |

3| 25| | | 0| |4| 23| | | 4| |5| 24| | | 1| |6| 25| | | 0| |7| 26| | | 1| |8| 25| | | 0| |

Here X = 25σ = 1.195We can define two-tailed statistics for above observations as follows: Null Hypothesis: H0: =24 vs. Ha: 24Rejection region:z < -z/2 or z > z/2Here significance level is 0.05,So, z0.025 = 1.96 (Using Statistical Ratio CalculatorNow,z = (X – μ) / σxWhere X is a normal random variable, μ is the mean, and σ is the standard deviation.

Where n is the sample size.Calculating t-test statistics:z = 25-24/(1.195/√8) = 1* √8/1.195= 2.828/1.195 = 2.366Calculating p-value:Degree of freedom = DF = 8-1 = 7Absolute value of calculated t-test statistics = |z| = |2.366| = 2.366 P(|z7| < 2.366) = 0.049899The p-value of 4.9% is very slightly less than significant level of 5%. Hence, we reject the null hypothesis and the average cycle length is not equal to 24 hours. (b) Sketch the distributions involved.

(c) Explain your answer to someone who has never taken a course in statistics. Statistical result shows that 24 hours is not the natural cycle in present case which is found to be slightly higher than expected mean. As there is only slight difference between statistical significance which can be practically ignored for validating the hypothesis.

Ch8 # 18H0: No difference in means of experimental and control groups H1: There is a difference.

n1=20 (number of cases in experimental group)n2=30 (number of cases in control group)mean 1 = 38s1 = 3mean 2 = 35s2 = 5

level of significance = 0.05

First, test for the equality of population variances between the two groups. F= s2^2/s1^2 = 5^2/3^2 = 25/9 = 2.78Compare with F(19,29) table value = 1.9446Since F-computed exceeds F-critical, the population variances are unequal.

Use the two sample t-test (assuming unequal variances)

Sample 1 size 20Sample 2 size 30Sample 1 mean 38Sample 2 mean 35Sample 1 S.D. 3Sample 2 S.D. 5t = (x-bar1 – x-bar2) / sqrt [ s1^2/n1+ s2^2/n2]sqrt [ s1^2/n1+ s2^2/n2] = 1.132843t = (38 – 35) = 3 / 1.132843t =2.6482

Degree of freedom =[s1^2/n1+s2^2/n2]^2 / [(s1^2/n1)^2/(n1-1)+(s2^2/n2)^2/(n2-1)] = [9/20 + 25/30]^2 / [(9/20)^2 /19+ (25/30)^2/29]Degree of freedom 47.5937

Compare the computed t 2.6482 with critical t (48 degrees of freedom and 0.05 level of significance) t-critical = 2.011

Computed t > critical tDecision: Reject the null hypothesis H0.The experimenter must conclude that there is a difference between theexperimental group and the control group. The instructional program has an effect

Ch.9 #17Ho: Ux=Uy=UzHa: not all are equal-compute the sum squares for between groups x,y,zSSB=sum(ni*(Xi-Xbar)^2) where ni=sample size for group i, Xi=sample average for group i, and Xbar=overall average. ni=25 for each group, Xx=5,Xy=4,Xz=6, Xbar=5SSB=25*(0^2+(-1)^2+1^2)=50degrees of freedom= k-1=3-1=2, k=number of groups-Compute the sum squares for the errorSSE=sum((ni-1)*Si^2)=24*(2+1.5+2.5)=144degrees of freedom= k(ni-1)=3(25-1)=3*24=72, k=number of groups –Compute Mean sum of squares for between groups and errorMSB=SSB/dof=50/2=25MSE=SSE/dof=144/72=2–compute F statistic= MSB/MSE=25/2=12.5–determine the critical F value for alpha 0.05, num dof=2, denom dof=72 Fcritical= 3.12–Since Fstatistic > Fcritical then reject Ho and concludeThe universities differ in their average time socializing