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13.8 Assume that, on average, healthy young adults dream90 minutes each night, as inferred from a number of measures, including rapid eye movement (REM) sleep. An investigator wishes to determine whether drinking coffee just before going to sleep affects the amount of dream time. After drinking a standard amount of coffee, dream time is monitored for each of28 healthy young adults in a random sample. Results showa sample mean,X, of 88minutes and asample standard deviation, s, of 9minutes.

N = 28, x = 90, xbar = 88, = 9, df= n-1, df = 28-1=27.

a) Use t to test the null hypothesis at the .05 level of significance

Test statisticst= (x-u)

/n

T= 88-90 =t = -2=-1.176

928 1.70

Df = 28 - 1 = 27

Excel formula: =t.dist.2t(t,df)

P = p(t> -1.176) =0.249860254 or rounded 0.25

P value = 0.25

Since alpha value of 0.05 is less than the p value of 0.25, then we would not reject theHo. Therefore, the sample data did not show significant evidence that drinking coffee beforegoing to sleep affects the amount of dreamtime

(b) If appropriate (because the null hypothesis has been rejected), construct a 95 per-cent confidence interval and interpret this interval.

The null hypothesis has been retained, so we cannot construct a 95% confidence interval. However, for practice purposes, see below

N = 28, x = 90, xbar = 88, = 9, df= n-1, df = 28-1=27,

df=27,=0.05)=2.05 (t-table)

xt

n

882.059=

28

883.487

883.487=84.51

88+3.487=91.49

C.I = 84.51, 91.49

This suggests that with 95 percent confidence, the average dream time after drinking coffee is between 84.51 and 91.49 minutes.

14.12 To determine whether training in a series of workshops on creative thinking increases IQ scores, atotal of 70 studentsare randomly divided into treatment and control groupsof 35 each. After two months of training, the sample mean IQ1Xfor the treatment group equals110, and thesample mean IQ 2Xfor the control groupequals 108. The estimatedstandard error equals 1.80.

(a) Using t, test the null hypothesis at the .01 level of significance

Ho: u1 = u2

Ha: u1>u2

Alpha = 0.01

Total = 70, N1 = 35, N2 = 35, u1= 110, u2 =108,

DF = 35 -1 = 34

Check T value table. alpha 0.01 and DF 34 = 2.441 

Critical value = 2.441 

t = (xbar1 - xbar2) = 110 - 108 =1.1111

S_e1.80

We reject Ho because t > critical value.

(b) If appropriate (because the null hypothesis has been rejected), estimate the stan-dardized effect size, construct a 99 percent confidence interval for the true population mean difference, and interpret these estimates.

Not appropriate since he null hypothesis (Ho) was not rejected.Based on the analysis we have done, we can conclude that creative thinking workshops do not have an effect on IQ scores.

Alpha = 0.99 - 1 = 0.01 alpha

DF = 34

Critical value 2.728

110- 108 plus minus 2.7281.80+1.80sq

35 35

2 + 1.1738 = 3.1738

2 - 1.1738 = 0.8262

C.I =0.8262, 3.1747