I. The following are readings of resting heart rates (beats per minute) of a sample of 40 year old males who participate in vigorous exercise at least three times per week.
51 71 42 69 74 61 60
49 75 55 54 47 52 55
66 65 70 58 77 63 54
57 58 58 55 61 37 66
65 41 46 47 53 60 48
52 67 70 60 58 53 56
43 67 72 67 62 50 39
Do all tests at the .05 level of significance.
1. Group the data into a frequency distribution using seven classes. For the grouped data formulas, calculate the mean, median, standard deviation, and 30th percentile.
MIN=
37
MAX=
77
K=
7
CLASS width=
5.714286
Row Labels
Count of DATA
37-42
4
43-48
5
49-54
9
55-60
12
61-66
8
67-72
8
73-78
3
Grand Total
49
MEAN = SUM(XI*Fi)/sum(Fi) = 2626/49 = 53.5918
Median = (n+1/2)th observation = 25th observation = 58
var = sum(Xi*Fi^2)/sum(Fi) - mean^2 = 153652/49-53.5918^2 = 263.6740748
sd = sqrt(var) = 16.23804406
39th percentile = 55 + (5/12)*(49*30/100-18) = 53.625
2. For the grouped data construct a histogram, a frequency polygon, and a “less than” ogive.
class interval
Fi
C Fi
37-42
4
4
43-48
5
9
49-54
9
18
55-60
12
30
61-66
8
38
67-72
8
46
73-78
3
49
Use the ungrouped data:
3. Find 95 percent confidence intervals for the mean and the standard deviation of the population.
95 percent confidence intervals for the mean
Mean = 57.88
Sd = sqrt(var) = 9.850
n = 49.00
Critical value, z(a/2) = z(0.05/2) = 1.960
CI = mean +- z(a/2,n-1)*(sd/sqrt(n))
Lower = 57.8776 - 1.96*(9.85019596476221/sqrt(49)) = 55.12
Upper = 57.8776 + 1.96*(9.85019596476221/sqrt(49)) = 60.64
4. Test the null hypothesis H0: µ = 60 versus HA: µ < 60. (µ is the population mean)
H0: µ = 60 versus HA: µ < 60
critical value, -z(a)
-z(0.05)
-1.645
test statistic, z = (mean-u)/(sd/sqrt(n))
= (57.8776-60)/(9.8502/sqrt(49))
-1.5083
p-value
P(Z
P(z<-1.5083)
=NORMSDIST(-1.5083)
0.065738882
With z=-1.5, p>5%, I fail to reject ho and conclude that u=60
5. Test the null hypothesis H0: p = .20 versus HA: p ≠ .20 where p is the proportion of the population with readings of more than 68.
X= 8
n= 49
p-hat = X/n= 0.163 =8/49
po= 0.200
test statistic, z = (phat-p)/sqrt(p*(1-p)/n)
=(0.163-0.2)/SQRT(0.2*(1-0.2)/49)
-0.648
p-value
2*(1-P(z<|z|)
2*(1-P(z
normsdist(abs(-0.6475))
0.5173
With z=-0.648, p>5%, I fail to reject Ho and conclude that p = .20
II. A group of ten 60-year old men with high blood pressure had their blood pressures checked. Recorded are the diastolic pressures (the second number of blood pressure readings). They then had six months of a controlled exercise...