.napter Section 4 Score: 0 of 5 pts 13 of 13 (11 complete) HW Score: 50%, 15 of
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.napter Section 4 Score: 0 of 5 pts 13 of 13 (11 complete) HW Score: 50%, 15 of 30 pe 7.4.32 Assigned Media l| Question Help A physician wants to develop criteria for determining whether a patient's pulse rate is atypical, and she wants to determine whether there are significant difterences between males and females. Use the sample pulse rates below Male Female 96 72 72 76 60 76 64 84 76 76 76 76 72 104 124 Click here to view a t distribution table Clck here to view page 1 of the standard normal distribution table. to view page 2 of the standard normal distribution table a. Construct a 95% confidence interval estimate of the mean pulse rate for males. (Round to one decimal place as needed.) b. Construct a 95% confidence interval estimate of the mean pulse rate for females. c. Compare the preceding results. Can we conclude that the population means for males and females are different? A. Yes, the population mean for females appears to be greater than the population mean for males O B. Yes, the population mean for males appears to be greater than the population mean for females c. Yes, because the two contidence intervals do not overtap, we can conclude that the two popuiation means are different the two confidence intervals overlap, we cannot conclude that the two population means are different O D. No, because Click to select your answer and then click Check Answer. All parts showing Clear All Check ndex.html DirecciónExplanation / Answer
Back-up Theory
Let X = pulse rate for male and Y = pulse rate for female. We assume that
Given X ~ N(M, M2), and Y ~ N(F, F2).
100(1 - ) % Confidence Interval for M, when M is not known is:
Xbar ± (tn- 1, /2)s1/n ……………………………………………………………………..(1)
where Xbar = sample mean, tn – 1, /2 = upper ( /2)% point of t-distribution with (n - 1) degrees of freedom, s1 = sample standard deviation and n = sample size.
Similarly, 100(1 - ) % Confidence Interval for F, when F is not known is:
YXbar ± (tn- 1, /2)s2/n …………………………………………………………………….(2)
where Ybar = sample mean, tn – 1, /2 = upper ( /2)% point of t-distribution with (n - 1) degrees of freedom, s2 = sample standard deviation and n = sample size.
Preparatory Work
Given n = 10, and = 5% [95% CI => = 5%],
tn- 1, /2 = t9, 0.025 = 2.262 [from Standard t-distribution Table] …………………………….(3)
Part (a)
Vide (1) and (3) above, 95% Confidence Interval for mean pulse rate of male is:
[68.4, 84.4] ANSWER
Details of calculations are given below:
=
0.05
n =
10
Xbar =
76.4
s =
11.2269517
tn-1, /2 =
2.262157158
95% CI for
Lower Bound =
68.3687226
Upper Bound =
84.4312774
Part (b)
Vide (2) and (3) above, 95% Confidence Interval for mean pulse rate of female is:
[74.8, 98.0] ANSWER
Details of calculations are given below:
=
0.05
n =
10
Xbar =
86.4
s =
16.24260516
tn-1, /2 =
2.262157158
95% CI for
Lower Bound =
74.7807403
Upper Bound =
98.0192597
Part (c)
Option (D) ANSWER
=
0.05
n =
10
Xbar =
76.4
s =
11.2269517
tn-1, /2 =
2.262157158
95% CI for
Lower Bound =
68.3687226
Upper Bound =
84.4312774
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