17.) A sample of 80 women is obtained and their heights (in Inches) and pulse rates (in beats per min.) are measured. the linear correlation coefficient is 0.202 and the equation of the regression line is y=17.6 + 0.940x where x represents height. the mean of 80 heights is 63.6 in. and the mean of the 80 pulse rates is 73.3 beats per min. find the best predicted pulse rate of a women who is 67 in. tall use a significance level of a=0.01

73.3 beats per minute

16.) (show Work) The eruption height and the time interval after eruption of a geyser were measured and are shown below.
Height 120,135,130,160,140,130,150,135
Interval after 75,82,104,97,60,104,102,111
(a) Find the value of the linear correlation coefficient r
(b) Find the critical values of r form the table drawing the critical values for the Pearson correlation coefficient using a=0.05. the critical values are
(c) Is there sufficient evidence to conclude that there is a linear correlation between the two variables

Work …
(a)
x / y / x^2 / y^2 / xy
120 / 75 / 14400 / 5625 / 9000
135 / 82 / 18225 / 6724 / 11070
130 / 104 / 16900 / 10816 / 13520
160 / 97 / 25600 / 9409 / 15520
140 / 60 / 19600 / 3600 / 8400
130 / 104 / 16900 / 10816 / 13520
150 / 102 / 22500 / 10404 / 15300
135 / 111 / 18225 / 12321 / 14985
n = / 8
Sums = / 1100 / 735 / 152350 / 69715 / 101315
SS(x) = Σ(x^2) - [(Σx)^2 /n] = 152350- (1100^2 /8) = 1100.00
SS(y) = Σ(y^2) - [(Σy)^2 /n] = 69715 - (735^2 /8) = 2186.88
SS(xy) = Σ(xy) - [(Σx)(Σy)/n] = 101315 - (1100 * 735/8) = 252.50
Correlation coefficient, r = SSxy / √[SS(x) * SS(y)] = 252.50/√(1100 * 2186.88) = 0.1628
(b) From the table, critical value of r for n = 8 and α = 0.05 is 0.707
(c) Since 0.1628 < 0.707, there is no sufficient evidence to conclude that there is a linear correlation between the two variables
Answers …
(a) 0.1628
(b) 0.707
(c) No

15.) (show Work) the heights were measured for nine supermodels they have a mean of 68.6 in. and a standard deviation of 2.3in. use the traditional method and a 0.01 significance level to test the claim that supermodels have heights with a mean that is greater than the mean of 63.6 in. for women from the general population

n = 9
μ = 63.6
s = 2.3
x-bar = 68.6
Hypotheses:
Ho: μ ≤ 63.6
Ha: μ > 63.6
Decision Rule:
α = 0.01
Degrees of freedom = 9 - 1 = 8
Critical t- score = 2.896459446
Reject Ho if t > 2.896459446
Test Statistic:
SE = s/n = 2.3/√9 = 0.766666667
t = (x-bar - μ)/SE = (68.6 - 63.6)/0.766666666666667 = 6.52173913
Decision (in terms of the hypotheses):
Since 6.52173913 > 2.896459446 we reject Ho and accept Ha
Conclusion (in terms of the problem):
The claim is valid. There is sufficient evidence that supermodels have heights with a mean that is greater than the mean of 63.6 in. for women from the general population

14.) (show Work) A clinical trail tests a method designed to increase the probability of conceiving a girl. In the study 354 babies were born and 200 of them were girls. Use the sample data with a 0.01 significance level to test the claim that with this method the probability of a baby being a girl is greater than 0.5. use this information to answer the following questions
(a) Which of the following is the hypothesis tests to be conducted
(b) What is the test statistic z=____
(c) what is the p-value
(d) what is the conclusion
(e) does the method appear to be effective?

Hypothesis test details …
n = 354
p = 0.5
p' = 200/354 = 0.5649718
Hypotheses:
Ho: p ≤ 0.5
Ha: p > 0.5
Decision Rule:
α = 0.01
Critical z- score = 2.326347874
Reject Ho if z > 2.326347874
Test Statistic:
SE = p (1 - p)/n= √(0.5 * (1 - 0.5)/√354) = 0.0265747
z = (p' - p)/SE = (0.564971751412429 - 0.5)/0.0265747001726367 = 2.444872416
p- value = 0.0072452
Decision (in terms of the hypotheses):
Since 2.4448724 > 2.326347874 we reject Ho
Conclusion (in terms of the problem):
The claim is valid. It appears that the probability of conceiving a baby girl by this method is greater than 0.5
Answers …
(a) Options are missing here?
The correct test to be conducted is Ho: p ≤ 0.5 and Ha: p > 0.5
(b) 2.445
(c) 0.0072
(d) Reject Ho
(e) Yes

13.) Identify the type 1 error and the type 2 error that corresponds to the given hypothesis, the percentage of adults who have a job is less than 88%

Type I error: Concluding that the percentage of adults who have a job is less than 88% while in reality it is 88% or more
Type II error: Failing to conclude that the percentage of adults who have a job is less than 88% while in reality it is less than 88%

11.) Using the simple random of weights of women form a data set we obtain these sample statistics n=40 and x=154.21lb. Research from other sources suggests that the population of weights of women has a standard deviation given by o=31.01lb.
(a) Find the best point estimate of the mean weight of all women
(b) Find a 99% confidence interval estimate of the mean weight of all women.

(a) 154.21 lb
(b) (141.58 lb, 166.84 lb)

10.) (how work)(a) with n=14 and p=0.7 find the binomial probability p(9) by using the binomial probability table
(b) if np≥5 and nq≥5 also estimate the indicated probability by using the normal distribution as an approximation to the binomial of np<5 or nq<5 then state that the normal approximation cannot be used

(a) Referring to the binomial probability table, for n = 14 and p = 0.7, P(9) = 0.196
(b) np = 14 * 0.7 = 9.8 and nq = 14 * 0.3 = 4.2
Since nq < 5, normal approximation can’t be used.

8.) Assume that thermometer readings are normally distribution with a mean of 0 degrees Celsius and a standard deviation of 1.00 degrees Celsius a thermometer is randomly selected and tested
Find the probability of the reading between -1.13 and 1.87

0.84

7.) A brand name has a 40% recognition rate. If the owner of the brand wants to verify that rate by beginning with a small sample of 15 randomly selected consumers find the probability that exactly 6 of the 15 consumers recognize the brand name also find the probability that the number who recognize the brand name is not 6.

P(x = 6) is 0.2066
P(x ≠ 6) is 0.7934