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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 ................
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