Stats Problems - Report Example

7-5 # 14 pg 372 Using Table 7-2 on page 371, n ≥ 47 The minimum sample size needed to be 95% confident that the sample standard deviation s is within20% of σ is 47. #18 pg 373 µ = 5.670 g, σ = 0.062 g, n = 24, s = 0.049 g Degree of freedom = n – 1 = 24 – 1 = 23 = 38.08, = 11.69 95% Confidence Interval Estimate of σ 0.038 g< σ < 0.069 g Based on this result, we have 95% confidence that the limits 0.038 g and 0.069 g contain the true value of σ, the standard deviation of the weights of quarters made with new equipment. This interval includes the standard deviation of 0.062 g of the weights of quarters currently minted. Therefore, the new equipment does not appear to produce a standard deviation that is clearly lower than the standard deviation of 0.062 g from the old equipment. Even though the standard deviation of the sample (o.o49 g) is less than the current standard deviation of o.062 g, it is not low enough to be significant. Based on that available data, the new equipment does not appear to be effective in reducing the variation of the weights. #24 pg 374  n = 10, s = 0.477 (Using Excel function STDEV) Degree of freedom = n – 1 = 10 – 1 = 9 = 19.023, = 2.700 95% Confidence Interval Estimate of σ 0.328 < σ < 0.871 Based on this result, we have 95% confidence that the limits 0.328 and 0.871 contain the true value of σ, the standard deviation of the waiting times (in minutes) of customers at the Jefferson Valley Bank. 8-2 #24 pg 404 The alternate hypothesis involves not equal to sign (≠), therefore the test is two-tailed (non-directional) test and the α = 0.005 is divided equally between the two tails of the critical region that is α/2 = 0.0025 The critical z values are ±2.81. #26 pg 405 = 152/580 = 0.262, p = 0.25, q = 1 – 0.25 = 0.75, n = 580 = = 0.67 # 40 pg 405  It was claimed that about 34% of smokers experience sleeping difficulty. However, data suggested that it is not equal to 34%. 8-3 #6 pg 414 = 0.1593, p = 0.20, q = 1 – 0.20 = 0.80, n = 703 (a) The test statistic is = = -2.698 (b) The critical value for left tailed test is z = -1.645 (c) The p-value is p-value = 0.0035 (d) Reject null hypothesis. (e) Based on the preceding results, we can conclude that 15.93% is significantly less than 20% for all such hypothesis tests. The reason for this is that p-value (0.0035) is less than 0.05 significance level. #14 pg 415 = 58/1520 = 0.0382, p = 0.058, q = 1 – 0.058 = 0.942, n = 1520 The Null and alternate hypothesis are The level of significance, α is .01. The test statistics is == -3.31 The p-value is 0.0005. Since the p-value (0.0005) is less than the 0.01 significance level. Therefore, The decision is to reject the null hypothesis. The result suggests that the claim that fewer job applicants now use drugs is true. #24 pg 417 = 0.183, p = 0.27, q = 1 – 0.27 = 0.73, n = 785 The Null and alternate hypothesis are The level of significance, α is .01. The test statistics is == -5.41 The p-value is less than 0.0001. Since the p-value (< 0.0001) is less than the 0.01 significance level. Therefore, The decision is to reject the null hypothesis. The result suggests that the claim that the rate of smoking among those with four years of college is less than the 27% rate for the general population is true. The college graduates smoke at a lower rate than the others may be because of awareness that smoking is injurious for health. 8-4 #14, pg 424 µ = 4.50, σ = 2.87, α = 0.05, = 1.753, n = 73 The Null and alternate hypothesis are The level of significance, α is .05. The test statistics is = = -8.18 The p-value is less than 0.0001. Since the p-value (< 0.0001) is less than the 0.05 significance level. Therefore, The decision is to reject the null hypothesis. Bonds claim that last digits come from a population with a mean of 4.50 is not true. Based on the results, it appears that the distances were accurately measured. #18 pg 425  µ = 1.8 g, σ = 0.30 g, α = 0.05, = 1.71, n = 15 The Null and alternate hypothesis are The level of significance, α is .05. The test statistics is = = -1.30 The p-value is 0.1946. Since the p-value (0.1946) is greater than the 0.05 significance level. Therefore, The decision is to retain the null hypothesis. The claim that bumblebee bats are from the same population with a known mean of 1.8 g is true. The bats appear to come from the same population of bumblebee bats. 8-5 #14 pg 432 µ = 7 years, s = 2.4 years, α = 0.05, = 6.8 years, n = 21 The Null and alternate hypothesis are years years The level of significance, α is .05. The degree of freedom is 20. The test statistics is = = -0.38 The p-value is 0.3533. Since the p-value (0.3533) is greater than the 0.05 significance level. Therefore, The decision is to retain the null hypothesis. The claim that the mean life span of desktop PCs is less than 7 years in not true. #20 pg 433 µ = 700, s = 95.9, α = 0.05, = 660.3, n = 18 The Null and alternate hypothesis are The level of significance, α is .05. The degree of freedom is 17. The test statistics is = = -1.76 The p-value is 0.0970. Since the p-value (0.0970) is greater than the 0.05 significance level. Therefore, The decision is to retain the null hypothesis. The claim that these credit ratings are from a population with a mean that is equal to 700 is true. No, the results do not indicate that everyone will be eligible for a car loan. In this case, that the Bank of Newport requires accredit rating of 700 or higher for a car loan the alternate hypothesis will be and the p-value for one tail test is 0.0485, which is less than the 0.05 significance level. Therefore, the decision will be to reject null hypothesis. This means that the claim that these credit ratings are from a population with a mean that is equal to 700 is not true. #26 pg 434  µ = 0.3, s = 0.168, α = 0.05, = 0.295, n = 16 The Null and alternate hypothesis are The level of significance, α is .05. The degree of freedom is 15. The test statistics is = = -0.12 The p-value is 0.4533. Since the p-value (0.4533) is greater than the 0.05 significance level. Therefore, The decision is to retain the null hypothesis. The claim of a cereal lobbyist that the mean for all cereals is less than 0.3 g is not true. 8-6 # 12, pg 442 σ = 43.7 ft, s = 52.3 ft, α = 0.05, n = 81 The Null and alternate hypothesis are The level of significance, α is .05. The degree of freedom is 80. The critical values are = 106.6, = 57.15 The test statistics is = = 114.59 Since the test statistic (114.59) is greater than the critical value = 106.6 (that is in the critical region). Therefore, The decision is to reject the null hypothesis. The claim that the new production line has errors with a standard deviation different from 43.7 ft is true. The new production method appears to be worse than the old method. #16 pg 443  σ = 2.5 in, s = 1.187 in, α = 0.05, n = 18 The Null and alternate hypothesis are The level of significance, α is .05. The degree of freedom is 17. The critical value is = 8.672 The test statistics is = = 3.83 Since the test statistic (3.83) is in the critical region. Therefore, The decision is to reject the null hypothesis. The claim that heights of female supermodels vary less than the heights of woman in general is true. . Read More