A 95% confidence interval for the difference between pulse rates for men and women is given by $-12.2 \lt \mu_{men} - \mu_{women} \lt -1.6$. What does this confidence interval suggest about pulse rates of men and women?
In a randomized controlled trial conducted with children suffering from viral croup, 46 children were treated with low humidity while 46 other children were treated with high humidity. The low humidity group had a mean Westley Croup Score of 0.98 with a standard deviation of 1.22, while the high humidity group had a mean score of 1.09 with a standard deviation of 1.11. Assume the samples are independent and taken from normally distributed populations. Test the claim that the means for these two groups are different using a 0.05 significance level, under both of the conditions below.
$H_0: \mu_1 = \mu_2$
$H_1: \mu_1 \neq \mu_2$
Test statistic: $t = -0.452$
Critical values: $t = \pm2.014$ (or with technology, $\pm1.987$)
Conclusion: Fail to reject null hypothesis, as test statistic is not in rejection region.
Inference: There is no significant evidence that suggests increased humidity results in a difference in the treatment of croup.
$H_0: \mu_1 = \mu_2$
$H_1: \mu_1 \neq \mu_2$
Test statistic: $t=-0.452$
Critical values: $t=\pm 1.987$
$p$-value $\gt 0.20$ (with technology, $=0.6521$)
Conclusion: Fail to reject null hypothesis, as test statistic is not in rejection reg ion.
Inference: There is no significant evidence that suggests increased humidity results in a difference in the treatment of croup.
The mean tar content of a simple random sample of 25 unfiltered king size cigarettes is 21.1 mg, with a standard deviation of 3.2 mg. The mean tar content of a simple random sample of 25 filtered 100 mm cigarettes is 13.2 mg with a standard deviation of 3.7 mg. Construct a 90% confidence interval estimate of the difference between the mean tar content of unfiltered king size cigarettes and the mean tar content of filtered 100 mm cigarettes.
$6.3 \textrm{ mg} \lt \mu_1 - \mu_2 \lt 9.6 \textrm{ mg}$. The confidence interval does not include a difference of 0, and thus suggests the mean tar content of unfiltered king size cigarettes is greater than the mean for filtered 100 mm cigarettes.
$6.3 \textrm{ mg} \lt \mu_1 - \mu_2 \lt 9.5 \textrm{ mg}$. The confidence interval is slightly smaller than in the previous case.
Below are costs (in dollars) for repairing front-end and rear-end damage in controlled low-speed crash tests for different types of cars. Construct a 95% confidence interval for the mean difference between repair costs for front-end and rear-end damage. Does there appear to be a difference?
$$\begin{array}{rccccccccc} & \textrm{Toyota} & \textrm{Mazda} & \textrm{Volvo} & \textrm{Saturn} & \textrm{Subaru} & \textrm{Hyundai} & \textrm{Honda} & \textrm{Volkswagon} & \textrm{Nissan}\\ \textrm{Front} & 936 & 978 & 2252 & 1032 & 3911 & 4312 & 3469 & 2598 & 4535\\ \textrm{Rear} & 1480 & 1202 & 802 & 3191 & 1122 & 739 & 2769 & 3375 & 1787 \end{array}$$This is paired data.
$-\$647.9 \lt \mu_d \lt \$2327.0$. The confidence interval includes the difference 0, so there does not appear to be a significant difference between the costs of repairing the front vs rear ends.
Below are fuel consumption ratings in miles/gal for 20 different cars measured under both the old rating system and a new rating system (listed in the same order). Test the claim that the old ratings are higher than the new ratings at a $0.01$ significance level.
$$\begin{array}{ccccccccccc} \textrm{old ratings} & 16 & 18 & 27 & 17 & 33 & 28 & 33 & 18 & 24 & 19\\ & 18 & 27 & 22 & 18 & 20 & 29 & 19 & 27 & 20 & 21\\\\ \textrm{new ratings} & 15 & 16 & 24 & 15 & 29 & 25 & 29 & 16 & 22 & 17\\ & 16 & 24 & 20 & 16 & 18 & 26 & 17 & 25 & 18 & 19\\ \end{array}$$This is paired data.
$H_0: \mu_d = 0$
$H_1: \mu_d \gt 0$
Test statistic: $t = 14.104$
Critical value: $t=2.539$
$p$-value $\lt 0.005$ (it's very small, according to technology)
Conclusion: Reject the null hypothesis, as the $p$-value is less than $\alpha = 0.01$
Inference: There is significant evidence that the old ratings are higher than the new ratings.
Listed below are times (seconds) that 10 animated Disney movies showed the use of tobacco and alcohol. Use a 0.05 significance level to test the claim that the mean of the differences is greater that 0 seconds, so that more time is devoted to showing tobacco than alcohol. Use the critical value method. $$\begin{array}{l|cccccccccc} \textrm{Tobacco} & 176 & 51 & 0 & 299 & 74 & 2 & 23 & 205 & 6 & 155\\\hline \textrm{Alcohol} & 88 & 33 & 113 & 51 & 0 & 3 & 46 & 73 & 5 & 74\\ \end{array}$$
Many studies have been conducted to test the effects of marijuana use on mental abilities. In one such study, groups of light and heavy users of marijuana in college were tested for memory recall, with the results given below.
$$\begin{array}{ccc} \textrm{Items sorted correctly by light marijuana users} & & \textrm{Items sorted correctly by heavy marijuana users}\\ n = 64,\, \overline{x} = 53.3,\, s = 3.6 & & n = 65,\, \overline{x} = 51.3,\, s = 4.5 \end{array}$$Use the P-value method with a 0.01 significance level to test the claim that the population of heavy marijuana users has a lower mean than the light users.
Construct a 98% confidence interval for the difference between the two population means. What does the confidence interval suggest about the equality of the two population means?
A researcher suggests that male nurses earn more than female nurses. A survey of 16 male nurses and 20 female nurses reports these data. Assume the distributions are normal and use $\alpha = 0.05$. $$\begin{array}{cc} \textrm{Female} & & \textrm{Male}\\ n = 20,\, \overline{x} = \$23570,\, s = \$250 & & n = 16,\, \overline{x} = \$23800,\, s = \$300 \end{array}$$
Are the variances equal?
Test the claim that male nurses earn more than female nurses.