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The braking distances of a simple random sample of cars has $n = 32$. and $\overline{x} = 132$ ft. Find the margin of error and 95% confidence interval for the braking distances of cars, if the necessary assumptions are met. Also, $\sigma$ is known to be 7 ft.
Check the assumptions:
$n \ge 30$, so the distribution of sample means is approximately normal.
$\sigma$ known, so $z$-interval is appropriate
Margin of Error:
$\displaystyle{E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.96 \frac{7}{\sqrt{32}} \doteq 2.425}$
95% Confidence Interval:
$\overline{x} \pm E \rightarrow [132 - 2.425,132 + 2.425] \rightarrow [129.575,134.425]$
A random sample of the birth weights of 186 babies has a mean of 3103 g and a standard deviation of 696 g. These babies are born to mothers who did not use cocaine during their pregnancy.
What is the best point estimate of the mean weight of babies born to mothers who did not use cocaine during their pregnancies?
Construct a 95% confidence interval estimate of the mean birth weight for all such babies.
Compare the confidence interval from part b) to this confidence interval obtained for birth weights (in grams) of babies born to mothers who used cocaine during pregnancy: $2608 \lt \mu \lt 2792$ Does cocaine use appear to affect the birth weight of a baby?
$\overline{x} = 3103$ g
Check assumptions:
$n \ge 30$, so the distribution of sample means is approximately normal.
$\sigma$ unknown, so $t$-interval is appropriate.
$\displaystyle{E = t_{\alpha/2} \frac{s}{\sqrt{n}} = 1.984 \frac{696}{\sqrt{186}} \doteq 100.025}$
(Note: the above $t_{\alpha/2}$ was taken from a table that listed such values for degrees of freedom $100$ and $200$, but nothing in between. Consequently, we used the $t_{\alpha/2}$ value corresponding to the lower degrees of freedom to be conservative.)
95% Confidence Interval:
$\overline{x} \pm E \rightarrow [3103 - 100.025, 3103 + 100.025] \rightarrow [3002.975,3203.025]$
There is no overlap at all between the two confidence intervals. Yes, it appears that cocaine use affects babies' birth weights.
How many cars must be randomly selected and tested in order to estimate the mean braking distance of registered cars in the United States. We want 99% confidence that the sample mean is within 2 ft of the population mean and the population standard deviation is 7 ft.
Use at least 82 cars.
In a study designed to test the effectiveness of acupuncture for treating migraine, 142 subjects were treated with acupuncture and 80 subjects were given a sham treatment. The numbers of migraine attacks for the acupuncture treatment group had a mean of 1.8 and a standard deviation of 1.4. The numbers of migraine attacks for the sham treatment group had a mean of 1.6 and a standard deviation of 1.2.
Construct the 95% confidence interval estimate of the mean number of migraine attacks for those treated with acupuncture.
Construct the 95% confidence interval estimate of the mean number of migraine attacks for those given a sham treatment.
Compare the two confidence intervals. What do the results suggest about the effectiveness of acupuncture?
$\displaystyle{E = 1.984 \frac{1.4}{\sqrt{142}} \doteq 0.2}$
$\displaystyle{1.8 \pm 0.2 \rightarrow}$ 95% confidence interval: $[1.6,2.0]$
$\displaystyle{E = 1.990 \frac{1.2}{\sqrt{80}} \doteq 0.3}$
$1.6 \pm 0.3 \rightarrow$ 95% confidence interval: $[1.3,1.9]$
The two confidence intervals are very similar. The acupuncture treatment does not appear to be effective.
A simple random sample of 50 adults is obtained, and each person’s red blood cell count (in cells per microliter) is measured. The sample mean is 5.23. The population standard deviation for red blood cell counts is 0.54. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 5.4, which is the value often used for the upper limit of the range of normal values. Use the P-value method.
Check assumptions:
$n \ge 30$, so the distribution of sample means is approximately normal.
$\sigma$ is known, so use a $z$-test.
Claim: $\mu \lt 5.4$
$H_0: \mu = 5.4$
$H_1: \mu \lt 5.4$
Test Statistic: $\displaystyle{z = \frac{\overline{x} - \mu}{\sigma/\sqrt{n}} = \frac{5.23 - 5.4}{0.54/\sqrt{50}} \doteq -2.226}$
$p$-value $= 0.013$
Conclusion: Fail to reject the null hypothesis as the $p$-value $\gt \alpha = 0.01$
Inference: There is no significant evidence that the sample comes from a population with a mean blood count less than 5.4 cells per microliter.
The average 1-ounce chocolate chip cookie contains 110 calories. A random sample of 15 different brands of cookies resulted in the following calorie amounts. At the 0.01 significance level, is there sufficient evidence that the average caloric content is greater than 110 calories? Use the critical value method.
$$\begin{array}{cccccccc} 100 & 125 & 150 & 160 & 185 & 125 & 155 & 145\\ 160 & 100 & 150 & 140 & 135 & 120 & 110 & \end{array}$$Check the assumptions of the test. Since $n \lt 30$, we need to test for normality.
$\overline{x} = 137.3$, $s = 24.1$, $n = 15$, $Q_2 = 140$
Check both outlier tests:
$\overline{x} \pm 3s \rightarrow (65,209.6)$
$(Q_1-1.5 \cdot IQR,Q_3+1.5 \cdot IQR) = (67.5,207.5)$
Thus, no outliers
Check for skewness:
$\displaystyle{I = \frac{3(137.3-140)}{24.1} \doteq -0.34}$
Thus, not significantly skewed.
With no evidence to the contrary, we assume the distribution is approximately normal.
Since $\sigma$ is unknown, we use the $t$-test.
Claim: $\mu \gt 110$
$H_0 : \mu = 110$
$H_1 : \mu \gt 110$
Test statistic: $\displaystyle{t = \frac{\overline{x} - \mu}{s/\sqrt{n}} = \frac{137.3 - 110}{24.1/\sqrt{15}} \doteq 4.39}$
Critical Value: $2.624$ ($d.f. = 14$, one-tailed test, $\alpha = 0.01$)
Conclusion: Reject the null hypothesis, because the test statistic is in the rejection region.
Inference: There is significant evidence that the average caloric content is greater than 110 calories.
In an analysis investigating the usefulness of pennies, the cents portions of 100 randomly selected credit card charges are recorded. The sample has a mean of 47.6 cents and a standard deviation of 33.5 cents. If the amounts from 0 to 99 cents are all equally likely, the mean is expected to be 49.5 cents. Use the confidence interval method at a 0.01 significance level to test the claim that the sample is from a population with a mean equal to 49.5 cents. What does the result suggest about the cents portions of credit card charges?
Check the assumptions.
$n \ge 30$, so the distribution of sample means is approximately normal.
$\sigma$ is unknown, so use the $t$-test.
Claim: $\mu = 49.5$
$H_0: \mu = 49.5$
$H_1: \mu \neq 49.5$
Confidence interval:
$\displaystyle{E = t_{\alpha/2} \frac{s}{\sqrt{n}} = 2.626 \frac{33.5}{\sqrt{100}} \doteq 8.8}$
$\overline{x} \pm E \rightarrow 47.6 \pm 8.8 \rightarrow [38.8,56.4]$
Conclusion: Fail to reject the null hypothesis, as $49.5$ is in the confidence interval.
Inference: There is no significant evidence that the mean is not 49.5.
Note, this means that the cents portion of credit card charges could be equally likely from 0 to 99.
A sample of 106 body temperatures has a mean of $98.20^{\circ}$F. Assume that $\sigma$ is known to be $0.62^{\circ}$F. Use a 0.05 significance level to test the claim that the mean body temperature of the population is equal to $98.6^{\circ}$F, as is commonly believed. Is there sufficient evidence to conclude that the common belief is wrong? Find the critical values, P-value, and confidence interval.
Check the assumptions.
$n \ge 30$, so the distribution of sample means is approximately normal.
$\sigma$ is known, so use the $z$-test.
Claim: $\mu = 98.6$
$H_0: \mu = 98.6$
$H_1: \mu \neq 98.6$
Test Statistic: $\displaystyle{z = \frac{98.20 - 98.6}{0.62/\sqrt{106}} \doteq -6.64}$
Critical Values: $\pm 1.96$
$p$-value $= 3.104 \times 10^{-11}$
Confidence Interval:
$\displaystyle{E = z_{\alpha/2} \frac{\sigma}{\sqrt{n}} = 1.96 \frac{0.62}{\sqrt{106}} \doteq -0.12}$
$\overline{x} \pm E \rightarrow [98.20 - 0.12,98.20 + 0.12] \rightarrow [98.08,98.32]$
Conclusion: Reject the null hypothesis for any of the following reasons: 1) the test statistic is in the rejection reason; 2) the $p$-value is less than $\alpha$; or 3) $98.6$ is not in the confidence interval.
Inference: There is significant evidence that the mean body temperature is not $98.6^{\circ}$ F.