1. Exercise 4.40 on p. 216 of OpenIntro Statistics. You can draw the distributions for part d by hand.
  1. What is the probability that a randomly chosen light bulb lasts more than 10,500 hours?
1-pnorm(1.5)
## [1] 0.0668072

\[ z = \frac{10,500 - 9,000}{1,00} = 1.5\\ P(z\geq1.5)=1-pnorm(1.5)=.0668 \]
b) Describe the distribution of the mean lifespan of 15 light bulbs.
The mean lifespan of a 15 lightbulb sample is also distributed normally with the mean \(9,000\) and with standard deviation \(\frac{1,000}{\sqrt{15}}\)

  1. What is the probability that the mean lifespan of 15 randomly chosen light bulbs is more than 10,500 hours?
1-pnorm(5.8)
## [1] 3.315746e-09

\[ P(\bar{X}\geq10,500)\\ z=\frac{10,500-9,000}{\frac{1,000}{\sqrt{15}}}=5.8\\ P(z\geq5.8)=1-pnorm(5.8)=0 \]
d) drawn by hand on back.
e) No. We cannot find a z score to find the probability that a randomly chosen variable will last more tha 10,500 hr. if the distribution is skewed and we cannot use the Central limit Thm. to say that the sample of 15 is normal because it is too small.

  1. Exercise 4.38 on p. 216 of OpenIntro Statistics
    Plot A-(3), Plot B-(1), Plot C-(2), larger sample sizes have greater spread around mean.

  2. You are using a sensitive scale to double-check the weight of a rock sample that should be 10.00 grams. Measurements with this scale are known to be normal with a standard deviation of \(\sigma=0.02\) grams. You decide to weigh the sample 3 times and find that the 3 measurements have mean 10.04 grams.
  1. Do you have statistically significant evidence to indicate that the rock sample does not really weigh 10.00 grams? Show all of the steps in your reasoning process.
pnorm(-3.464)+(1-pnorm(3.464))
## [1] 0.0005322065

\[ H_o: \mu=10\\ H_a: \mu\ne10\\ Test Statistic: z=\frac{10.04-10}{\frac{.02}{\sqrt{3}}}= 3.464\\ \\ p-value: p=(z\leq-3.464)+P(z\geq3.464)=.000532\\ \]
Assuming level of significance .05, since p-value is less than .05 we reject \(H_o\) in favor of \(H_a\). We have statistically significant evidence to indicate that the rock sample does not really weigh 10 grams.

  1. Construct and interpret a 90% confidence interval for the true weight of the rock sample.
    \((\bar{x}-z*\frac{\sigma}{\sqrt{n}},\bar{x}+z*\frac{\sigma}{\sqrt{n}})\),\(\bar{x}=10.04\), \(n=3\), \(\sigma=.02\) for 90% confidence we use \(z=1.645\), so We are 90% confident that the population mean rock weight lies between (10.021, 10.0589).
  1. Exercise 4.14 on p. 208 of OpenIntro Statistics
  1. false; CI gives us interval for population mean
  2. false; since we have a large enough sample size we can use Central Limit Thm. so it is valid
  3. false; CI tells about population mean not sample mean
  4. true;correct interpretation of CI
  5. false; it would be broader
  6. false; we would need a sample \(3^2\) times larger
  7. true; \(\frac{89.11-80.31}{2}=4.4\)
  1. Exercise 5.12 on p. 259 of OpenIntro Statistics
    a)Write down the hypotheses that would be appropriate for testing if the police ocers appear to have been exposed to a higher concentration of lead. \[ \mu_p = \text{the mean concentration of lead in the police officers}\\ H_o: \mu_p=35\\ H_a: \mu_p>35\\ \]
  1. The conditions necessary for inference on these data are:
    independence, which is satisfied if the sample sample size tested is less than 10% of the population size (with \(n_p\)=52, independence is satisfied if there are at least 520 police officers subjected to constant inhalation of automobile exhaust fumes in that urban environment)
    normality, which is satisfied if most observations lie around 2.5 s.d. from the mean.

c)Test the hypothesis that the downtown police ocers have a higher lead exposure than the group in the previous study. Interpret your results in context.

1-pt(17.066,51)
## [1] 0

\[ H_o: \mu_p=35\\ H_a: \mu_p>35\\ t=\frac{124.32-35}{\frac{37.74}{\sqrt{52}}}=17.066\\ p-value=1-pt(17.066,51)=0 \] We reject the \(H_o\) in favor of the \(H_a\). There is significant statistical evidence that the mean concentration of lead of downtown police officer blood stream is greater than that of the previous study,35.

  1. Based on your preceding result, without performing a calculation, would a 99% confidence interval for the average blood concentration level of police officers contain 35?
    Since our p-value is so close to 0, a 99% confidence interval would not include 35.
  1. Exercise 5.20 on p. 261 of OpenIntro Statistics
  1. Is there a clear difference in the average reading and writing scores? No, the histogram only shows that there are slightly more negative differences so there may slightly higher writing scores but it is not clear.
  2. Are the reading and writing scores of each student independent of each other? No.
  3. Create hypotheses appropriate for the following research question: is there an evident difference in the average scores of students in the reading and writing exam? let \(\mu_d\) = mean reading score - mean writing score, then $H_o: _d = 0, H_a: _d0\ $
  4. Check the conditions required to complete this test. Independence, satisfied with sample (200 students) being less than 10% of population of high school seniors. normality, satisfied because we have a large enough sample of 200
  5. The average observed difference in scores is x_read-write = 0.545, and the standard deviation of the differences is 8.887 points. Do these data provide convincing evidence of a difference between the average scores on the two exams?
-.545/(8.887/sqrt(200))
## [1] -0.867274
 1-pt(-.867,199)
## [1] 0.8065069

\[ t=\frac{-.545-0}{\frac{8.887}{\sqrt{200}}}=-.867\\ p-value=1-pt(-.867,199)=.806 \]
We would fail to reject the null hypotheses, there is not convincing evidence of a difference between the average scores on the two exams.
f) What type of error might we have made? Explain what the error means in the context of the application. We might have made a Type II error by failing to reject the null if it is actually false and there is a difference between the average scores on the two exams.
g) Based on the results of this hypothesis test, would you expect a confidence interval for the average difference between the reading and writing scores to include 0? Explain your reasoning. Yes because we did not reject the null hypothesis so the interval would be around zero.