1.) Exercise 4.40 on p. 216 of OpenIntro Statistics. You can draw the distributions for part d by hand.
1.a.)

1 - pnorm(1.5)
## [1] 0.0668072

1.b.) When picking out 15 lightbulbs the distribution of the mean lifespan should be similer to the distribution of the population mean 9,000 with a lower standard deviation of 258.1989.

1.c.) The probability of the mean lifespan being greater than 10,500 hours in a sample of 15 lightbulbs is extremely close to zero.

1.d.) By hand.

1.e.) For part a.) No without a normal distribution the probability for an observation cannot be determined because there is no way to tell which side is skewed, or the amount of skewed the data is.

2.) Exercise 4.38 on p. 216 of OpenIntro Statistics
The plot A best represents the distribution 3.
The plot B best represents the distribution 1.
The plot C best represents the distribution 2.

3.) 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 0.02 grams. You decide to weigh the sample 3 times and find that the 3 measurements have a sample mean of 10.04 grams.

3.a.) 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.

First, set the value for alpha to be 0.05 to test our p-value against.

\(H_o\): \(\mu = 10\)

\(H_a\): \(\mu \neq 10\)
Z-score must be computed: \[Z = \frac{10.04-10.00}{0.02/\sqrt{3}} = 3.464\]
The test is two side, both sides of the distribution must be accounted for the p-value.
P-value:

2*pnorm(-3.464)
## [1] 0.0005322065

The calculated p-value was less than alpha, thus the null hypothesis was rejected, and there is statistically significant evidence that the rock samples weight was not 10.00 grams.

3.b.) Construct and interpret a 90% confidence interval for the true weight of the rock sample.

Z-score for 45% on one side of the distribution: 1.645.
\(SE = 0.02/\sqrt{3} = 0.11547\) Confidence interval \((a, b)\), where: \[a = \bar{x} - z*(SE) \\ b = \bar{x} + z*(SE) \] Thus the interval for 90% confidence is:(10.021, 10.059).

4.) Exercise 4.14 on p. 208 of OpenIntro Statistics

4.a.) False. The average of these American’s is already know to be $84.71.
4.b.) False. The confidence interval still holds even though the data is skewed. The Central Limit Theorem states that the limiting distribution of the sample mean is normal regardless of the distribution of the original data, as long as the sample size is large enough the CLT applys and 436 is large enough.
4.c.) True. the confidence interval tells us that 95% of samples will have a mean in the range of 80.31 dollars to 89.11 dollars.
4.d.) True. Since any sample of Americans taken has 95% confidence to be within 80.31 dollars to 89.11 dollars. 4.e.) True. Since we are not as confident the estimate might not fall within the new interval.
4.f.) False. The sample size would have to be increased by a size of 9 times larger.
4.g.) True. The margin of error is half the range of the confidence interval.

5.) Exercise 5.12 on p. 259 of OpenIntro Statistics

5.a.) \(H_a\): \(\mu > 35\) micrograms per liter; The downtown police officers are exposured to greater than 35 micrograms per liter of lead.

5.b.) Conditions: \(n = 52\) and \(df = 51\); \(\bar{x}_1 = 124.32\) and \(\mu = 35\); sample \(sd = 37.74\) and Standard Error = 5.234; alpha = 0.05; One sample t-test will be performed

5.c.) T-test:
\[t = \frac{(\bar{x} - \mu)}{SE} \\ = \frac{(124.32 - 35)}{5.234} \\ = 17.065 \]
P-value:

1 - pt(17.065, 51)
## [1] 0

The resulting p-value is equal to zero which is less than our alpha. Thus, our hypothesis of The downtown police officers being exposured to greater than 35 micrograms per liter of lead is accepted rather than claiming that it is the same as 35 micrograms per liter.

5.d.) No 35 would not fall in a 99% confidence interval because the p-value of 0 is less than the alpha of 0.01.

6.) Exercise 5.20 on p. 261 of OpenIntro Statistics
6.a.) The box plot shows that the median of the reading scores is similar to the mean of the writing score. The histogram suggests that the mean of the difference (reading - writing) is slightly below zero. Thus there is no clear difference between the too means.

6.b.) Each reading score has a writing to go along with it, thus they are paired not independent.

6.c.) \(H_a\): \(\mu_{diff} \neq 0\) indicating that there is a difference between a student’s reading score and writing score. A Paired T-test has to be performed for evidence.

6.d.) Conditions: Sample size \(n = 200\), which is less then 10% of total population and high enough to assume normality. \(df = 199\), this value is approximated, \(\bar{x}_1 = -0.545\) and \(\mu = 0\); sample \(sd = 8.887\); alpha = 0.05; Paired sample, two-tailed t-test will be performed.

6.e.) With a sample mean and a sample standard deviation a t-test can be performed on the paired data. The t-statistic is then computed to be: \[ t = \frac{-0.545}{\frac{8.887}{\sqrt{200}}} = \frac{-0.545}{0.6284} = -0.867\\ \]
With the test statistic calculated R can be used to compute the p-value for the sample. two-tailed means the p-value will need to be multiplied by two to cover for both sides of the distribution.

2*pt(-.867,199)
## [1] 0.3869862

With a p-value less than alpha, alpha = 0.05, our \(H_a\) is not accepted and we fail to reject the null:
\(H_0\): \(\mu_{diff} = 0\).

6.f.) Type II error could have been made which is when we failed to reject the null hypothesis when it is actually false in reality. Thus, there was actually a difference between the scores.

6.g.) Yes the confidence interval would include 0 since we fail to reject that the mean difference between the samples were different than 0. \(\mu_{diff} = 0\).