Question
The sizes, in square feet, of the 20 rooms in a student residence hall at a certain university are summarized in the following histogram.
(a) Based on the histogram, write a few sentences describing the distribution of room size in the residence hall.
(b) Summary statistics for the sizes are given in the following table.
Determine whether there are potential outliers in the data. Then use the following grid to sketch a boxplot of room size.
(c) What characteristic of the shape of the distribution of room size is apparent from the histogram but not from the boxplot?
▶️Answer/Explanation
Ans:
The distribution room sizes in the residence hall seems to be centered around 200 to 300 sift since the median room size is between 200 and \(300 \mathrm{sg} f t\),
There do not seem to be any outliers in the distribution. The distribution seems to be bimodal and fairly symmetric. The IQR of the distribution is between 50 and 150 .
(b)
$
\begin{aligned}
& U F=292+\operatorname{IQR}(1.5)=242+177=469 \\
& L F=174-177=-3
\end{aligned}
$
There are potential outliers in the data since ell data values fall within the upper and lower fences.
The boxplat shows that the distribution of som sizes is skewed to the left. This is not shown in the histogram.
Question
Researchers are investigating the effectiveness of using a fungus to control the spread of an insect that destroys trees. The researchers will create four different concentrations of fungus mixtures: 0 milliliters per liter \((\mathrm{ml} / \mathrm{L})\), \(1.25 \mathrm{ml} / \mathrm{L}, 2.5 \mathrm{ml} / \mathrm{L}\), and \(3.75 \mathrm{ml} / \mathrm{L}\). An equal number of the insects will be placed into 20 individual containers. The group of insects in each container will be sprayed with one of the four mixtures, and the researchers will record the number of insects that are still alive in each container one week after spraying.
(a) Identify the treatments, experimental units, and response variable of the experiment.
Treatments:
Experimental units:
Response variable:
(b) Does the experiment have a control group? Explain your answer.
(c) Describe how the treatments can be randomly assigned to the experimental units so that each treatment has the same number of units.
▶️Answer/Explanation
Ans:
Treatments:
The treatments are the \(0 \mathrm{~m} / \mathrm{h}, 1.25 \mathrm{~mL} / \mathrm{L}, 2.5 \mathrm{ml} / \mathrm{L}\), and \(3.75 \mathrm{~m} / \mathrm{c}\) fungus mixtures
Experimental units: :
The experimental units ave, the groups of connects on each consumer
Response variable:
The response variable is the number of insects alive in each container one week after spraying.
(b) Yes, the experiment has a control group
The control group are the containers that are Sprayed with \(0 \mathrm{nu} / \mathrm{L}\) fungus mixture concentration.
(c)First, write out the number 1 : on 5 sheets of paper. Write the number 2 on another 5, the number 3 on another 5 , and the number 4 on 5 sheets. Place the total of 20 sweets of paper in a hat and for each container, take out a piece of paper. If you take out a sheet that has a “1” on it, it’ll relieve the \(0 \mathrm{ml} / \mathrm{L}\) concentration of fungus mixture. If it has \(a^{\prime} 2\), it will be the \(1.25 \mathrm{ml} / \mathrm{L}\), a 3 will be the \(2.5 \mathrm{ml} / \mathrm{c}\), and \(a\) will be the \(3.75 \mathrm{mV} / \mathrm{c} .900\) hot place the shoots af paper back into the half once its been taken out.
Question
A medical researcher surveyed a large group of men and women about whether they take medicine as prescribed. The responses were categorized as never, sometimes, or always. The relative frequency of each category is shown in the table.
(a) One person from those surveyed will be selected at random.
(i) What is the probability that the person selected will be someone whose response is never and who is a woman?
(ii) What is the probability that the person selected will be someone whose response is never or who is a woman?
(iii) What is the probability that the person selected will be someone whose response is never given that the person is a woman?
(b) For the people surveyed, are the events of being a person whose response is never and being a woman independent? Justify your answer.
(c) Assume that, in a large population, the probability that a person will always take medicine as prescribed is 0.54 . If 5 people are selected at random from the population, what is the probability that at least 4 of the people selected will always take medicine as prescribed? Support your answer.
▶️Answer/Explanation
Ans:
a(i) \(P(\) never \(n\) woman \()=0.0636\)
The probability that some ane who’s neper is never and is a werner ir 0.0636
a(ii)$
\begin{aligned}
& P \text { (never woman) }-P(\text { never })+P(\text { woman })-P(\text { never } n \text { woman })=0.12+0.53-0.013 \\
& =0.5864 \text { The probability that you Pl\&c someone who said never } \\
& \text { or is a woman is } 0 . \$ 864 \\
&
\end{aligned}
$
a(iii) $
P(\text { never lawman })=\frac{P(\text { nevernowan })}{P \text { (woman) }}=\frac{0.0636}{0.53}=0.12
$
The probability that you pick sombre who said newer given that they were a woman on is 0.12 .
(b) Yes they are independent since the probability of someone saying never given that they are a women and the probability of som one saying never saying never should be the same in order for the two events to be indipadent. In this cases both probabilities equal 0,12 , hence the events of saying never and being \(a\). women are in de pendent.
(c)\(P(\) at least 4 people take their medicine \()=1-P(\) up to 3 people take their medicine \(=1-\) Binempdf \((5,0.54,3)=\{-0.759=0.241\)
The probability of at least 4 people taking their medicine as prescribed is 0.241,
Question
Tumbleweed, commonly found in the western United States, is the dried structure of certain plants that are blown by the wind. Kochia, a type of plant that turns into tumbleweed at the end of the summer, is a problem for farmers because it takes nutrients away from soil that would otherwise go to more beneficial plants. Scientists are concerned that kochia plants are becoming resistant to the most commonly used herbicide, glyphosate. In 2014, 19.7 percent of 61 randomly selected kochia plants were resistant to glyphosate. In 2017, 38.5 percent of 52 randomly selected kochia plants were resistant to glyphosate. Do the data provide convincing statistical evidence, at the level of \(\alpha=0.05\), that there has been an increase in the proportion of all kochia plants that are resistant to glyphosate?
▶️Answer/Explanation
Ans:
2-Sample Proportion 2 -test
\(\pi_A\) : true proportion of kochi plants resistant to glyphosate in 2014
\(\pi_B\) : true proportion of kochi plants resistant to glyphosate in 2017
$
\begin{aligned}
& H_0: \pi_A=\pi_B \\
& H_A: \pi_A<\pi_B .
\end{aligned}
$
Conditions:
- Random sample plants were randomly selected
- Independence: chosen from a large population
- Normal sampling Distribution:
$
\begin{array}{ll}
n_A P_A:(61)(.197)=12.017 \geqslant 10 \mathrm{~V} & n_B P_B=(52)(.385)=20.02 \geqslant 10 \mathrm{~V} \\
n_A\left(1-P_A\right)=(61)(.803)=48.983 \geqslant 10 \mathrm{~V} & n_B\left(1-P_B\right)=(52)(.615)=31.98 \geqslant 10 \mathrm{~V}
\end{array}
$
Z-scare:-2.2095 p-value:. \(0136 \quad \alpha=.05\)
Because the \(p\)-value is smaller than \(\alpha\), we will reject \(H_0\). There is sufficient evidence to support \(H_A\).
Question
A company that manufactures smartphones developed a new battery that has a longer life span than that of a traditional battery. From the date of purchase of a smartphone, the distribution of the life span of the new battery is approximately normal with mean 30 months and standard deviation 8 months. For the price of \(\$ 50\), the company offers a two-year warranty on the new battery for customers who purchase a smartphone. The warranty guarantees that the smartphone will be replaced at no cost to the customer if the battery no longer works within 24 months from the date of purchase.
(a) In how many months from the date of purchase is it expected that 25 percent of the batteries will no longer work? Justify your answer.
(b) Suppose one customer who purchases the warranty is selected at random. What is the probability that the customer selected will require a replacement within 24 months from the date of purchase because the battery no longer works?
(c) The company has a gain of \(\$ 50\) for each customer who purchases a warranty but does not require a replacement. The company has a loss (negative gain) of \(\$ 150\) for each customer who purchases a warranty and does require a replacement. What is the expected value of the gain for the company for each warranty purchased?
▶️Answer/Explanation
(a) $
\begin{aligned}
z_0 .75 & =0.6745 \\
0.6745 & =\frac{x-30}{8}
\end{aligned}
$
\(x=35.3959\) months from date of purchase
(b)\(x=\) life span of newbatcery)
$
y \sim N(30,8)
$
$
P(x<24)=0.2266
$
(c) Events Gain probability
buy warrant but do not replacement 50 0.7734
buy warrant do not replacement -150 0.2266
\(50(0.7734)-150(0.2266)=\$ 4.6745\).
Question
Emma is moving to a large city and is investigating typical monthly rental prices of available one-bedroom apartments. She obtained a random sample of rental prices for 50 one-bedroom apartments taken from a Web site where people voluntarily list available apartments.
(a) Describe the population for which it is appropriate for Emma to generalize the results from her sample.
The distribution of the 50 rental prices of the available apartments is shown in the following histogram.
(b) Emma wants to estimate the typical rental price of a one-bedroom apartment in the city. Based on the distribution shown, what is a disadvantage of using the mean rather than the median as an estimate of the typical rental price?
(c) Instead of using the sample median as the point estimate for the population median, Emma wants to use an interval estimate. However, computing an interval estimate requires knowing the sampling distribution of the sample median for samples of size 50 . Emma has one point, her sample median, in that sampling distribution. Using information about rental prices that are available on the Web site, describe how someone could develop a theoretical sampling distribution of the sample median for samples of size 50 .
Because Emma does not have the resources to develop the theoretical sampling distribution, she estimates the sampling distribution of the sample median using a process called bootstrapping. In the bootstrapping process, a computer program performs the following steps.
- Take a random sample, with replacement, of size 50 from the original sample.
- Calculate and record the median of the sample.
- Repeat the process to obtain a total of 15,000 medians.
Emma ran the bootstrap process, and the following frequency table is the bootstrap distribution showing her results of generating 15,000 medians
The bootstrap distribution provides an approximation of the sampling distribution of the sample median. A confidence interval for the median can be constructed using a percentage of the values in the middle of the bootstrap distribution.
(d) Use the frequency table to find the following.
(i) Value of the 5th percentile:
(ii) Value of the 95 th percentile:
(e) Find the percentage of bootstrap medians in the table that are equal to or between the values found in part (d).
(f) Use your values from parts (d) and (e) to construct and interpret a confidence interval for the median rental price.
▶️Answer/Explanation
Ans:
(a) The population of one-bedroom apartments on the website she went an.
(b)Since the distribution of prices is skewed to the right, that means the mean is greater than the median of the distribution. So the disadvantage would be getting a higher estimate of the typical rental price so Emma would be misinformed on what the typical price should be (thinking it’s higher) and spending more than she heeds to:
(c) By using the website, some one could randomly rick one bedroom apartments from the net site \(\hat{50}\) and we those to develop a distribution of the sample median, through calculations, and you uni regear by obtaining multiple samples of size 50 and calculating the median for each.
\((.05)(15000)=750\) medians
=(2500)
\((.95)(15000)=14250\) medians
\(2950)\)
(e) b/w 2000 and 2950 the there are 14404 medians so
$
\frac{14404}{15000}=96.027 \%
$
(f) \((2500,2950)\)
96. 027\% of intervals constructed have a median that lies b/w 2500 and 2950 .