You collected some ecological samples in a field experiment in west Texas. Collected were soil samples to determine the amount of nutrients that would need to be applied to sustain a desired level of crop growth. The observation was the amount of nutrient the soil contained. Samples were taken from 100 different locations. It was found that the nutrient data are normally distributed. The mean value of the samples was 16 units (the exact name for the units do not really matter in this problem, so just call them “units”). The sample standard deviation of the samples is 3.5 units.

a. estimate the probability range for the Confidence Interval of the Mean (95% confidence) for the parameters given above.

b. What is the interpretation of the Confidence Interval of the Mean?

c. Consider the experiment was performed again in a second year, however, this time only 15 sites were sampled. The mean value in year two was found to be 14.5 units and the sample standard deviation was found to be 2.2 units. Using equation 3.16, estimate the probability ranges for the 90%, 95%, and 99% confidence intervals of the mean.. You will need to estimate THREE probability ranges to answer this question. One range for 90%; one range for 95%, and one range for 99%.

d. Describe how the probability range changes as the confidence interval increases. e. What is the probability range for the 50% confidence?

f. If you were to repeat the experiment described initially (100 samples per year) for twenty years, what percentage of the years would you expect the true mean value to be bracketed by the 50% confidence interval? Additionally, what percentage of the years would you expect the true mean value to be bracketed by the 95% confidence interval

Edit: Please show work on how to calculate Confidence intervals.

In the National AIDS Behavioral Surveys sample of 2723 adult heterosexuals, 8 respondents had both received a blood transfusion and had a sexual partner from a group at high risk of AIDS. You should not use the large-sample confidence interval for the proportion p in the population who share these two risk factors. The plus four method adds four observations, two successes and two failures.

What is the plus four estimate pˆp^ (±0.0001) of p? p~ = ……...

Give the plus four 93% confidence interval (±0.0001) for p:...… ⩽ p ⩽ ……….

the percentage of deaths due to disease is 12.5%. A medical student wants to test whether the percentage of deaths due to disease is less than 12.5%. She collected data on deaths due to disease over 8 states. She found out that there were 102 patients diagnosed with disease and 11 patients died. Carry out a test of significance by answering the questions below in as much detail as possible. Use a 10% level of significance. What is the value of the test statistic? (Hint: Round sample proportion to four decimal places.) Question 2 options: A) -0.53 B) 0.53 C) 0.01 D) -0.01 E) -0.05

An oil company purchased an option on land in Alaska. Preliminary geologic studies assigned the following prior probabilities.

If required, round your answers to two decimal places.

A restaurant wants to estimate the average bill paid by a party of two for dinner (service hours after 5 p.m.) They take a random sample of 100 bills and find that they average $80 with a standard deviation of $40. In making an 80% confidence interval for the parameter of interest, they should get a margin of error of about(Answer)?

5. Meier and Frank is a chain of department stores located in the Northwest. The company has issued its own credit cards for a number of years. As a new employee in the finance department, imagine that you have been assigned to a major investigation of the firm’s credit policies designed to reduce bad debt losses. You are requested to investigate the relationship between marital status at the time the card is issued and the subsequent payment record of the individual. You randomly select 100 credit cards issued three years previously. Of the 70 individuals that were married at the time the cards were issued, 11 have defaulted on a payment. Of the 30 individuals who were single when the cards were issued, 7 have defaulted. Do these data present sufficient evidence to indicate that marital status affects credit-worthiness? Use Alpha = 0.05. Solve problem using Minitab as the tool.

. Insurance companies are interested in knowing the mean weight of cars currently licensed in the United States; they believe it is 3000 pounds. To see if the estimate is correct, they check a random sample of 80 cars. For that group, the mean weight was 2910 pounds with a standard deviation of 532 pounds. Is this strong evidence that the mean weight of all cars is not 3000 pounds? a. What is the population of interest? b. Describe in words. d. Perform the test using a significance level of 0.05 ( = 0.05) ii. P-value Due to the different methods, I rounded out to 2 decimal places. Select the best possible answer. iii. Conclusion in context h. Find a 90% confidence interval for . Due to the different methods, select the best possible answer. k. What sample size would allow us to increase our confidence level to 95% while reducing the margin of error to only 50 pounds?

3. Let X, Y, and Z be independent unit exponential random variables, with common density f(t) = e^(-t) for t > 0.

Let T_1 = min (X, Y, Z )

T_2 = middle value of the three numbers X, Y, Z

T_3 = max (X, Y, Z )

(a) Find P( T_1 > t ) for t >0.

(b) Find P( T_3 < t ) for t > 0.

(c) Find P( T_2 >  t ) for t > 0.

HINT: T_2 > t happens when how many of X and Y and Z are greater than t ?

(d)  Find E ( T_3 - T_2 ) = expected difference between T_3 and T_2 .

HINT: One way to do part (d) is obviously to find the densities of T_2 and T_3 from the answers to parts (b) and (c) and then to use those densities to calculate E( T_2 ) and E(T_3 ). You could also integrate the survival functions (See page 332, under "Expectation from the survival function"). A much easier way is to just write down the answer, which you can do if you use the memoryless property of exponential distributions. Think about 3 light bulbs with independent unit exponential lifetimes. As long as such a bulb is working, its future behavior is exactly the same as the future behavior of a new bulb.

(e) Find E(T_3) and var(T_3).

HINT: There is almost no work involved in doing part (e) if you figured out the clever way to do part (d) and you use the equality

T_3 = T_1 + (T_2 - T_1) + (T_3 - T_2 ).

. Let X be the soil strength of a sample taken at 35mm depth in a large field. Assume X is normally distributed with an UNKNOWN standard deviation. We are interested in computing confidence intervals for μ, the unknown population mean soil strength at 35mm in this particular field. Four soil samples are randomly chosen from the field with the measurements (15, 21, 30, 34).

       a.     compute the 80% CI.

       b.     compute the 95% CI.

       c.     compare the two confidence intervals from a. and b.

(Which one is more precise? Which one is more accurate

Gun Murders - Texas vs California - Significance Test: California had stricter gun laws than Texas. However, California had a greater proportion of gun murders than Texas. Here we test whether or not the proportion was significantly greater in California. A significant difference is one that is unlikely to be a result of random variation.

The table summarizes the data for each state. The p̂'s are actually population proportions but you should treat them as sample proportions.

The standard error (SE) is given to save calculation time if you are not using software. Data Summary number of total number Proportion State gun murders (x) of murders (n) p̂

California 1220 1786 0.68309

Texas 699 1084 0.64483

SE = 0.01812

The Test: Test the claim that the proportion of gun murders was significantly greater in California than Texas in 2011. Use a 0.05 significance level.

(a) Letting p̂1 be the proportion of gun murders in California and p̂2 be the proportion from Texas, calculate the test statistic using software or the formula z = (p̂1 − p̂2) − δp SE where δp is the hypothesized difference in proportions from the null hypothesis and the standard error (SE) is given with the data. Round your answer to 2 decimal places. z = To account for hand calculations -vs- software, your answer must be within 0.01 of the true answer.

(b) Use software or the z-table to get the P-value of the test statistic. Round to 4 decimal places. P-value =

(c) What is the conclusion regarding the null hypothesis? reject H0 fail to reject H0

(d) Choose the appropriate concluding statement.

The data supports the claim that the proportion of gun murders was significantly greater in California than Texas.

While the proportion of gun murders in California was greater than Texas, the difference was not great enough to be considered significant.

We have proven that the stricter gun laws in California actually increased the proportion of gun murders

A pharmaceutical company has a new drug which relieves headaches. However, there is some indication that the drug may have the side effect of increasing blood pressure. Suppose the drug company conducts a hypothesis test to determine whether the medication raises blood pressure. The hypotheses are:

H₀: The drug does not increase blood pressure.
H_a: The drug increases blood pressure.

Answer the following questions completely:

1. Do you think that for doctors and patients it is more important to have a small α probability or a small β probability? Why?

2. Do you think that the pharmaceutical company would prefer to have a small α probability or a small β probability? Why?

Question 16 options:

1a. Suppose x has a distribution with μ = 21 and σ = 16.

(a) If a random sample of size n = 39 is drawn, find μ_x, σ_x and P(21 ≤ x ≤ 23). (Round σ_x to two decimal places and the probability to four decimal places.)

(b) If a random sample of size n = 55 is drawn, find μ_x, σ_x and P(21 ≤ x ≤ 23). (Round σ_x to two decimal places and the probability to four decimal places.)

A politician claims that medical insurance companies do not cover a majority of the cost and the average patient has to pay more than $10,000 in hospital bills. Using the critical value method, test this claim at the 5% level of significance. Why might having high hospital charges be an issue for patients?

Mean= 13266.97

Standard Deviation= 12110.01