Suppose x has a distribution with μ = 17 and σ = 13.

(a) If a random sample of size n = 42 is drawn, find μ_x, σ_x and P(17 ≤ x ≤ 19). (Round σ_x to two decimal places and the probability to four decimal places.)

(b) If a random sample of size n = 67 is drawn, find μ_x, σ_x and P(17 ≤ x ≤ 19). (Round σ_x to two decimal places and the probability to four decimal places.)

1) These numbers are measurements of the left and right claw length (cm) from 8 crabs. On average, is left claw length different from right claw length?

Recall that Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Now suppose you are the auditor for a very large corporation. The revenue file contains millions of numbers in a large computer data bank. You draw a random sample of n = 229 numbers from this file and r = 88 have a first nonzero digit of 1. Let p represent the population proportion of all numbers in the computer file that have a leading digit of 1.

(i) Test the claim that p is more than 0.301. Use α = 0.05.

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: p = 0.301; H₁: p < 0.301

H₀: p > 0.301; H₁: p = 0.301  

   H₀: p = 0.301; H₁: p ≠ 0.301

H₀: p = 0.301; H₁: p > 0.301

(b) What sampling distribution will you use?

The standard normal, since np < 5 and nq < 5.

The Student's t, since np < 5 and nq < 5.

     The standard normal, since np > 5 and nq > 5.

The Student's t, since np > 5 and nq > 5.

What is the value of the sample test statistic? (Round your answer to two decimal places.)


(c) Find the P-value of the test statistic. (Round your answer to four decimal places.)

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level α?

At the α = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.

At the α = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.

    At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant.

At the α = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.

(e) Interpret your conclusion in the context of the application.

There is sufficient evidence at the 0.05 level to conclude that the true proportion of numbers with a leading 1 in the revenue file is greater than 0.301.

There is insufficient evidence at the 0.05 level to conclude that the true proportion of numbers with a leading 1 in the revenue file is greater than 0.301.    

(ii) If p is in fact larger than 0.301, it would seem there are too many numbers in the file with leading 1's. Could this indicate that the books have been "cooked" by artificially lowering numbers in the file? Comment from the point of view of the Internal Revenue Service. Comment from the perspective of the Federal Bureau of Investigation as it looks for "profit skimming" by unscrupulous employees.

No. There does not seem to be too many entries with a leading digit 1.

Yes. There seems to be too many entries with a leading digit 1.

    No. There seems to be too many entries with a leading digit 1.

Yes. There does not seem to be too many entries with a leading digit 1.

(iii) Comment on the following statement: If we reject the null hypothesis at level of significance α , we have not proved H₀ to be false. We can say that the probability is α that we made a mistake in rejecting H_o. Based on the outcome of the test, would you recommend further investigation before accusing the company of fraud?

We have not proved H₀ to be false. Because our data lead us to accept the null hypothesis, more investigation is not merited.

We have not proved H₀ to be false. Because our data lead us to reject the null hypothesis, more investigation is not merited.

    We have proved H₀ to be false. Because our data lead us to reject the null hypothesis, more investigation is not merited.

We have not proved H₀ to be false. Because our data lead us to reject the null hypothesis, more investigation is merited.

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 67 inches and standard deviation 6 inches.

(a) What is the probability that an 18-year-old man selected at random is between 66 and 68 inches tall? (Round your answer to four decimal places.)


(b) If a random sample of thirteen 18-year-old men is selected, what is the probability that the mean height x is between 66 and 68 inches? (Round your answer to four decimal places.)


(c) Compare your answers to parts (a) and (b). Is the probability in part (b) much higher? Why would you expect this?

The probability in part (b) is much higher because the mean is smaller for the x distribution.The probability in part (b) is much lower because the standard deviation is smaller for the x distribution.    The probability in part (b) is much higher because the mean is larger for the x distribution.The probability in part (b) is much higher because the standard deviation is smaller for the x distribution.The probability in part (b) is much higher because the standard deviation is larger for the x distribution.

1a. Given: mean difference between men and women = 2.2; standard error of mean difference = 0.8; n men = 61, n women = 61.

Test the following null hypothesis at a=.05

Ho: u men = u women

Ha: u men > u women

You need to compute the t statistic, look up t critical value and make decision including likelihood of appropriate error. You don't have to compute the standard error of mean difference because it's given.

1b. Compute the appropriate confidence interval for the two-sample t-test of the previous question. Make sure you look at the alternative hypothesis before computing this interval.

Please upload into a Microsoft Word file and show work for answers

For 1b. please show work by doing step by step. line margins

The air temperature is measured every 6 hours for one week. The temperatures are given in the data table. Use the data to complete parts a through e below. _# Temperature 1 45.4 2 56.7 3 49.2 4 45.2 5 43.2 6 53.6 7 48.6 8 43.7 9 54.2 10 53.8 11 50.2 12 44.9 13 53.4 14 57.5 15 56.8 16 49.7 17 41.2 18 58.7 19 55.2 20 42.2 21 49.3 22 57.1 23 58.8 24 41.5 25 40.4 26 51.4 27 55.8 28 45.4

a. Draw a systematic sample consisting of 2​ temperatures, and then calculate the sampling error for the sample. The sampling error for the sample is

B. Draw a systematic sample consisting of 4 temperatures, and then calculate the sampling error for the sample. The sampling error for the sample is ___ F. (Round to two decimal places as needed.)

c. Draw a systematic sample consisting of 7 temperatures, and then calculate the sampling error for the sample. The sampling error for the sample is ___ F. (Round to two decimal places as needed.)

d. Compare the sampling error for parts a, b, and c and explain the reason for the differences. Choose the correct answer below. In general, increasing the sample size has no effect on the sampling error. In general, increasing the sample size makes the magnitude of the sampling error larger. The sample obtained in part b does not follow this trend, however. In general, increasing the sample size makes the magnitude of the sampling error smaller. The sample obtained in part c does not follow this trend, however. In general, increasing the sample size makes the sampling error increasingly negative.

e. What problems might be encountered with the sample obtained in part c? The presence of periodicity in the data might impact the sample obtained in part c. The lack of periodicity in the data might impact the sample obtained in part c. The larger sample obtained in part c might lead to a larger diversion from the da

A tire manufacturer claims that the life span of its tires is 52,000 miles. Assume the life spans of the tire are normally distributed. You selected 16 tires at random and tested them. The mean life span of the sample is 50,802 miles. The tires had a population standard deviation, σ = 800. Use the .05 level of significance. a) Which distribution would be indicated? b) Explain why you chose that distribution. c) Construct a confidence interval for the mean using the above data. Use the .05 level of significance. d) Re-compute the confidence interval if “n” is increased to 125 with the same mean and standard deviation. e) Re-compute the confidence interval if level of significance .01 with the same mean and standard deviation with the original n.

Five years ago the average university student owed $19,000 in student-loan debt at the time of graduation. With all the cuts in funding, it is suspected that this amount has gone up. A survey of 45 recent university graduates revealed an average student-loan debt of $20,000. Assume that the population standard deviation is $2,500.

a) Define the parameter of interest (in words), and then formulate the null hypothesis and the alternative hypotheses.

b) Find the p-value and make a conclusion in the context of the question. Use a level of significance of 5% (i.e. α = 0.05).

Jobs are sent to a printer at an average rate of 6 jobs per hour.

(a) What is the expected time between jobs? [Note: Give the exact answer either in hours or in minutes.]

(b) What is the probability that the next job is sent within 4 minutes? [Note: Round the answer to four decimal places.]

Obesity in adult males is associated with lower levels of sex hormone. A study investigated a possible link between obesity and testosterone levels in adolescent males between the ages of 14 and 20 years. The study compared 25 obese adolescent males and 25 adolescent males with a healthy weight. Their plasma testosterone concentrations (in nanomoles per liter, nmol/l) are provided:

To access the complete data set, click the link for your preferred software format:

Excel  Minitab  JMP  SPSS  TI  R  Mac-TXT  PC-TXT  CSV  CrunchIt!

We suspect that obese adolescent males have a lower plasma testosterone level, on average. Do the data support this suspicion?

(a) Select the null and alternative hypotheses.

?0:??=?? versus ??:??<?? .

?0:??=?? versus ??:??>?? .

?0:??=?? versus ??:??≠?? .

?0:??=?? versus ??:??≥?? .

Obtain these values. (Enter your answers rounded to three decimal places.)

?⎯⎯⎯?=

?⎯⎯⎯?=

??=

??=

?=

The test ?‑value is

?>0.05

0.05>?>0.01

?<0.001

0.01>?>0.001

Choose the correct conclusion.

The study found strong, significant evidence that obese adolescent males have higher testosterone levels, on average, than adolescent males with a healthy weight.

The study found strong, significant evidence that obese adolescent males have lower testosterone levels, on average, than adolescent males with a healthy weight.

The study failed to find evidence that obese adolescent males have different testosterone levels, on average, than adolescent males with a healthy weight.

The study found strong, significant evidence that obese adolescent males have the same testosterone levels, on average, than adolescent males with a healthy weight.

(b) Give a 95% confidence interval for the difference in mean plasma testoterone level between adolescent males who are obese (?1) and those who have a healthy weight (?2). (Enter your answers rounded to 3 decimal places.)

lower bound:

upper bound:

A graduate student in the School of Education is interested in whether families of students in the Chicago Public Schools are for or against the new legislation proposing school uniform requirements. She surveys 600 students and finds that 480 are against the new legislation. Compute a 90 and 98 percent confidence interval for the true proportion who are for the new legislation.

Average pizza delivery times to the West Loop are normally distributed with an unknown population mean and a population standard deviation of six minutes. Students in 270 at UIC collect a random sample of 28 pizza deliveries to the West is taken and they find sample mean delivery time of 36 minutes. Find a 60 and 95 percent confidence interval for µ.

The McCollough Corporation, a producer of various kinds of batteries, has been producing "D" size batteries with a life expectancy of 87 hours. Due to an improved production process, the management believes that there has been an increase in the life expectancy of their "D" size batteries. A sample of 36 batteries showed an average life of 88.5 hours. Assume from past information that it is known that the standard deviation of the population is 9 hours.

a. Use a 0.01 level of significance to test for the evidence of improvement in the life expectancy of the batteries. Use the critical value criterion in your decision making.

[Hint: You need to (1) write the null and alternative hypotheses (2) know the significance level of the test (3) write the appropriate formula for the teststatistic (4) using the sample data, compute the numerical value of the teststatistic (5) look up the critical value, and finally (6) draw your conclusion].

b. What is the p-value of the test? Draw your conclusion using the p-value criterion.

Sheets of aluminum from a supplier have a thickness that is normally distributed with a mean of 50 mm and a standard deviation of 4 mm (call this random variable X). Your company compresses the aluminum with a tool that is normally distributed with a mean of 20 mm and a standard deviation of 3 mm (call this random variable Y). You are interested in the random variable V = X – Y, the random variable V is the final aluminum thickness. 3.1. What is the probability that the outputted aluminum (that is, V), will be between 25 mm and 32 mm? 3.2. What is the probability that the outputted aluminum (that is, V), will be between 26.5 mm and 33.5 mm? 3.3. If the company had the choice between compressing aluminum to between 25-32 mm or 26.5-33.5 mm, then which is preferred (or neither)? 3.4. In one or two sentences, why is one preferred (if either) over the other [continuation of 3.3]; if neither are preferred, then why?