The heights of 1000 students are normally distributed with a mean of 177.5 centimeters and a standard deviation of 6.7 centimeters. Assuming that the heights are recorded to the nearest​ half-centimeter, how many of these students would be expected to have heights

​(a) less than 167.0 centimeters?

​(b) between 173.5 and 185.0 centimeters​ inclusive?

​(c) equal to 180.0 ​centimeters?

​(d) greater than or equal to 191.0 ​centimeters?

Let x = age in years of a rural Quebec woman at the time of her first marriage. In the year 1941, the population variance of x was approximately σ² = 5.1. Suppose a recent study of age at first marriage for a random sample of 51 women in rural Quebec gave a sample variance s² = 2.6. Use a 5% level of significance to test the claim that the current variance is less than 5.1. Find a 90% confidence interval for the population variance.

(a) What is the level of significance?


State the null and alternate hypotheses.

H_o: σ² = 5.1; H₁: σ² < 5.1H_o: σ² = 5.1; H₁: σ² > 5.1    H_o: σ² < 5.1; H₁: σ² = 5.1H_o: σ² = 5.1; H₁: σ² ≠ 5.1

(b) Find the value of the chi-square statistic for the sample. (Round your answer to two decimal places.)


What are the degrees of freedom?


What assumptions are you making about the original distribution?

We assume a normal population distribution.We assume a exponential population distribution.    We assume a uniform population distribution.We assume a binomial population distribution.

(c) Find or estimate the P-value of the sample test statistic.

P-value > 0.1000.050 < P-value < 0.100    0.025 < P-value < 0.0500.010 < P-value < 0.0250.005 < P-value < 0.010P-value < 0.005

(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?

Since the P-value > α, we fail to reject the null hypothesis.Since the P-value > α, we reject the null hypothesis.    Since the P-value ≤ α, we reject the null hypothesis.Since the P-value ≤ α, we fail to reject the null hypothesis.

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

At the 5% level of significance, there is insufficient evidence to conclude that the variance of age at first marriage is less than 5.1.At the 5% level of significance, there is sufficient evidence to conclude that the that the variance of age at first marriage is less than 5.1.    

(f) Find the requested confidence interval for the population variance. (Round your answers to two decimal places.)

Interpret the results in the context of the application.

We are 90% confident that σ² lies outside this interval.

We are 90% confident that σ² lies above this interval.    

We are 90% confident that σ² lies below this interval.

We are 90% confident that σ² lies within this interval.

Disadvantage groups, notably Blacks and Hispanics. Have had smaller high school graduation rates and so less access to college than Whites. Among those with college degrees is an educational success beyond college similarly affected? To address this question use the data below. sample of 30-year-old Americans with college degrees.

Highest Degree. Whites Black Hispanic Row Totals

College 5030 549 412 5991

Advanced 1324    117 99 1540

Columm totals 6354 668 511 7634 (grand)

a) state appropriate hypotheses

b) Find the degree of freedom

Compute the expected values for the entries for Hispanics. Compute the corresponding contributions to x^2. To save time here are the contributions of x² from the cells of white and blacks: 0.12, 0.7, 0.47, and 2.70.

d) conduct the appropriate test and give (an estimate of) p-value.

e) give an appropriate conclusion in statistical and everyday language.

Annual per capita consumption of milk is 21.6 gallons (Statistical Abstract of the United States: 2006). Being from the Midwest, you believe milk consumption is higher there and wish to support your opinion. A sample of 16 individuals from the midwestern town of Webster City showed a sample mean annual consumption of 24.1 gallons with a standard deviation of 4.8.

Compute the value of the test statistic. (Round to two decimal places)

What is the p-value? (Round to three decimal places).

At α=0.05, what is your conclusion?

Using all the data below, construct an empirical model using a computational tool (matlab, or R, any preferred). explain your model.

Data Description: These data are from a NIST study involving calibration of ozone monitors. The response variable (y) is the customer's measurement of ozone concentration and the predictor variable (x) is NIST's measurement of ozone concentration. MATLAB Row Vectors: xLst = [0.2, 337.4, 118.2, 884.6, 10.1, 226.5, 666.3, 996.3, 448.6, 777.0, 558.2, 0.4, 0.6, 775.5, 666.9, 338.0, 447.5, 11.6, 556.0, 228.1, 995.8, 887.6, 120.2, 0.3, 0.3, 556.8, 339.1, 887.2, 999.0, 779.0, 11.1, 118.3, 229.2, 669.1, 448.9, 0.5];

yLst = [0.1, 338.8, 118.1, 888.0, 9.2, 228.1, 668.5, 998.5, 449.1, 778.9, 559.2, 0.3, 0.1, 778.1, 668.8, 339.3, 448.9, 10.8, 557.7, 228.3, 998.0, 888.8, 119.6, 0.3, 0.6, 557.6, 339.3, 888.0, 998.5, 778.9, 10.2, 117.6, 228.9, 668.4, 449.2, 0.2];

A small computer shop is selling laser and Deskjet printers of the customers purchasing printers, 60% purchase a Laser printer. Let the number among the next 10 purchasers who select to buy a Laser printer. If the store currently has in stock 8 laser printers and 8 Deskjet printers, what is the probability that the requests of these 10 customers can all be met from existing stock (to 4 decimals)?

A shareholders’ group, in lodging a protest, claimed that the mean tenure for a chief executive office (CEO) was at least nine years. A survey of 85 companies reported in The Wall Street Journal found a sample mean tenure of 7.27 years for CEOs with a standard deviation of 6.38 years (The Wall Street Journal, January 2, 2007).

Compute the value of the test statistic. (Round to two decimal places)

What is the p-value? (Round to three decimal places).

At α=0.01, what is your conclusion?

Large Sample Proportion Problem. A survey was conducted on high school marijuana use. Of the 2266 high school students surveyed, 970 admitted to smoking marijuana at least once.  A study done 10 years earlier estimated that 45% of the students had tried marijuana. We want to conduct a hypothesis test to see if the true proportion of high school students who tried marijuana is now less than 45%.   Use alpha = .01.
What is the critical value for this test?

Group of answer choices

-1.96

-2.576

-2.33

2.33

1.) A boat capsized and sank in a lake. Based on an assumption of a mean weight of 148 ​lb, the boat was rated to carry 60 passengers​ (so the load limit was 8 comma 880 ​lb). After the boat​ sank, the assumed mean weight for similar boats was changed from 148 lb to 170 lb. Complete parts a and b below.

a.) Assume that a similar boat is loaded with 60 ​passengers, and assume that the weights of people are normally distributed with a mean of 178.3 lb and a standard deviation of 40.9 lb. Find the probability that the boat is overloaded because the 60 passengers have a mean weight greater than 148 lb.

b. The boat was later rated to carry only 15 ​passengers, and the load limit was changed to 2 comma 550 lb. Find the probability that the boat is overloaded because the mean weight of the passengers is greater than 170 ​(so that their total weight is greater than the maximum capacity of 2 comma 550 ​lb). The probability is nothing. ​(Round to four decimal places as​ needed.) Do the new ratings appear to be safe when the boat is loaded with 15 ​passengers? Choose the correct answer below.

A. Because there is a high probability of​ overloading, the new ratings appear to be safe when the boat is loaded with 15 passengers.

B. Because there is a high probability of​ overloading, the new ratings do not appear to be safe when the boat is loaded with 15 passengers.

C. Because the probability of overloading is lower with the new ratings than with the old​ ratings, the new ratings appear to be safe.

D. Because 178.3 is greater than 170​, the new ratings do not appear to be safe when the boat is loaded with 15 passengers.

2.) An engineer is going to redesign an ejection seat for an airplane. The seat was designed for pilots weighing between 120 lb and 171 lb. The new population of pilots has normally distributed weights with a mean of 130 lb and a standard deviation of 29.4 lb.

a. If a pilot is randomly​ selected, find the probability that his weight is between 120 lb and 171 lb.

The probability is approximately ​(Round to four decimal places as​ needed.)

b. If 33 different pilots are randomly​ selected, find the probability that their mean weight is between 120 lb and 171 lb.

The probability is approximately. ​(Round to four decimal places as​ needed.)

c. When redesigning the ejection​ seat, which probability is more​ relevant?

A. Part​ (b) because the seat performance for a single pilot is more important.

B. Part​ (a) because the seat performance for a single pilot is more important.

C. Part​ (a) because the seat performance for a sample of pilots is more important.

D. Part​ (b) because the seat performance for a sample of pilots is more important.

The following data represent the pH of rain for a random sample of 12 rain dates. A normal probability plot suggests the data could come from a population that is normally distributed. A boxplot indicates there are no outliers. Complete parts​ (a) through​ (d) below. 5.05 5.72 4.38 4.80 5.02 4.68 4.74 5.19 4.61 4.76 4.56 5.54 ​(a) Determine a point estimate for the population mean. A point estimate for the population mean is 4.92. ​(Round to two decimal places as​ needed.) ​(b) Construct and interpret a 95​% confidence interval for the mean pH of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice. ​(Use ascending order. Round to two decimal places as​ needed.) A. There is 95​% confidence that the population mean pH of rain water is between nothing and nothing. B. There is a 95​% probability that the true mean pH of rain water is between nothing and nothing. C. If repeated samples are​ taken, 95​% of them will have a sample pH of rain water between nothing and nothing. ​(c) Construct and interpret a 99​% confidence interval for the mean pH of rainwater. Select the correct choice below and fill in the answer boxes to complete your choice. ​(Use ascending order. Round to two decimal places as​ needed.) A. There is a 99​% probability that the true mean pH of rain water is between nothing and nothing. B. There is 99​% confidence that the population mean pH of rain water is between nothing and nothing. C. If repeated samples are​ taken, 99​% of them will have a sample pH of rain water between nothing and nothing. ​(d) What happens to the interval as the level of confidence is​ changed? Explain why this is a logical result. As the level of confidence​ increases, the width of the interval ▼ decreases. increases. This makes sense since ▼ including fewer numbers for consideration makes it more likely one of them is correct. including more numbers for consideration makes it more likely one of them is correct. all confidence intervals of a given level of confidence have the same width.

The financial impact of IT systems downtime is a concern of plant operations management today. A survey of manufacturers examined the satisfaction level with the reliability and availability of their manufacturing IT applications. The variables of focus are whether the manufacturer experienced downtime in the past year that affected one or more manufacturing IT applications, the number of downtime incidents that occurred IT applications, the number of downtime incidents that occurred in the past year, and the approximate cost of a typical downtime incident. The results from a sample of 200 manufacturers are as follows: Sixty-two experienced downtime this year that affected one of more manufacturing applications. Number of downtime incidents: ̄x = 3.5, S=2.0 Cost of downtime incidents: ̄x = $18,000 , S= $3,000.

a) construct a 90 % confidence interval estimate for the population proportion of manufacturers who experienced downtime in the past year that affected one or more manufacturing IT applications.

b) Construct a 95% confidence interval estimate for the population mean number of downtime incidents experienced by manufacturers in the past year.

c) Construct a 95% confidence interval estimate for the population mean cost of downtime incidents.

Show your work. Carry out all calculations to at least 3 significant digits.

Marketing strategists like to study the differences (in, e.g., age and income) between buyers and non-buyers of a product. In an earlier study of the purchasers and non-purchasers of a product sold by the AAA Company, demographic data were collected. Their age profiles (in years) are summarized and reported as follows:

Purchasers
Sample size 900

Sample mean 43.8

Sample standard deviation 14.6

Non-Purchasers

Sample size 800

sample Mean 40.3

Sample standard deviation15.4

A. Based on the market demographic data obtained, do we have strong enough evidence to conclude that buyers of the company’s product are on average more than two years older than non-buyers are? Test using α = 0.05.

B. Construct a 95% confidence interval for the difference in average age between purchasers and non-purchasers of the product. Interpret the significance of the age difference.

3 The ten most recent cars sold at a dealership have an average price of $24,525 with a standard deviation of s = $450. Assuming that the prices of all cars sold by the dealership have an approximately normal distribution, find the 98% confidence interval for the mean price of all cars sold by the dealership, to the nearest dollar.

Conclusion: We can be ___________confident that the mean price of all cars sold by the dealership is between $ ________________ and $ ___________________3

For the data set (−1,−1),(2,0),(6,5),(9,6),(12,11), find interval estimates (at a 95% significance level) for single values and for the mean value of ? corresponding to ?=1.

Note: For each part below, your answer should use interval notation.

Interval Estimate for Single Value =

Interval Estimate for Mean Value =