Real Fruit Juice (Raw Data, Software Required):
A 32 ounce can of a popular fruit drink claims to contain 20% real fruit juice. Since this is a 32 ounce can, they are actually claiming that the can contains 6.4 ounces of real fruit juice. The consumer protection agency samples 30 such cans of this fruit drink. The amount of real fruit juice in each can is given in the table below. Test the claim that the mean amount of real fruit juice in all 32 ounce cans is 6.4 ounces. Test this claim at the 0.05 significance level.

According to a​ study, the proportion of people who are satisfied with the way things are going in their lives is 0.76. Suppose that a random sample of 100 people is obtained. Complete parts​ (a) through​ (e) below.

​(a) Suppose the random sample of100 people is​ asked, "Are you satisfied with the way things are going in your​ life?" Is the response to this question qualitative or​quantitative? Explain.

A.The response is quantitative because the responses can be measured numerically and the values added or​ subtracted, providing meaningful results.

B.The response is quantitative because the responses can be classified based on the characteristic of being satisfied or not.

C.The response is qualitative because the responses can be measured numerically and the values added or​ subtracted, providing meaningful results.

D.The response is qualitative because the responses can be classified based on the characteristic of being satisfied or not.

​(b) Explain why the sample​ proportion, ModifyingAbove p with caretp​, is a random variable. What is the source of the​ variability?

A. The sample proportion ModifyingAbove p with caretp is a random variable because the value of ModifyingAbove p with caretp represents a random person included in the sample. The variability is due to the fact that different people feel differently regarding their satisfaction.

B. The sample proportion ModifyingAbove p with caretp is a random variable because the value of ModifyingAbove p with caretp represents a random person included in the sample. The variability is due to the fact that people may not be responding to the question truthfully.

C. The sample proportion ModifyingAbove p with caretp is a random variable because the value of ModifyingAbove p with caretp varies from sample to sample. The variability is due to the fact that people may not be responding to the question truthfully.

D. The sample proportion ModifyingAbove p with caretp is a random variable because the value of ModifyingAbove p with caretp varies from sample to sample. The variability is due to the fact that different people feel differently regarding their satisfaction.

​(c) Describe the sampling distribution of ModifyingAbove p with caretp​, the proportion of people who are satisfied with the way things are going in their life. Be sure to verify the model requirements.

Since the sample size is no

more

less

than​ 5% of the population size and np(1minus−​p)equals=nothinggreater than or equals≥​10, the distribution of ModifyingAbove p with caretp is

▼

skewed left

approximately normal

skewed right

uniform

with mu Subscript ModifyingAbove p with caret Baseline equalsμp=(BLANK)

and

sigma Subscript ModifyingAbove p with caret Baseline equalsσp=(BLANK).

​(Round to three decimal places as​ needed.)

​(d) In the sample obtained in part​ (a), what is the probability that the proportion who are satisfied with the way things are going in their life exceeds

0.7878​?

The probability that proportion who are satisfied with the way things are going in their life exceeds 0.7878 is BLANK

​(Round to four decimal places as​ needed.)

​(e) Using the distribution from part​ (c), would it be unusual for a survey of 100 people to reveal that 70 or fewer people in the sample are satisfied with their​ lives?The probability that

70 or fewer people in the sample are satisfied is BLANK, which

▼

is not

is

unusual because this probability

▼

is

is not

less than

▼

5

0.05

0.5

50

​%.

​(Round to four decimal places as​ needed.)

Previous studies have shown that playing video games can increase visual perception abilities on tasks presented in the gaming zone of the screen (within 5 degrees of the center). A graduate student is interested in whether playing video games increases peripheral visual perception abilities or decreases attention to peripheral regions because of focus on the gaming zone. For her study, she selects a random sample of 64 adults. The subjects complete a difficult spatial perception task to determine baseline levels of their abilities. After playing an action video game (a first-person combat simulation) for 1 hour a day over 10 days, they complete the difficult perception task for a second time.

Before playing the action video game, the mean score in their accuracy on the spatial task was 0.42. After playing the action video game, the mean score was -0.08. The mean of the differences between each person’s pre- and post- scores was 0.5, with a standard deviation of the differences equal to 2.4.

The graduate student has no presupposed assumptions about whether playing video games increases peripheral visual perception abilities or decreases attention to peripheral regions because of focus on the gaming zone, so she formulates the null and alternative hypotheses as:

She uses a repeated-measures t test. Because the sample size is large, if the null hypothesis is true as an equality, the test statistic follows a t-distribution with n – 1 = 64 – 1 = 63 degrees of freedom.

This is a one or two tailed test?

The critical score(s) (the value(s) for t that separate(s) the tail(s) from the main body of the distribution, forming the critical region) is/are ( ) ?   

To calculate the test statistic, you first need to calculate the estimated standard error under the assumption that the null hypothesis is true. The estimated standard error is (    ) ?

The test statistic is t = ( )?

The t statistic lies or does not lie in the critical region for a two-tailed hypothesis test?

Therefore, the null hypothesis is rejected or not rejected?

The graduate student can or can not conclude that playing video games alters peripheral visual perception?

The graduate student repeats her study with another random sample of the same size. This time, instead of the treatment being playing the combat simulation video game, the treatment is playing a soccer video game. Suppose the results are very similar. After playing a soccer video game, the mean score was still 0.5 lower, but this time the standard deviation of the difference was 1.8 (vs. the original standard deviation of 2.4). This means..... playing a soccer video game OR playing the combat simulation game has the more consistent treatment effect?

This difference in the standard deviation also means that a 95% confidence interval of the mean difference would be narrower or wider for the original study, when the treatment was playing the combat simulation video game, than the 95% confidence interval of the mean difference for the second study, when the treatment was playing a soccer video game.

Finally, this difference in the standard deviation means that when the graduate student conducts a hypothesis test testing whether the mean difference is zero for the second study, she will be more or less likely to reject the null hypothesis than she was for the hypothesis test you completed previously for the original study?

As a​ follow-up to a report on gas​ consumption, a consumer group conducted a study of SUV owners to estimate the mean mileage for their vehicles. A simple random sample of 80 SUV owners was​ selected, and the owners were asked to report their highway mileage. The results that were summarized from the sample data were x overbarequals19.6 mpg and sequals6.6 mpg. Based on these sample​ data, compute and interpret a 95​% confidence interval estimate for the mean highway mileage for SUVs. The 95​% confidence interval is mpg---mpg

A random sample of leading companies in South Korea gave the following percentage yields based on assets.

Use a calculator to verify that s² ≈ 2.034 for these South Korean companies.

Another random sample of leading companies in Sweden gave the following percentage yields based on assets.

Use a calculator to verify that s² ≈ 0.407 for these Swedish companies.

Test the claim that the population variance of percentage yields on assets for South Korean companies is higher than that for companies in Sweden. Use a 5% level of significance. How could your test conclusion relate to an economist's question regarding volatility of corporate productivity of large companies in South Korea compared with those in Sweden?

(a) What is the level of significance?

State the null and alternate hypotheses.

H_o: σ₁² = σ₂²; H₁: σ₁² > σ₂²H_o: σ₁² > σ₂²; H₁: σ₁² = σ₂²    H_o: σ₂² = σ₁²; H₁: σ₂² > σ₁²H_o: σ₁² = σ₂²; H₁: σ₁² ≠ σ₂²

(b) Find the value of the sample F statistic. (Use 2 decimal places.)


What are the degrees of freedom?

What assumptions are you making about the original distribution?

The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent chi-square distributions. We have random samples from each population.    The populations follow independent normal distributions.The populations follow independent normal distributions. We have random samples from each population.

(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 < 0.0250.001 < p-value < 0.010p-value < 0.001

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

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

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

Fail to reject the null hypothesis, there is sufficient evidence that the variance in percentage yields on assets is greater in the South Korean companies.Reject the null hypothesis, there is insufficient evidence that the variance in percentage yields on assets is greater in the South Korean companies.    Reject the null hypothesis, there is sufficient evidence that the variance in percentage yields on assets is greater in the South Korean companies.Fail to reject the null hypothesis, there is insufficient evidence that the variance in percentage yields on assets is greater in the South Korean companies.

Exam Grades

Exam grades across all sections of introductory statistics at a large university are approximately normally distributed with a mean of 72 and a standard deviation of 11. Use the normal distribution to answer the following questions.

(a) What percent of students scored above a 91 ?

(b) What percent of students scored below a 63 ?

(c) If the lowest 8% of students will be required to attend peer tutoring sessions, what grade is the cutoff for being required to attend these sessions?
(d) If the highest 9% of students will be given a grade of A, what is the cutoff to get an A?

Assume that a simple random sample has been selected from a normally distributed population and test the given claim. Identify the null and alternative​ hypotheses, test​ statistic, P-value, critical​ value(s), and state the final conclusion that addresses the original claim. A safety administration conducted crash tests of child booster seats for cars. Listed below are results from those​ tests, with the measurements given in hic​ (standard head injury condition​ units). The safety requirement is that the hic measurement should be less than 1000 hic. Use a 0.01 significance level to test the claim that the sample is from a population with a mean less than 1000 hic. Do the results suggest that all of the child booster seats meet the specified​ requirement? 690     739     1248     635     589     515

What are the​ hypotheses?

Identify the test statistic.

Identify the critical​ value(s).

State the final conclusion that addresses the original claim.

What do the results suggest about the child booster seats meeting the specified​ requirement?

1. Two cards are selected at random from a standard deck of 52 cards without replacement. What is the probability that both cards are diamonds? Submit your answer as a decimal rounded to the nearest thousandth.

1. A jar contains 14 blue candies, 10 green candies, and 5 yellow candies. Three candies are selected at random without replacement. What is the probability that the first is yellow, the second is blue, and the third is yellow? Submit your answer as a decimal rounded to the nearest thousandth.

1. In a group of 400 employees (142 men and 258 women), 29 of the men and 39 of the women work in accounting.  If an accountant is selected at random, what is the probability the accountant is a woman? Round your answer to the nearest thousandth.

1. In a group of 400 employees (142 men and 258 women), 29 of the men and 39 of the women work in accounting. If a male employee is selected at random, what is the probability that he is an accountant? Round your answer to the nearest thousandth.

The Toylot company makes an electric train with a motor that it claims will draw an average of only 0.8 ampere (A) under a normal load. A sample of nine motors was tested, and it was found that the mean current was x = 1.30 A, with a sample standard deviation of s = 0.42 A. Do the data indicate that the Toylot claim of 0.8 A is too low? (Use a 1% level of significance.)

(a) What is the level of significance?


State the null and alternate hypotheses.

H₀: μ = 0.8; H₁:  μ ≠ 0.8H₀: p = 0.8; H₁:  p ≠ 0.8     H₀: p = 0.8; H₁:  p > 0.8H₀: μ = 0.8; H₁:  μ > 0.8H₀: μ ≠ 0.8; H₁:  μ = 0.8H₀: p ≠ 0.8; H₁:  p = 0.8

(b) What sampling distribution will you use? What assumptions are you making?

The standard normal, since we assume that x has a normal distribution with known σ.The Student's t, since we assume that x has a normal distribution with known σ.     The standard normal, since we assume that x has a normal distribution with unknown σ.The Student's t, since we assume that x has a normal distribution with unknown σ.

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


(c) Find (or estimate) the P-value.

P-value > 0.2500.125 < P-value < 0.250     0.050 < P-value < 0.1250.025 < P-value < 0.0500.005 < P-value < 0.025P-value < 0.005

1. The provost at the University of Chicago claimed that the entering class this year is larger than the entering class from previous years but their mean SAT score is lower than previous years. He took a sample of 20 of this year’s entering students and found that their mean SAT score is 1,501 with a standard deviation of 53. The University’s record indicates that the mean SAT score for entering students from previous years is 1,520. He wants to find out if his claim is supported by the evidence at a 5% level of significance. Round final answers to two decimal places. Solutions only.

(C) State the null hypothesis for this study.

(D) State the alternative hypothesis for this study.

(E) What critical value should the president use to determine the rejection region?

(H) The lowest level of significance at which the null hypothesis can still be rejected is ___.

1. A researcher is interested in testing two different dosages of a new sleeping medication in a phase II clinical trial against a control group receiving a placebo. Suppose the following table observes the hours of sleep reported the preceding evening for subjects randomly assigned to each of the three groups. α = .05

*Continue as though all assumptions for ANOVA are met.

A) Calculate the MSB and MSW for a one-way analysis of variance procedure.

B) Using the information above, calculate an F statistic, and provide an interpretation of your p-value

On the distant planet Cowabunga, the weights of cows have a normal distribution with a mean of 400 pounds and a standard deviation of 43 pounds. The cow transport truck holds 12 cows and can hold a maximum weight of 5076.

If 12 cows are randomly selected from the very large herd to go on the truck, what is the probability their total weight will be over the maximum allowed of 5076? (This is the same as asking what is the probability that their mean weight is over 423.)

The National Assessment of Educational Progress (NAEP) gave a test of basic arithmetic and the ability to apply it in everyday life to a sample of 840 men 21 to 25 years of age. Scores range from 0 to 500; for example, someone with a score of 325 can determine the price of a meal from a menu. The mean score for these 840 young men was x¯¯¯x¯ = 272. We want to estimate the mean score μμ in the population of all young men. Consider the NAEP sample as an SRS from a Normal population with standard deviation σσ = 60.

(a) If we take many samples, the sample mean x¯¯¯x¯ varies from sample to sample according to a Normal distribution with mean equal to the unknown mean score μμ in the population. What is the standard deviation of this sampling distribution?
(b) According to the 95 part of the 68-95-99.7 rule, 95% of all values of x¯¯¯x¯ fall within _______ on either side of the unknown mean μμ. What is the missing number?
(c) What is the 95% confidence interval for the population mean score μμ based on this one sample? Note: Use the 68-95-99.7 rule to find the interval.

2.

The National Institute of Standards and Technology (NIST) supplies "standard materials" whose physical properties are supposed to be known. For example, you can buy from NIST a liquid whose electrical conductivity is supposed to be 5. (The units for conductivity are microsiemens per centimeter. Distilled water has conductivity 0.5.) Of course, no measurement is exactly correct. NIST knows the variability of its measurements very well, so it is quite realistic to assume that the population of all measurements of the same liquid has the Normal distribution with mean μμ equal to the true conductivity and standard deviation σσ = 0.2. Here are 6 measurements on the same standard liquid, which is supposed to have conductivity 5:

5.32   4.88   5.10   4.73   5.15   4.75

NIST wants to give the buyer of this liquid a 96% confidence interval for its true conductivity. What is this interval?

3.

Here are the IQ test scores of 31 seventh-grade girls in a Midwest school district:

These 31 girls are an SRS of all seventh-grade girls in the school district. Suppose that the standard deviation of IQ scores in this population is known to be σσ = 15. We expect the distribution of IQ scores to be close to Normal. Estimate the mean IQ score for all seventh-grade girls in the school district, using a 98% confidence interval.

3.

How heavy a load (pounds) is needed to pull apart pieces of Douglas fir 4 inches long and 1.5 inches square? Here are data from students doing a laboratory exercise:

Suppose that the strength of pieces of wood like these follows a Normal distribution with standard deviation 3000 pounds. Give a 99% confidence interval for the mean load required to pull the wood apart.

to  lb

A researcher wishes to estimate the proportion of adults who have​ high-speed Internet access. What size sample should be obtained if she wishes the estimate to be within 0.04 with 90​% confidence if ​

(a) she uses a previous estimate of 0.38​? ​n=

(b) she does not use any prior​ estimate? n=