Calculate raw scores from the following z-scores. The mean of the original dataset was 10, and the standard deviation was 3.

1. What is the raw score for a z-score of -2.89?

2. What is the raw score for a z-score of +0.74?

3. What is the raw score for a z-score of +1.18?

4. What is the raw score for a z-score of -0.94?

5. What is the raw score for a z-score of -1.26?

12.52 Christmas Drink.(P576)

In a 2014 nationwide Harris Poll survey about beverage preferences, 195 1 American adults were asked about their preferred beverages during different holidays throughout the year. Overall, 47% of the adults voted for table wine as their preferred beverage for Christmas. Determine and interpret a 95% confidence interval for the proportion, p, of all American adults who prefer wine during holiday

Empirical studies have provided support for
the belief that an ordinary share’s annual
rate of return is approximately normally
distributed. Suppose that you have invested in
the shares of a company for which the annual
return has an expected value of 16% and a
standard deviation of 10%.
a Find the probability that your one-year
return will exceed 30%.
b Find the probability that your one-year
return will be negative.
c Suppose that this company embarks on
a new, high-risk, but potentially highly
profi table venture. As a result, the return
on the share now has an expected value
of 25% and a standard deviation of 20%.
Answer parts (a) and (b) in light of the
revised estimates regarding the share’s
return.
d As an investor, would you approve of the
company’s decision to embark on the new
venture?

Describe when ANOVA is used and why it is necessary.

The college Physical Education Department offered an Advanced First Aid course last summer. The scores on the comprehensive final exam were normally distributed, and the zscores for some of the students are shown below.

(d) If the mean score was μ = 146 with standard deviation σ = 23, what was the final exam score for each student? (Round your answers to the nearest whole number.)

The data below represents a sample of 22 people and how they voted on a referendum in a recent election. You are a social scientist who is interested in investigating age differences in voting patterns.

1. State the kind of statistical test you will perform (i.e., chi-square or an ANOVA).

HINT: Look at the types of variables available to you, which test is best suited for these types of variables with this many categories?

2. State the research and null hypotheses using words and symbols when applicable.

3. Report the degrees of freedom and the corresponding critical value for your test.

4. Compute the test statistic (i.e., chi-square or an ANOVA).

HINT: Make a computation table to keep track of your work.

5. Using your critical and obtained values, make a decision regarding your null hypothesis.

6. What conclusions can you draw about the voting behaviors of older and younger folks? Even if you don’t have enough information to compute the measure of association, which one would be appropriate for this test? What do you think it would indicate in terms of strength and direction of the observed relationship?

HINT: Make sure you respond to all of these questions using sentences that make sense!

Search on the following three phrases:

• Insertion Anomaly
• Update Anomaly
• Deletion Anomaly

Determine whether the BEST, most common interpretation of the given statement is:

     TRUE - Select 1

     FALSE - Select 2

Question 1 options

When the population standard deviation sigma is assumed known, a confidence interval can assume NORMALITY of the SAMPLE MEAN if the sample size is greater than 30.

INCREASING the confidence level of a confidence interval from 90% to 99% makes the interval SHORTER.

A CONFIDENCE INTERVAL can be interpreted as the single best ESTIMATE of a population parameter.

As the sample size INCREASES for computing a confidence interval, the width of the confidence interval DECREASES.

For a PROBABILITY DENSITY FUNCTION, the area between two values aand b is the probability a randomly selected individual will have a value between a and b.

A Z-SCORE can be interpreted for a value as the value's number of standard deviation above or below the mean.

A NORMAL distribution will have an approximately SYMMETRIC histogram.

As the STANDARD DEVIATION decreases for a normal distribution, the values become LESS concentrated around the MEAN.

The goal when using confidence intervals is to have WIDE INTERVALS to be assured that the interval contains the population parameter.

A SYMMETRIC histogram implies the plotted variable is NORMALLY distributed.

A sample of n=18 observations is drawn from a normal population with μ=1040 and σ=230. Find each of the following:

A. P(X¯>1159)

Probability =

B. P(X¯<953)

Probability =

C. P(X¯>996)

Probability =

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?