1. A survey of the mean number of cents off that coupons give was conducted by randomly surveying one coupon per page from the coupon sections of a recent San Jose Mercury News. The following data were collected: 20¢; 75¢; 50¢; 65¢; 30¢; 55¢; 40¢; 40¢; 30¢; 55¢; $1.50; 40¢; 65¢; 40¢. Assume the underlying distribution is approximately normal. You wish to conduct a hypothesis test (α = 0.05 level) to determine if the mean cents off for coupons is less than 50¢.
   1. State the null and alternate hypotheses clearly.
   2. Conduct the hypothesis test based on the test statistic and critical value(s). Clearly indicate each.
   3. What is the p-value? Use the p-value to conduct the same test
   4. Report your conclusion in words, in the context of the problem.
   5. What is the power of the for an alternative hypothesis value of 49¢?

Four fair dice, colored red, green, blue, and white, are tossed.

(a) Determine the probability of getting all four face values equal to 3.

(b) After tossing, a quick glance at the outcome indicated that two of the face values were 3 but no other information (about their color or the values of the remaining two faces) was noted. Now determine the probability of getting all four face values equal to 3.

(c) After tossing it was further noted that, out of the two observed face values of 3, one was red in color. Now determine the probability of getting all four face values equal to 3.

(d) After tossing, it was finally confirmed that the two observed face values of 3 were red and white. Now determine the probability of getting all four face values equal to 3.

(e) You should get distinct answers for the above four probabilities. Qualitatively explain why the above four probabilities make sense.

These are supposed to be the correct answers: (a) 1/1296 (b) 1/171, (c) 1/91, and (d) 1/36.

Each of the first 6 letters of the alphabet is printed on a separate card. The letter “a” is printed twice. What is the probability of drawing 4 cards and getting the letters f, a, d, a in that order? Same question if the order does not matter.

Suppose that a category of world class runners are known to run a marathon (26 miles) in an average of 149 minutes with a standard deviation of 12 minutes. Consider 49 of the races.
Let X = the average of the 49 races.

a.) X ~ N (149, ? )

b.Find the probability that the runner will average between 148 and 151 minutes in these 49 marathons. (Round your answer to four decimal places.)

c. Find the 80th percentile for the average of these 49 marathons. (Round your answer to two decimal places.)

d. Find the median of the average running times.

You wish to test the following claim (Ha) at a significance level of α=0.002.

      Ho:p1=p2
      Ha:p1>p2

You obtain a sample from the first population with 43 successes and 267 failures. You obtain a sample from the second population with 85 successes and 581 failures. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) α

• greater than α

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the first population proportion is greater than the second population proportion.
• There is not sufficient evidence to warrant rejection of the claim that the first population proportion is greater than the second population proportion.
• The sample data support the claim that the first population proportion is greater than the second population proportion.
• There is not sufficient sample evidence to support the claim that the first population proportion is greater than the second population proportion.

Question 2)

You wish to test the following claim (Ha) at a significance level of α=0.10.

      Ho:p1=p2
      Ha:p1≠p2

You obtain 12.3% successes in a sample of size n1=759 from the first population. You obtain 8.8% successes in a sample of size n2=646 from the second population. For this test, you should NOT use the continuity correction, and you should use the normal distribution as an approximation for the binomial distribution.

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =

What is the p-value for this sample? (Report answer accurate to four decimal places.)
p-value =

The p-value is...

• less than (or equal to) α

• greater than α

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the first population proportion is not equal to the second population proprtion.
• There is not sufficient evidence to warrant rejection of the claim that the first population proportion is not equal to the second population proprtion.
• The sample data support the claim that the first population proportion is not equal to the second population proprtion.
• There is not sufficient sample evidence to support the claim that the first population proportion is not equal to the second population proprtion.

Questiom 3)

Test the claim that the proportion of men who own cats is smaller than the proportion of women who own cats at the .10 significance level.

The null and alternative hypothesis would be:

H0:μM=μF

H1:μM>μF

H0:pM=pF

H1:pM>pF

H0:pM=pF

H1:pM≠pF

H0:μM=μF

H1:μM<μF

H0:μM=μF

H1:μM≠μF

H0:pM=pF

H1:pM<pF

The test is:

right-tailed

left-tailed

two-tailed

Based on a sample of 20 men, 40% owned cats
Based on a sample of 80 women, 65% owned cats

The test statistic is: (to 2 decimals)

The p-value is: (to 2 decimals)

Based on this we:

• Reject the null hypothesis
• Fail to reject the null hypothesis

Groups of dolphins were systematically observed off the coast of Iceland near Keflavik in 1998. Each observation included the main activity of a dolphin group (Activity) and the time of day the group was observed (Time). The groups varied in size, with feeding or socialising groups usually including more dolphins than travelling groups, but no information about group size was included with the data. The observations are summarised in the following table: No. of groups, summarised by activity and time. Time Morning Noon or Afternoon Evening Activity Travelling 6 20 13 Feeding 28 4 56 Socialising 38 14 10

(a) In looking for an association between Activity and Time, which variable would be the predictor and which the response? Justify your answer.

(Is this correct),My answer is-

1. The predictor variable is the main activity of a dolphin group

         (Activity) and the response variable the time of day the group was

         observed (Time). The Activity of the dolphins decides what time of day

         it is.

(b) How strong is the evidence that dolphin activity typically varies during the day? Test at a 1% significance level.If you conclude that there is a relationship, describe it.

Does this look like I am on the right path?

(My answer)

1. H0: There is no association between Dolphin Activity and the Time of day.

H1: There is some association.

     Significance Level: α= 0.01

         The test requires for the sample to be randomly selected and all the

         expected observations to be ≥5.

(My question)-It says the dolphins are systematically observed(does that mean it is not a random sample?)

Assume that a certain batch of 200 castings contains 5 defectives. Calculate the probability that of three castings selected, exactly one will be defective. Answer: 0.0720 (Show work and reasoning!)

*

*****Complete #3 two ways*********

            i. Same as in Example above

            ii. using the Excel NORMDIST and NORMINV functions

                        demonstrated in the Prob Dist podcast

NOTE: Excel functions (just as the table) may not directly give the answer you are looking for, you must understand what they return and how to use it.

3. The average amount parents spent per child on back-to-school clothes in Fall 2019 was $635. Assume the standard deviation is $150 and the amount spent is normally distributed.

a. What is the probability that the amount spent on a randomly selected child is more than $800?

b. What is the probability that the amount spent on a randomly selected child is more than $500

c. 95% of the parents will spend more than what amount?

Lying to a teacher. One of the questions in a survey of high school students asked about lying to teachers. The following table gives the number of students who said that they lied to a teacher as least once during the past year, classified by sex:

A. Add the marginal totals to the table

B. Calculate appropriate percents to describe the results of this question

C. Summarize your findings in a short paragraph.

D. Test the null hypothesis that there is no association between sex and lying to teachers. Give the test statistics and the p-value with a sketch similar to the one on page 535 and summarize your conclusion. Be sure to include numerical and graphical summaries.

E. The survey asked student if they lied, but we do not know if they answered the question truthfully. How does this fact affect the conclusions that you can draw from this data?

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, 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 0.05 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?

724    660     1157     575     552     442

Identify test statistic

identify P value

State conclusion

coffee tea juice

Do a One-way ANOVA by hand (at least once in your life!) …Is there a difference in attention for those who drink coffee, tea, or juice during an 8 a.m. class? Utilize the five steps of hypothesis testing to analyze the following data (p<.01).

Attention Ratings (1=no attention- 5=full attention)

The data set represents the number of movies that a sample of 20 people watched in a year.

121 148 94 142 170 88 221 106 18 67 149 28 60 101 134 168 92 154 53 66

a.) construct a frequency distribution for the data set using six classes. Include class limits, midpoints, frequencies, relative frequencies, and cumulative frequencies. b.) Display the data using a frequency histogram (Must use EXCEL) c.) Describe the shape of the distribution as symmetric, uniform, skewed left, skewed right or none of these and give an interpretation of this data.

9.9. Is gender independent of education level? A random sample of people was surveyed and

each person was asked to report the highest education level they obtained. Perform a hypothesis

test. Include all 5 steps.

10.9. Compute and interpret the correlation coefficient for the following grades of 6 students

selected at random.

1. The primary rule of subject selection for experimental designs is the comparability of experimental and control groups. Ideally, the control group would be identical to the experimental group if it had not been exposed to the experimental stimulus. Therefore, the experimental and control groups should be as similar as possible. Probability sampling, randomization, and matching are several different methods for achieving this similarity.

One technique for ensuring an appropriate control group is the process of selecting multiple samples from a population using a method that is subject to chance rather than the bias of the experimenter and then assigning each sample to either an experimental or a control group. This technique is called A.PROBABILITY SAMPLING B. RANDOMIZATION

C. MATCHING

2. Nikhil is interested in finding out whether reading a comic strip about euthanasia will change people’s opinions on the subject. Nikhil decides to recruit a group of subjects and then divide the group into subgroups of older and younger students of each ethnicity. He then assigns half of each subgroup to be part of the experimental group and half to the control group, so that the experimental and control groups have the same makeup in terms of age and ethnicity. Nikhil administers a pretest to everyone to measure their opinions about euthanasia. Then the experimental group is asked to read the comic strip and the control group is not. Finally, Nikhil administers a posttest to everyone to see if their opinions about euthanasia have changed.

Which technique is Nikhil employing to ensure that the experimental and control groups are equivalent? A. Matching B. Randomization C. Probability sampling

3. The three different techniques have different advantages and disadvantages. In which of the following ways is randomization a better choice than matching? Check all that apply.

A. You do not necessarily know in advance which variables will be the most relevant for the study.

B.Randomization allows for conscious stratification.

C. Most of the statistics used to analyze the results of experiments assume randomization.