Assume that the traffic to the web site of Smiley’s People, Inc., which sells customized T-shirts, follows a normal distribution, with a mean of 4.56 million visitors per day and a standard deviation of 820,000 visitors per day.

Consider the data set below of the 14” pizzas at Little Italy. There are two pizza chefs (Anna and Lorenzo, Factor A), two work shifts (day and evening, Factor B), and two dough companies (Sysco and PFG Holdings, Factor C). For each combination, there are four pizzas made (n = 4). The diameter of each pizza is measured.

                                                                        Measurements                                                

Pizza                                       Anna                                                   Lorenzo

Dough                Shift             1          2        3        4                    1            2        3        4

Sysco                   Day             14.1     14.1     14.2     14.1                 14.2     14.1     14.1     13.9

Sysco                   Evening       14.1     14.2     14.3     14.2                 14.3     14.2     14.3     14.1

PFG Holdings     Day             13.9     14.0     14.0     14.1                 13.9     13.9     13.8     13.9

PFG Holdings     Evening       13.8     13.7     13.9     14.0                 14.3     14.2     14.2     14.1

Using the main effects and the interactions effects and the contrasts you calculated in part 1:

1. Calculate the SS of each factor, the factor interactions, the error, and the total.
2. Construct the ANOVA table showing the MS of each factor, interaction and error, along with the F values.
3. Which main effects and interactions are significant at 95% confidence?

Use the given information to find the number of degrees of​ freedom, the critical values

chi Subscript Upper L Superscript χ2L

and

chi Subscript Upper R Superscript χ2R​,

and the confidence interval estimate of

sigmaσ.

It is reasonable to assume that a simple random sample has been selected from a population with a normal distribution.

Platelet Counts of Women

95​%

​confidence;

n=40​,

s=65.4.

A random sample of 64 students at a university showed an average age of 20 years and a sample standard deviation of 4 years. The 90% confidence interval for the true average age of all students in the university is

A survey of 170 commuters shows that on average people travel 17.4 km with standard deviation 8.4 km. Find 99% confidence interval for the population mean commuting distance. State your conclusion.

What is the minimal sample size needed for a 95% confidence interval to have a maximal margin of error of 0.1 in the following scenarios? (Round your answers up the nearest whole number.)

(a) a preliminary estimate for p is 0.15


(b) there is no preliminary estimate for p

Question 8
Hypothesis Test - Terminology
Match terms to descriptions Question 8 options:

a) The hypothesis expressing a claim involving one of =, ≤ (at most), or ≥ (at least) and requiring no (null) action.
b) The hypothesis expressing a claim involving one of ≠, >, or < and requiring action.
c) The sign of the critical value of a a 1-tail test with upper reject region is
d) Greek letter denoting the population standard deviation
e) For an upper tail test, the probability of an equal or greater test statistic.
f) Greek letter denoting the population mean
g) The sign of the critical value of a 1-tail test with lower reject region is
h) Rejecting H0 when H0 is actually false
i) The risk as a probability that we are willing to take of rejecting H0 when it is actually true.
j) Greek letter denoting the population proportion
k) Failure to reject H0 when H0 is actually false
l) Rejection of the null hypothesis when H0 is actually true
M) The value of the test statistic where the pvalue = significance level α.

Match with the following:
1) H0
2) HA
3) pvalue
4) alpha α
5) Critical value
6) Type 1 error
7) Type 2 error
8) Not an error
9) Negative
10) Positive
11) μ
12) π
13) σ
14) No answer fits

MK 332 Homework April 1, 2020

Q2. A company wants to know how long newly hired MBAs remain on their first jobs. A sample of 17 MBAs has an average of 3 years, with a standard deviation of 2. Create a Confidence Interval with Confidence level of 95%. Interpret.

Q1. A Pizza Hut store manager believes that the average number of customers who return a pizza or ask for a refund is 20 per day. The store records the number of returns and exchanges for the 25 days it was open during a given month. Are the returns different than 20 per day?

Sample mean= 25, s=5, n=25. Test at α = .01.

H0: µ = 20

H1: µ ≠ 20

Test by both methods, interpret and conclude.

17) Not everyone pays the same price for the same model of a car. For a particular new car the mean price is $18,750 with a standard deviation of $690.   ROUND VALUES CORRECTLY TO DOLLAR.

a) Make a sketch of the normal distribution, labeling the center and ± 1, 2, 3SD intervals with the values.

b) If a car is picked at random, what is the probability that the car price is between $17,800 and $19,100?

c) A car in the top 12% will have a price of at least how much? (To nearest dollar)

d) We take many samples of 30 cars of this model. Find the standard deviation of the sample to nearest dollar. (no sentence)

e) Sketch the sampling distribution if we took many samples of 30 of these models of car. (no sentence)

f) What is the probability that a sample of 30 cars, this model, will have a mean price cost of $18,400 or less?

g) A sample of 30 cars, this model, that has an average price in the top car in the bottom 5% will average price at most how much? (To nearest dollar)

Before the furniture store began its ad campaign, it averaged 193 customers per day. The manager is investigating if the average is smaller since the ad came out. The data for the 12 randomly selected days since the ad campaign began is shown below:

200, 181, 185, 205, 198, 201, 173, 165, 180, 198, 199, 200

Assuming that the distribution is normal, what can be concluded at the αα = 0.05 level of significance?

1. For this study, we should use Select an answer z-test for a population proportion t-test for a population mean
2. The null and alternative hypotheses would be:

H0:H0:  ? p μ  Select an answer < > = ≠       

H1:H1:  ? p μ  Select an answer < > = ≠    

3. The test statistic ? z t  =  (please show your answer to 3 decimal places.)
4. The p-value =  (Please show your answer to 4 decimal places.)
5. The p-value is ? > ≤  αα
6. Based on this, we should Select an answer reject accept fail to reject  the null hypothesis.
7. Thus, the final conclusion is that ...
   • The data suggest the populaton mean is significantly less than 193 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is less than 193.
   • The data suggest the population mean is not significantly less than 193 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of customers since the ad campaign began is equal to 193.
   • The data suggest that the population mean number of customers since the ad campaign began is not significantly less than 193 at αα = 0.05, so there is insufficient evidence to conclude that the population mean number of customers since the ad campaign began is less than 193.
8. Interpret the p-value in the context of the study.
   • There is a 25.3512784% chance that the population mean number of customers since the ad campaign began is less than 193.
   • There is a 25.3512784% chance of a Type I error.
   • If the population mean number of customers since the ad campaign began is 193 and if you collect data for another 12 days since the ad campaign began, then there would be a 25.3512784% chance that the sample mean for these 12 days would be less than 190.4.
   • If the population mean number of customers since the ad campaign began is 193 and if you collect data for another 12 days since the ad campaign began, then there would be a 25.3512784% chance that the population mean number of customers since the ad campaign began would be less than 193.
9. Interpret the level of significance in the context of the study.
   • If the population mean number of customers since the ad campaign began is 193 and if you collect data for another 12 days since the ad campaign began, then there would be a 5% chance that we would end up falsely concuding that the population mean number of customers since the ad campaign began is less than 193.
   • If the population mean number of customers since the ad campaign began is less than 193 and if you collect data for another 12 days since the ad campaign began, then there would be a 5% chance that we would end up falsely concuding that the population mean number of customers since the ad campaign is equal to 193.
   • There is a 5% chance that there will be no customers since everyone shops online nowadays.
   • There is a 5% chance that the population mean number of customers since the ad campaign began is less than 193.

A team of researchers would like to find out if Country X residents, who favor a low fat diet, have significantly lower cholesterol levels compared to that of the US population. They collected data from a large sample of Country X residents and the Z statistic is calculated to be -1.34 in a one-tailed test with an alpha level of .05. What is the researchers' conclusion of the hypothesis test?

Group of answer choices

The researchers fail to reject the null hypothesis.

The researchers reject the alternative hypothesis.

The researchers reject the null hypothesis.

More information is needed to reach a conclusion.

A study was conducted on the effect of money on life satisfaction by comparing a group of people who are financially wealthy to the general population. Based on the statistical result, the researchers failed to reject the null hypothesis, so it can be concluded that ____.

Group of answer choices

the probability is high that the life satisfaction of people who are wealthy is different from the general population

life satisfaction of people who are wealthy has been proven to be the same as the general population

life satisfaction of people who are wealthy has been proven to be different from the general population

the probability is high that the life satisfaction of people who are wealthy is the same as the general population

Based on the national statistics on MPG ratings, the national average rating for sedans is 25 with a standard deviation of 5. My car has an MPG of 29, what is the percentage of sedans performing better than my car in MPG?

Group of answer choices

50 + 28.81% - 78.81%

50% - 28.81% = 21.19%

100% - 34.13% = 65.87%

100% - 34.13% = 65.86%

When using the Z table for a two-tailed hypothesis test with a preset alpha (significance) level, what is the correct sequence of the following steps (it is possible not all steps will be used):

1) Look for the corresponding Z value(s)
2) Multiply the percentage of alpha by 2 to be the "tail area percentage"
3) Look in the "tail" column for the "tail area percentage”
4) Divide the percentage of alpha by 2 to be the "tail area percentage"
5) Convert the alpha to percentage

Group of answer choices

5, 4, 3, 1

5, 2, 3, 1

5, 4, 3, 1, 2

1, 2, 3

How does a two-tailed test compared to a one-tailed test when given a sample statistic and a fixed alpha level?

Group of answer choices

The critical value to be compared to the statistic would be less extreme with a two-tailed test.

The critical value to be compared to the statistic would be more extreme with a two-tailed test.

The total significance area in the comparison distribution is larger in a two-tailed test.

The total significance area in the comparison distribution is smaller in a one-tailed test.

15% of all Americans suffer from sleep apnea. A researcher suspects that a higher percentage of those who live in the inner city have sleep apnea. Of the 315 people from the inner city surveyed, 63 of them suffered from sleep apnea. What can be concluded at the level of significance of αα = 0.01? Round numerical answers to 3 decimal places

1. For this study, we should use Select an answer z-test for a population proportion t-test for a population mean
2. The null and alternative hypotheses would be:
   Ho: ? p μ  Select an answer < > = ≠   (please enter a decimal)   
   H₁: ? μ p  Select an answer > ≠ = <   (Please enter a decimal)

3. The test statistic ? z t  =  (please show your answer to 3 decimal places.)
4. The p-value =  (Please show your answer to 4 decimal places.)
5. The p-value is ? ≤ >  αα
6. Based on this, we should Select an answer accept fail to reject reject  the null hypothesis.
7. Thus, the final conclusion is that ...
   • The data suggest the population proportion is not significantly larger than 15% at αα = 0.01, so there is not sufficient evidence to conclude that the population proportion of inner city residents who have sleep apnea is larger than 15%.
   • The data suggest the population proportion is not significantly larger than 15% at αα = 0.01, so there is sufficient evidence to conclude that the population proportion of inner city residents who have sleep apnea is equal to 15%.
   • The data suggest the populaton proportion is significantly larger than 15% at αα = 0.01, so there is sufficient evidence to conclude that the population proportion of inner city residents who have sleep apnea is larger than 15%
8. Interpret the p-value in the context of the study.
   • If the sample proportion of inner city residents who have sleep apnea is 20% and if another 315 inner city residents are surveyed then there would be a 0.65% chance of concluding that more than 15% of all inner city residents have sleep apnea.
   • There is a 0.65% chance that more than 15% of all inner city residents have sleep apnea.
   • If the population proportion of inner city residents who have sleep apnea is 15% and if another 315 inner city residents are surveyed then there would be a 0.65% chance that more than 20% of the 315 inner city residents surveyed have sleep apnea.
   • There is a 0.65% chance of a Type I error.
9. Interpret the level of significance in the context of the study.
   • There is a 1% chance that aliens have secretly taken over the earth and have cleverly disguised themselves as the presidents of each of the countries on earth.
   • If the population proportion of inner city residents who have sleep apnea is larger than 15% and if another 315 inner city residents are surveyed then there would be a 1% chance that we would end up falsely concluding that the proportion of all inner city residents who have sleep apnea is equal to 15%.
   • There is a 1% chance that the proportion of all inner city residents who have sleep apnea is larger than 15%.
   • If the population proportion of inner city residents who have sleep apnea is 15% and if another 315 inner city residents are surveyed then there would be a 1% chance that we would end up falsely concluding that the proportion of all inner city residents who have sleep apnea is larger than 15%.

Instructions:

• Show all work, including every calculation and every formula, with proper symbols. Formulas must be those used in the required textbook for the course.
• Use the t-table in your book to find the critical t value(s).
• Calculators are fine; computer software of any kind is NOT. (You are welcome to use SPSS to check your work, but all values generated/reported below must come from hand calculations or the t table in your book.)

A therapist was interested in determining whether patients experiences reduced anxiety following diaphragmatic breathing exercises. She includes 9 participants in her brief study. Each patient provides a rating for current anxiety, on a scale of 1 (least anxiety) to 10 (extreme anxiety). She then instructs them on a 45-minute diaphragmatic breathing exercise. Following the exercise, each patient again rates his/her anxiety on the same 1-10 scale. (Note that this is a repeated measures study because each patient/participant is measured twice, once before the treatment and once after the treatment.)

1. Can the therapist conclude that there was a significant change in anxiety levels after treatment? Use a two-tailed test with α = .05. (Note that you will need to use – and clearly show – all 4 steps of hypothesis testing to answer this question.)

2. Compute the value of r² (percentage of variance accounted for) for these data.

3. Write a sentence showing how the outcome of the hypothesis test and the measure of effect size would appear in a research report (i.e., in APA format). Note that you can find an example of APA format for a repeated measures t-test both in your book, in the “In the Literature” section of chapter 11, and in the example I provided in this week’s resources. APA format is slightly different depending on the type of test used, so be sure to look for an example of a repeated-measures t-test!

TV sets: According to the Nielsen Company, the mean number of TV sets in a U.S. household in 2013 was 2.24 . Assume the standard deviation is 1.2 . A sample of 95 households is drawn. Use the Cumulative Normal Distribution Table if needed.

What is the probability that the sample mean number of TV sets is greater than 2? Round your answer to four decimal places.

What is the probability that the sample mean number of TV sets is between 2.5 and 3? Round your answer to four decimal places.

Find the 80th percentile of the sample mean. Round your answer to two decimal places.

Would it be unusual for the sample mean to be less than 2? Round your answer to four decimal places.