# FileHN3

Question 1
Many high school students take the AP tests in different subject areas.  In 2007, of the 211,693 students who took the calculus AB exam 102,598 of them were female and 109,095 of them were male (“AP exam scores,” 2013).  Estimate the difference in proportion of female students taking the calculus AB exam versus male students taking the calculus AB exam using a 90% confidence level.
(i)  Enter the level of significance α used for this test:
(ii)  Determine  p ^  female exam takers   and  p ^  male exam
(iii)  Determine Z score corresponding to desired confidence level
(iv)  Determine error bound of the proportion:
(v)  Let “p1” refer to proportion of children diagnosed with ASD in Pennsylvania.  Let “p2” represent proportion of children diagnosed with ASD in Utah.  Determine the confidence interval of the proportion difference p1 – p2:
(vi)  Using the confidence interval, select the correct description of the result of the survey:
A.  We estimate with 90% confidence that the true proportion of female students taking the calculus AB exam is anywhere from 2.78% to 3.31% higher than the proportion of male students taking the calculus AB exam.
B.  We estimate with 90% confidence that the true proportion of male students taking the calculus AB exam is anywhere from 2.78% to 3.31% lower than the proportion of female students taking the calculus AB exam.
C.  We estimate with 90% confidence that the true mean population of female students taking the calculus AB exam is anywhere from 2.78% to 3.31% higher than the true mean population of male students taking the calculus AB exam.
Question 2
Are there more children diagnosed with Autism Spectrum Disorder (ASD) in states that have larger urban areas over states that are mostly rural?  In the state of Pennsylvania, a fairly urban state, there are 245 eight yearolds diagnosed with ASD out of 18,440 eight year olds evaluated.  In the state of Utah, a fairly rural state, there are 45 eight yearolds diagnosed with ASD out of 2,123 eight year olds evaluated (“Autism and developmental,” 2008).  Is there enough evidence to show that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion in Utah?  Test at the 1% level.
(i)  Let p1 = proportion of children diagnosed with ASD in Pennsylvania and p2 = proportion of children diagnosed with ASD in Utah.  Which of the following statements correctly defines the null hypothesis HO?
(ii)  Let p1 = proportion of children diagnosed with ASD in Pennsylvania and p2 = proportion of children diagnosed with ASD in Utah.  Which of the following statements correctly defines the alternate hypothesis HA?
(iii)  Enter the level of significance α used for this test:
(iv)  Determine  p ˆ  Pennsylvania   and  p ˆ  Utah
(v)  Find pooled sample proportion p ¯ ¯   Let p ˆ  Pennsylvania  = p ˆ  1  and p ˆ  Utah  = p ˆ  2
Enter in decimal form to nearest ten-thousandth.  Examples of correctly entered answers:
(vi)  Calculate and enter test statistic
(vii)  Using tables, calculator, or spreadsheet:  Determine and enter p-value corresponding to test statistic.
(viii)  Comparing p-value and α value, which is the correct decision to make for this hypothesis test?
(ix)  Select the statement that most correctly interprets the result of this test:
A.  The result is not statistically significant at .01 level of significance.  Evidence supports the claim that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion of children diagnosed with ASD in Utah.
B.  The result is statistically significant at .01 level of significance.  There is not enough evidence to show that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion of children diagnosed with ASD in Utah.
D.  The result is statistically significant at .01 level of significance.  Evidence supports the claim that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion of children diagnosed with ASD in Utah.
Question 3
All Fresh Seafood is a wholesale fish company based on the east coast of the U.S.  Catalina Offshore Products is a wholesale fish company based on the west coast of the U.S.  Table #9.2.5 contains prices from both companies for specific fish types (“Seafood online,” 2013) (“Buy sushi grade,” 2013).  Data is in following table:

Fish
All Fresh Seafood Prices
Catalina Offshore Products Prices
Cod
19.99
17.99
Tilapi
6.00
13.99
Farmed Salmon
19.99
22.99
Organic Salmon
24.99
24.99
Grouper Fillet
29.99
19.99
Tuna
28.99
31.99
Swordfish
23.99
23.99
Sea Bass
32.99
23.99
Striped Bass
29.99
14.99
Do the data provide enough evidence to show that a west coast fish wholesaler is more expensive than an east coast wholesaler?  Test at the 5% level.
(i)  Let μ1 = mean wholesale prices from west coast fishery.  Let μ2 = mean wholesale prices from east coast fishery .  Which of the following statements correctly defines the null hypothesis HO?
(iii)  Enter the level of significance α used for this test:
(iv)  Determine sample mean of differences  x ¯  d
(v)  Determine sample standard deviation of differences sd
(vi)  Calculate and enter test statistic

Enter value in integer form.  Examples of correctly entered answers:
(vii)  Determine degrees of freedom for the sample of differences dfd:
(viii)  Using tables, calculator, or spreadsheet:  Determine and enter p-value corresponding to test statistic.
(ix)  Comparing p-value and α value, which is the correct decision to make for this hypothesis test?
A.  Reject Ho
B.  Fail to reject Ho
C.  Accept Ho
D.  Accept HA
(x)  Select the statement that most correctly interprets the result of this test:
A.  The result is not statistically significant at .05 level of significance.  Sufficient evidence exists to support the claim that west coast fish wholesalers are more expensive than east coast wholesalers.
B.  The result is not statistically significant at .05 level of significance.  There is not enough evidence to support the claim that west coast fish wholesalers are more expensive than east coast wholesalers.
C.  The result is statistically significant at .05 level of significance.  There is not enough evidence to support the claim that west coast fish wholesalers are more expensive than east coast wholesalers.
D.  The result is statistically significant at .05 level of significance.  Sufficient evidence exists to support the claim that west coast fish wholesalers are more expensive than east coast wholesalers.
Question 4
The British Department of Transportation studied to see if people avoid driving on Friday the 13th.   They did a traffic count on a Friday and then again on a Friday the 13th at the same two locations (“Friday the 13th,” 2013).  The data for each location on the two different dates is in following table:
Let μ1 = mean traffic count on Friday the 6th.  Let μ2=  mean traffic count on Friday the 13th.  Estimate the mean difference in traffic count between the 6th and the 13th using a 90% level.
(i)  Determine the sample mean of differences  x ¯  d
(ii)  Determine sample standard deviation of differences sd
(iii)  Enter the level of significance α used for this test:
(iv)  Determine degrees of freedom for the sample of differences dfd:
(v)  Determine t – score associated with critical value:  tc
(vi)  Determine “error bound of the mean of difference” E
(vii)  Determine confidence interval of the mean difference μd
(viii)  Using the confidence interval, select the correct description of the result of the survey:
A.  We estimate with 90% confidence that the true population mean traffic count between Friday the 6th and Friday the 13th is between 1154.1 and 2517.5.
B.  We estimate with 90% confidence that the true mean difference in traffic counts between Friday the 6th and Friday the 13th falls outside 1154.1 and 2517.5.
C.  We estimate with 90% confidence that the true sample mean traffic counts between Friday the 6th and Friday the 13th is between 1154.1 and 2517.5.
D.  We estimate with 90% confidence that the true mean difference in traffic counts between Friday the 6th and Friday the 13th is between 1154.1 and 2517.5.
Question 5
The income of males in each state of the United States, including the District of Columbia and Puerto Rico, are given in Table 1, and the income of females is given in Table 2 (“Median income of,” 2013):
Table 1: Data of Income for Males
(i)  Let μ1 = mean income of a male in 2013.  Let μ2=  mean income of a female in 2013.  Which of the following statements correctly defines the null hypothesis HO?
A.  μ1 − μ2 = 0  (μ1 = μ2)
B.  μ1 – μ2<0  (μ1<μ2) C.  μ1 − μ2>0  (μ1> μ2)
D.  μ1 + μ2= 0

(ii)   Let μ1 = mean income of a male in 2013.  Let μ2=  mean income of a female in 2013.  Which of the following statements correctly defines the alternate hypothesis HA?
A.  μ1 − μ2 = 0  (μ1 = μ2)
B.  μ1 – μ2<0  (μ1<μ2) C.  μ1 − μ2>0  (μ1> μ2)
D.  μ1 + μ2= 0
(iii)  Enter the level of significance α used for this test:
(iv)  For sample from population with mean = μ 1  :          Determine sample mean x ¯  1  and sample standard deviation s 1
(v)  For sample from population with mean = μ 2  :       Determine sample mean x ¯  2  and sample standard deviation s
(vi)  Determine degrees of freedom df :
Enter value in decimal form rounded down to nearest whole number.  Examples of correctly entered answers:
(vii)  Determine test statistic:
Enter value in decimal form rounded to nearest thousandth.  Examples of correctly entered answers:
(viii)  Using tables, calculator, or spreadsheet:  Determine and enter p-value corresponding to test statistic.
(ix)  Comparing p-value and α value, which is the correct decision to make for this hypothesis test?
A.  Reject Ho
B.  Fail to reject Ho
C.  Accept Ho
D.  Accept HA
(x)  Select the statement that most correctly interprets the result of this test:
A.  The result is not statistically significant at .01 level of significance.  Sufficient evidence exists to support the claim that the mean income of males is more than of females. .
B.  The result is not statistically significant at .01 level of significance.  There is not enough evidence to support the claim that the mean income of males is more than of females.
C.  The result is statistically significant at .01 level of significance.  Sufficient evidence exists to support the claim that the mean income of males is more than of females.
D.  The result is statistically significant at .01 level of significance.  There is not enough evidence to support the claim that the mean income of males is more than of females.
Question 6
A study was conducted that measured the total brain volume (TBV) (in mm3) of patients that had schizophrenia and patients that are considered normal.  Table #1 contains the TBV of the normal patients and Table #2 contains the TBV of schizophrenia patients (“SOCR data Oct2009,” 2013).
Is there enough evidence to show that the patients with schizophrenia have less TBV on average than a patient that is considered normal?  Test at the 10% level.
(i)  Let μ1 = mean TBV of patients that are considered normal.  Let μ2 = mean TBV of patients that had schizophrenia.  Which of the following statements correctly defines the null hypothesis HO?
(ii)   Let μ1 = mean TBV of patients that are considered normal.  Let μ2 = mean TBV of patients that had schizophrenia.  Which of the following statements correctly defines the alternate hypothesis HA?
(iii)  Enter the level of significance α used for this test:
(iv)  For sample from population with mean = μ1 : Determine sample mean ˉx1 and sample standar
(v)  For sample from population with mean = μ2 :       Determine sample mean x¯2 and sample standard deviation s2
(vi)  Determine degrees of freedom df :
(vii)  Determine test statistic:
(viii)  Using tables, calculator, or spreadsheet:  Determine and enter p-value corresponding to test statistic.
(ix)  Comparing p-value and α value, which is the correct decision to make for this hypothesis test?
A.  Reject Ho
B.  Fail to reject Ho
C.  Accept Ho
D.  Accept HA
(x)  Select the statement that most correctly interprets the result of this test:
(ii)  For the sample from the population with mean = μ 2  :      Determine sample mean x ¯  2  and sample standard deviation s 2
(i)  For the sample from population with mean = μ 1  :      Determine sample mean x ¯  1  and sample standard deviation s 1  i  For the sample from population with mean = μ1 :      Determine sample mean
(iii)  Enter the level of significance α used for this test:
(iv)  Determine degrees of freedom df :
(v)  Determine t – score associated with critical value:  tc
(vi)  Determine “error bound of the mean” E
(vii) Determine confidence interval estimate of the difference μ1 – μ2:
(viii)  Using the confidence interval, select the most correct description of the result of the survey:
A.  We estimate with 90% confidence that the mean TBV of people considered normal is anywhere from 51565 mm3 more to 75677 mm3 less than the mean TBV for people with schizophrenia.
B.  We estimate with 90% confidence that the true proportional TBV of people considered normal is anywhere from 51565 mm3 less to 75677 mm3 more than the mean TBV for people with schizophrenia.
C.  We estimate with 90% confidence that the sample mean TBV of people considered normal is anywhere from 51565 mm3 less to 75677 mm3 more than the mean TBV for people with schizophrenia.
D.  We estimate with 90% confidence that the true mean TBV of people considered normal is anywhere from 51565 mm3 less to 75677 mm3 more than the mean TBV for people with schizophrenia.
Question 8
The number of cell phones per 100 residents in countries in Europe is given in Table #1 for the year 2010.  The number of cell phones per 100 residents in countries of the Americas is given in Table #2 also for the year 2010 (“Population reference bureau,” 2013).
Let μ1 = mean number of cell phones per 100 residents in countries of Europe.  Let μ2 = mean number of cell phones per 100 residents in countries of the Americas.  Find the 98% confidence interval for the difference in mean number of cell phones per 100 residents in Europe and the Americas.
(i)  For the sample from population with mean = μ 1  :      Determine sample mean x ¯  1  and sample standard deviation s 1
Enter sample mean in decimal form to nearest ten-thousandth, then comma, then sample standard deviation in decimal form to nearest ten-thousandth.  Examples of correctly entered answers:
(ii)  For the sample from the population with mean = μ2 :      Determine sample mean x¯2 and sample standard deviation s2
(iii)  Enter the level of significance α used for this test:
(iv)  Determine degrees of freedom df :
(v)  Determine t – score associated with critical value:  tc
(vi)  Determine “error bound of the mean” E
(vii) Determine confidence interval estimate of the difference μ1 – μ2:

(viii)  Using the confidence interval, select the most correct description of the result of the survey:
A.  We estimate with 98% confidence that the sample mean number of cell phones per 100 residents in countries of Europe is anywhere from 4.3641 to 37.5276 more than the mean number of cell phones per 100 residents in countries of the Americas.
B.  We estimate with 98% confidence that the true mean number of cell phones per 100 residents in countries of Europe is anywhere from 4.3641 to 37.5276 more than the mean number of cell phones per 100 residents in countries of the Americas.
C.  We estimate with 98% confidence that the true proportion of cell phones per 100 residents in countries of Europe is anywhere from 4.3641 to 37.5276 more than the mean number of cell phones per 100 residents in countries of the Americas.
D.  We estimate with 98% confidence that the mean number of cell phones per 100 residents in countries of Europe is anywhere from 4.3641 to 37.5276 less than the mean number of cell phones per 100 residents in countries of the Americas.
Question 9
Levi-Strauss Co manufactures clothing.  The quality control department measures weekly values of different suppliers for the percentage difference of waste between the layout on the computer and the actual waste when the clothing is made (called run-up).  The data is in the following table, and there are some negative values because sometimes the supplier is able to layout the pattern better than the computer (“Waste run up,” 2013).
(ii)  Which of the following statements correctly defines the alternate hypothesis HA?
A.  All five mean percentage differences are equal
B.  Two of the mean percentage differences are not equal
C.  At least four of the mean percentage differences are equal
D.  At least two of the mean percentage differences are not equal
(iii)  Enter the level of significance α used for this test:
(iv)  Calculate sample mean and sample standard deviation for Plant 1 sample
(v)  Calculate sample mean and sample standard deviation for Plant 2 sample
(vi)  Calculate sample mean and sample standard deviation for Plant 3 sample
(vii)  Calculate sample mean and sample standard deviation for Plant 4 sample
(viii)  Calculate sample mean and sample standard deviation for Plant 5 sample
(ix)  Using technology, determine F ratio test statistic and corresponding p-value.
Use “CTRL-click” to access link.  Enter test statistic to nearest hundredth, then enter comma, then enter p-value to nearest thousandth.  Examples of correctly entered responses:
(x)  Comparing p-value and α value, which is the correct decision to make for this hypothesis test?
A.  Reject Ho
B.  Fail to reject Ho
C.  Accept Ho
D.  Accept HA
Enter letter corresponding to correct answer.
(xi)  Select the statement that most correctly interprets the result of this test:
A.  The result is not statistically significant at .01 level of significance.  Sufficient evidence exists to support the claim that there is a difference between some of the suppliers.
B.  The result is statistically significant at .01 level of significance.  There is not enough evidence to support the claim that there is a difference between some of the suppliers.
C.  The result is statistically significant at .01 level of significance.  Sufficient evidence exists to support the claim that there is a difference between some of the suppliers.
D.  The result is not statistically significant at .01 level of significance.  There is not enough evidence to support the claim that there is a difference between some of the suppliers.
Question 10
A study was undertaken to see how accurate food labeling for calories on food that is considered “reduced calorie”.  The group measured the amount of calories for each item of food and then found the percent difference between measured and labeled food. The group also looked at food that was nationally advertised, regionally distributed, or locally prepared.  The data is in the following table (“Calories data file,” 2013).
(i)  Which of the following statements correctly defines the null hypothesis HO?
A.  None of the three mean percentage differences are equal
B.  At least two of the mean percentage differences are not equal
C.  At least two of the mean percentage differences are equal
D.  All three mean percentage differences are equal
(ii)  Which of the following statements correctly defines the alternate hypothesis HA?
A.  None of the three mean percentage differences are equal
B.  At least two of the mean percentage differences are not equal
C.  At least two of the mean percentage differences are equal
D.  All three mean percentage differences are equal
(iii)  Enter the level of significance α used for this test:
(iv)  Calculate sample mean and sample standard deviation for sample percent differences between measured and labeled food that is advertised nationally.
(v)  Calculate sample mean and sample standard deviation for for sample percent differences between measured and labeled food that is distributed regionally.
(vi)  Calculate sample mean and sample standard deviation for sample percent differences between measured and labeled food that is prepared locally.
(vii)  Determine F ratio test statistic and corresponding p-value.
Use “CTRL-click” to access link.  Enter test statistic to nearest hundredth, then enter comma, then enter p-value to nearest ten-thousandth.  Examples of correctly entered responses:
(viii)  Comparing p-value and α value, which is the correct decision to make for this hypothesis test?
(ix)  Select the statement that most correctly interprets the result of this test:
A.  The result is not statistically significant at .10 level of significance.  Sufficient evidence exists to support the claim that at least two of the mean percent differences between the three groups are different.
B.  The result is statistically significant at .10 level of significance.  There is not enough evidence to support the claim that at least two of the mean percent differences between the three groups are different.
C.  The result is statistically significant at .10 level of significance.  Sufficient evidence exists to support the claim that at least two of the mean percent differences between the three groups are different.
D.  The result is not statistically significant at .10 level of significance.  There is not enough evidence to support the claim that at least two of the mean percent differences between the three groups are different.
A claim is made that when parents use the XSORT method of gender selection during in-vitro fertilization, the proportion of baby girls is greater than 0.5.  The latest results show that among 945 babies born to couples using the XSORT method of gender selection, 879 were girls.  Identify the null and alternative hypothesis for this test.
Question 12
A 0.05 significance level is used for a hypothesis test of the claim that when parents use the XSORT method of gender selection, the proportion of baby girls is different from 0.5.  Assume that the data consists of 55 girls born in 100 births.  Identify the null and alternative hypothesis for this test.

H0: p = 0.5 ; HA: p > 0.5

H0: p = 0.5 ; HA: p < 0.5 H0: p = 0.5 ; HA: p ≠ 0.5 None of these Answers H0: p ≤ 0.5 ; HA: p > 0.5

Answer: H0: p = 0.5 ; HA: p ≠ 0.5
Question 13
A 0.05 significance level is used for a hypothesis test of the claim that when parents use the XSORT method of gender selection, the proportion of baby girls is different from 0.5.  Assume that the data consists of 55 girls born in 100 births.  Identify the null and alternative hypothesis for this test.  What is the value of α?

α = 0.05

α = 0.01

α = 0.95

α = 0.99
Question 14
A 0.05 significance level is used for a hypothesis test of the claim that when parents use the XSORT method of gender selection, the proportion of baby girls is different from 0.5.  Assume that the data consists of 55 girls born in 100 births.   Is this a two-tailed, left-tailed, or right-tailed test?
A 0.05 significance level is used for a hypothesis test of the claim that when parents use the XSORT method of gender selection, the proportion of baby girls is different from 0.5.  Assume that the data consists of 55 girls born in 100 births.   What is the value of the test statistic associated with this hypothesis test?

Question 16
A 0.05 significance level is used for a hypothesis test of the claim that when parents use the XSORT method of gender selection, the proportion of baby girls is different from 0.5.  Assume that the data consists of 55 girls born in 100 births.   What is the p-value associated with this test?
Question 17
Vincent randomly selected a bag of M&Ms from a randomly selected gas station as he was driving down the interstate.  Being bored while riding in the car, he decided to evaluate his green M&Ms.  From this simple random sample, there were 19 green M&Ms with a sample mean weight of 0.8635 grams and a sample standard deviation of 0.0570 grams.  Use a 0.05 significance level to test the claim that the mean weight of all M&Ms is equal to 0.8535 (which is the mean weight required so that M&Ms have the weight printed on the packaged label).
What is the null and alternative hypotheses for this test?
Question 18
Vincent randomly selected a bag of M&Ms from a randomly selected gas station as he was driving down the interstate.  Being bored while riding in the car, he decided to evaluate his green M&Ms.  From this simple random sample, there were 19 green M&Ms with a sample mean weight of 0.8635 grams and a sample standard deviation of 0.0570 grams.  Use a 0.05 significance level to test the claim that the mean weight of all M&Ms is equal to 0.8535 (which is the mean weight required so that M&Ms have the weight printed on the packaged label).
Question 19
Vincent randomly selected a bag of M&Ms from a randomly selected gas station as he was driving down the interstate.  Being bored while riding in the car, he decided to evaluate his green M&Ms.  From this simple random sample, there were 19 green M&Ms with a sample mean weight of 0.8635 grams and a sample standard deviation of 0.0570 grams.  Use a 0.05 significance level to test the claim that the mean weight of all M&Ms is equal to 0.8535 (which is the mean weight required so that M&Ms have the weight printed on the packaged label).
Question 20
Vincent randomly selected a bag of M&Ms from a randomly selected gas station as he was driving down the interstate.  Being bored while riding in the car, he decided to evaluate his green M&Ms.  From this simple random sample, there were 19 green M&Ms with a sample mean weight of 0.8635 grams and a sample standard deviation of 0.0570 grams.  Use a 0.05 significance level to test the claim that the mean weight of all M&Ms is equal to 0.8535 (which is the mean weight required so that M&Ms have the weight printed on the packaged label).
What is the P-value associated with this hypothesis test?
Question 21
Vincent randomly selected a bag of M&Ms from a randomly selected gas station as he was driving down the interstate.  Being bored while riding in the car, he decided to evaluate his green M&Ms.  From this simple random sample, there were 19 green M&Ms with a sample mean weight of 0.8635 grams and a sample standard deviation of 0.0570 grams.  Use a 0.05 significance level to test the claim that the mean weight of all M&Ms is equal to 0.8535 (which is the mean weight required so that M&Ms have the weight printed on the packaged label).
What is the critical value associated with this hypothesis test?
Vincent randomly selected a bag of M&Ms from a randomly selected gas station as he was driving down the interstate.  Being bored while riding in the car, he decided to evaluate his green M&Ms.  From this simple random sample, there were 19 green M&Ms with a sample mean weight of 0.8635 grams and a sample standard deviation of 0.0570 grams.  Use a 0.05 significance level to test the claim that the mean weight of all M&Ms is equal to 0.8535 (which is the mean weight required so that M&Ms have the weight printed on the packaged label).
Question 23
The Mars Candy Company claims that 13% of its M&M candies are brown, but a simple random sample of 100 M&M candies has only 8 brown M&M candies.  Suppose we want to use the sample data to construct a 98% confidence interval estimate of the proportion of brown M&M candies.
Which test statistic is used to compute the margin of error in this situation?
Question 24
The Mars Candy Company claims that 13% of its M&M candies are brown, but a simple random sample of 100 M&M candies has only 8 brown M&M candies.
Suppose we want to use the sample data to construct a 98% confidence interval estimate of the proportion of brown M&M candies.  Compute the Margin of Error for a 98% confidence interval estimate of the proportion of brown M&M candies.
Question 25
The Mars Candy Company claims that 13% of its M&M candies are brown, but a simple random sample of 100 M&M candies has only 8 brown M&M candies.
use the sample data to construct a 98% confidence interval estimate of the proportion of brown M&M candies.