DO THE CHI SQUARED TEST
Q: n terms of the model parameters, state the null hypothesis that, after controlling for sales and…
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DO THE CHI SQUARED TEST
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- A chi square test for homogeneity applies to: Select one: a. data from a single random sample where the sample is classified by values of three or more categorical variables. b. data from samples from two or more populations where each sample is classified by values of two categorical variables. c. data from samples from two or more populations where each sample is classified by values of a single categorical variable. d. data from a single random sample where the sample is classified by values of two categorical variables.b) The number of planes to arrive at the Dubai International Airport over N days were recorded and inputted into MINITAB for analysis. Exhibit 1 was subsequently generated. Exhibit 1 Stem-and-Leaf Display: No. of Arrival Leaf Unit = 1 N =* 5 2 01123 10 2 56899 ** 3 0001123 (***) 3 6666678888 12 4 01134 7 4 57789 2 5 00 Calculate he missing values *, ** and ***The number of planes to arrive at the Dubai International airport over N days were recorded and imputted into MINITAB for analysis. Exibit 1 was subsequently generated. Exhibit 1 stem - and leaf display: no. of arrival leaf unit = 1 N =* 5 2 01123 10 2 56899 * * 3 0001123 ( * * * ) 3 6666678888 12 4 01134 7 4 57789 2 5 00 i. Calculate the missing *, ** and *** ii Find the value of the a. Median b. Highest value c. Mode Using the stem and leaf diagram in Exibit 1, i. Form a frequency table by identifying the class intervals implicitly Using the table created in 2ci) above ii. calculate the sample mean iii. calculate the sample variance NB. please show all workings, preferably typewritten answer!
- A new drug is released to market and it has been determined that the drug is responsible for causing hypertension among those individuals who take the drug. If 30,276 individuals were prescribed the drug in its first year on the market and 17,620 individuals developed hypertension, what is the point prevalence of hypertension among those individuals prescribed the drug, after the drug’s first year on the market? 0.00582 0.05820 0.58200 5.82000Are the characteristics female and pathologist independent? Prove mathematically. Pathology Pediatrics Psychiatry Totals male 12,575 33,020 27,803 73,398 female 5604 33,351 12,292 51,247 totals 18,179 66,371 40,095 124,645Age Social media user Non-social media user Total Younger (18−22)(18−22)left parenthesis, 18, minus, 22, right parenthesis 787878 444 828282 Middle (23−27)(23−27)left parenthesis, 23, minus, 27, right parenthesis 494949 212121 707070 Older (28(28left parenthesis, 28 and up))right parenthesis 212121 464646 676767 Total 148148148 717171 219219219 According to a survey of college students, the use of social media varies widely according to age. Based on the survey, what is the approximate probability that a non-social media user is an older student aged 282828 and up?
- 1.) This is an Econ - Econometrics question: Let arr86 be a binary variable equal to 1 if a man was arrested during 1986, and zerootherwise. You have a dataset of young men from California who were born in 1960 or1961 and have at least one arrest prior to 1986. You also have data on the proportion ofprior arrests that led to a conviction (pcnv), the average sentence served from priorconvictions in months (avgsen), months spent in prison since age 18 prior to 1986(tottime), months spent in prison in 1986 (ptime86), and the number of quarters (0 to 4)that the man was legally employed in 1986 (qemp86). You estimate the following linearprobability model (standard errors are in parentheses): arr86 = 0.441 - .162pcnv + .0061avgsen - .0023tottime(.017) (.021) (.0065) (.005) - .022ptime86 - .043qemp86(.005) (.005) n= 2725R2=.0474 a) Interpret the coefficient on avgsen, commenting on both magnitude and significance. Based on these results, do longer sentences deter future crime? b) What is the…21 Data Mining A frequent itemset X is closed if there exists no frequent super-itemset Y such that . True FalseA researcher is studying the hatching rates of two groups of Pacific tree frog eggs. One group of eggs was shaded from the sun’s ultraviolet light, whereas the second group was not shaded from the sun’s ultraviolet light. The researcher believes he can demonstrate that the hatching rate for the shaded group is higher than the hatching rate for the exposed group. Sun Shaded group Unshaded Group Total in Group 75 90 Total Hatched 47 41 Let p1 denote the hatching rate for the sun shaded group, and let p2 denote the hatching rate for the unshaded group. The researcher will test the hypotheses H0: p1-p2=0 vs Ha: p1 >p2 at a 5%level of significance. Find the p-value.
- Find the: lmd (lower boundary of the median class) n/2 (half of the total frequency) <cfb (cumulative frequency before the median class) fmd (frequency of the median class) c (class size) lmo (lower boundary of the modal class) fmo (frequency of the modal class) f1 (frequency before the median class) f2 (frequency after the median class)2.62 For the period 2001–2008, the Bristol-Myers Squibb Company, Inc. reported the following amounts (in billions of dollars) for (1) net sales and (2) advertising and product promotion. The data are also in the file XR02062. Source: Bristol-Myers Squibb Company, Annual Reports, 2005, 2008. Year Net Sales Advertising/Promotion 2001 $16.612 $1.201 2002 16.208 1.143 2003 18.653 1.416 2004 19.380 1.411 2005 19.207 1.476 2006 16.208 1.304 2007 18.193 1.415 2008 20.597 1.550 For these data, construct a line graph that shows both net sales and expenditures for advertising/product promotion over time. Some would suggest that increases in advertising should be accompanied by increases in sales. Does your line graph support this?Moviegoers The average “moviegoer” sees 8.5 moviesa year. A moviegoer is defined as a person who sees atleast one movie in a theater in a 12-month period.A random sample of 40 moviegoers from a largeuniversity revealed that the average number of moviesseen per person was 9.6. The population standarddeviation is 3.2 movies. At the 0.05 level ofsignificance, can it be concluded that this represents adifference from the national average?