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# Hypothesis testing using ANOVA, Chi Square

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10. The manager of a computer software company wishes to study the number of hours top executives spend at their computer terminals by type of industry. The manager selected a sample of five executives from each of three industries. At the .05 significance level, can she conclude there is a difference in the mean number of hours spent at a terminal per week by industry?

Banking Retail Insurance
12 8 10
10 8 8
10 6 6
12 8 8
10 10 10

54. The chief of security for the Mall of the Dakotas was directed to study the problem of missing goods. He selected a sample of 100 boxes that had been tampered with and ascertained that for 60 of the boxes, the missing pants, shoes, and so on were attributed to shoplifting. For 30 other boxes employees has stolen goods, and for the remaining 10 boxes he blamed poor inventory control.

In his report to the mall management, can he say that shoplifting is twice as likely to be the cause of the loss as compared with either employee theft or poor inventory control and that the employee theft and poor inventory control are equally likely? Use the .02 level.

65. In a particular market there are three commercial television stations, each with its own evening news program from 6:00 to 6:30 PM. According to a report in this morning's local newspaper, a random sample of 150 viewers last night revealed 53 watched the news on WNAE (channel 5), 64 watched on WRRN (channel 11), and 33 on WSPD (channel 13). At a .05 significance level, is there a difference in the proportion of viewers watching the three channels?

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10. The manager of a computer software company wishes to study the number of hours top executives spend at their computer terminals by type of industry. The manager selected a sample of five executives from each of three industries. At the .05 significance level, can she conclude there is a difference in the mean number of hours spent at a terminal per week by industry?

Banking Retail Insurance
12 8 10
10 8 8
10 6 6
12 8 8
10 10 10

Solution:
We will use ANOVA (Analysis of Variance ) to solve the problem

Null Hypothesis: H0: &#956;1=&#956;2=&#956;3 There is no difference between the means
Alternative Hypothesis: H1: &#956;1&#8800;&#956;2&#8800;&#956;3 There is difference between the means

Banking Retail Insurance
12 8 10
10 8 8
10 6 6
12 8 8
10 10 10

Mean=&#956; 10.8 8 8.4
Standard deviation=sj 1.1 1.41 1.67
Sample size=nj= 5 5 5

k=no of samples= 3
Grand mean of all the three samples= 9.07 =(5 * 10.8 + 5 * 8 + 5 * 8.4 ) / (5+5+5)

First estimate of the population variance=Varaince among the sample means
df=k-1= 2 df=degrees of freedom

Varaince among the sample means=&#931;nj(xsample-xgrand)2/k-1 =
=(5 *( 10.8 - 9.07) ^2 + 5 *( 8 - 9.07) ^2 + 5 *( 8.4 - 9.07) ^2 )/ 2= 11.47

Second estimate of the population variance=Within column variance
df=(&#931;nj)- k=nT - k= 12 = 5 + 5 + 5 - 3 df=degrees of ...

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