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Homework answers / question archive / In a particular market there are three commercial television stations, each with its own evening news program from 6:00 to 6:30 P
In a particular market there are three commercial television stations, each with its own evening news program from 6:00 to 6:30 P.M. 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 the .05 significance level, is there a difference in the proportion of viewers watching the three channels?
Question (2)
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 had stolen the 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 employee theft and poor inventory control are equally likely? Use the .02 significance level.
Question (3)
The manager of a computer software company wishes to study the number of hours senior executives spend at their desktop computers 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 per week by industry(For further information on the question please refer the attached question file)
Question (4):
Given the following sample information, test the hypothesis that the treatment means are equal at the .05 significance level( For further information on the question, please refer the attached question file)
1. State the null hypothesis and the alternate hypothesis.
2. What is the decision rule?
3. Compute SST, SSE, and SS total.
4. State your decision regarding the null hypothesis.
5. If H0 is rejected, can we conclude that treatment 1 and treatment 2 differ? Use the 95 percent level of confidence.
6. Complete an ANOVA table.
Chi-Square Distribution and Hypothesis Test.
For step by ste detailed solutions, please download the attached solution file. Following are just a few steps from the attachment.
Sollution (1)
Null Hypothesis H0 : there is no difference in the proportion of viewers watching
the channels.
Alternate Hypothesis H1 : viewers watching the three channels are in different
proportion.
Level of significance =0.05
Hence, we conclude that distribution of viewers across the three categories is not proportionate.
That is, viewers watching the three channels are in different proportion.
( For step by step detailed solution, please refer the attached solution file)
Solution (2)
This result is confirmed from the fact that the probability of getting a chi-square value of 12 is less than 0.005. Hence we can not support the cause of the loss to be in the proportion of 2:1:1.
That is, cause of the loss is not in the proportion of 2:1:1.
( For complete step by step solution, please see the attached solution file)
Solution (3)
SST = 46.93 , SSA = 22.93 , SSW=24 , MSA=11.465 , MSW=2 ,
Test Statistic ( F) = 5.73
Since F* = 5.73 > 3.89 , we reject the Null Hypothesis H0, and accept the Alternate Hypothesis H1
Since calculated F* > table value of F (3.89) , hence we conclude that according the sample information we have, there is significant difference in the mean number of hours spent per week by each industry.
(For complete step by step solution, please see the attached solution file)
Solution (4)
Decision Rule : Reject H0 if calculated Test statistic (F*) > ( F0.05 , 2 , 9 = 4.26 )
Step - III
Now we compute SST ( total sum of the squares) , SSA ( Sum of the squares among columns ) and SSW ( sum of the squares within the columns), MSA, MSW and test statistic.
SST=116.67 , SSA=107.2 , SSW=9.47 , MSA=53.6 , MSW=1.05
Hence treatment 1 and treatment 2 differ.( For complete step by step solution, please see the attached solution file)
