H0:H0:  ? μ p  Select an answer = > ≠ <         H1:H1:  ? μ p  Select an answer ≠ < = >     The test statistic ? t z  =  (please show your answer to 3 decimal places.) The p-value =  (Please show your answer to 4 decimal places.) The p-value is ? > ≤  αα Based on this, we should Select an answer accept fail to reject reject  the null hypothesis. Thus, the final conclusion is that ... The data suggest the population mean is not significantly less than 10 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of paintings that are in teachers' houses is equal to 10. The data suggest the populaton mean is significantly less than 10 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of paintings that are in teachers' houses is less than 10. The data suggest that the population mean number of paintings that are in teachers' houses is not significantly less than 10 at αα = 0.05, so there is insufficient evidence to conclude that the population mean number of paintings that are in teachers' houses is less than 10. Interpret the p-value in the context of the study. There is a 1.34% chance that the population mean number of paintings that are in teachers' houses is less than 10. If the population mean number of paintings that are in teachers' houses is 10 and if you survey another 13 teachers, then there would be a 1.34% chance that the sample mean for these 13 teachers would be less than 9.08.  There is a 1.34% chance of a Type I error. If the population mean number of paintings that are in teachers' houses is 10 and if you survey another 13 teachers, then there would be a 1.34% chance that the population mean number of paintings that are in teachers' houses would be less than 10. Interpret the level of significance in the context of the study. If the population mean number of paintings that are in teachers' houses is less than 10 and if you survey another 13 teachers, then there would be a 5% chance that we would end up falsely concuding that the population mean number of paintings that are in teachers' houses is equal to 10. If the population mean number of paintings that are in teachers' houses is 10 and if you survey another 13 teachers, then there would be a 5% chance that we would end up falsely concuding that the population mean number of paintings that are in teachers' houses is less than 10. There is a 5% chance that teachers are so poor that they are all homeless. There is a 5% chance that the population mean number of paintings that are in teachers' houses is less than 10.

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Author:Amos Gilat
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Chapter1: Starting With Matlab
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The average house has 10 paintings on its walls.  Is the mean smaller for houses owned by teachers? The data show the results of a survey of 13 teachers who were asked how many paintings they have in their houses. Assume that the distribution of the population is normal.

8, 10, 11, 9, 9, 9, 9, 7, 10, 7, 11, 8, 10

What can be concluded at the  αα = 0.05 level of significance?

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

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

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

  1. The test statistic ? t z  =  (please show your answer to 3 decimal places.)
  2. The p-value =  (Please show your answer to 4 decimal places.)
  3. The p-value is ? > ≤  αα
  4. Based on this, we should Select an answer accept fail to reject reject  the null hypothesis.
  5. Thus, the final conclusion is that ...
    • The data suggest the population mean is not significantly less than 10 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of paintings that are in teachers' houses is equal to 10.
    • The data suggest the populaton mean is significantly less than 10 at αα = 0.05, so there is sufficient evidence to conclude that the population mean number of paintings that are in teachers' houses is less than 10.
    • The data suggest that the population mean number of paintings that are in teachers' houses is not significantly less than 10 at αα = 0.05, so there is insufficient evidence to conclude that the population mean number of paintings that are in teachers' houses is less than 10.
  6. Interpret the p-value in the context of the study.
    • There is a 1.34% chance that the population mean number of paintings that are in teachers' houses is less than 10.
    • If the population mean number of paintings that are in teachers' houses is 10 and if you survey another 13 teachers, then there would be a 1.34% chance that the sample mean for these 13 teachers would be less than 9.08.
    •  There is a 1.34% chance of a Type I error.
    • If the population mean number of paintings that are in teachers' houses is 10 and if you survey another 13 teachers, then there would be a 1.34% chance that the population mean number of paintings that are in teachers' houses would be less than 10.
  7. Interpret the level of significance in the context of the study.
    • If the population mean number of paintings that are in teachers' houses is less than 10 and if you survey another 13 teachers, then there would be a 5% chance that we would end up falsely concuding that the population mean number of paintings that are in teachers' houses is equal to 10.
    • If the population mean number of paintings that are in teachers' houses is 10 and if you survey another 13 teachers, then there would be a 5% chance that we would end up falsely concuding that the population mean number of paintings that are in teachers' houses is less than 10.
    • There is a 5% chance that teachers are so poor that they are all homeless.
    • There is a 5% chance that the population mean number of paintings that are in teachers' houses is less than 10.

 

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Expert Solution
Given

Working:

x (x-xbar)^2
8 1.173611
10 0.840278
11 3.673611
9 0.006944
9 0.006944
9 0.006944
7 4.340278
10 0.840278
7 4.340278
11 3.673611
8 1.173611
10 0.840278

 

The mean is given by

\overline{x} = \frac{1}{n}\sum x

\overline{x} =9.0833

 

The standard deviation is given by

\sigma =\sqrt{\frac{1}{n-1}\sum (x - \overline{x})^2}

\sigma =1.37895

 



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