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True [87]
3 years ago
6

You are testing a claim and incorrectly use the normal sampling distribution instead of the​ t-sampling distribution. Does this

make it more or less likely to reject the null​ hypothesis? Is this result the same no matter whether the test is​ left-tailed, right-tailed, or​ two-tailed? Explain your reasoning.
Mathematics
2 answers:
Keith_Richards [23]3 years ago
8 0

Answer:

From the given data, the typical appropriation is utilized rather than t distribution for testing. The thicknesses of the curve similarly influence both normal distribution and territory of t distribution. The normal distribution test and t test give a similar dismissal for the invalid theory for any tail of the test. Besides, the tests normal sampling distribution and t sampling distribution give a similar dismissal to null hypothesis.  

Along these lines, the outcomes got by utilizing normal distribution and t distribution are distinguished as same and for each situation, the tail thickness doesn't influence the event of the basic qualities.

egoroff_w [7]3 years ago
6 0

Answer:

Check the explanation

Step-by-step explanation:

The vital t-values are most of the time more extreme than the resultant critical z-values, hence it is expected to be less likely to refuse the null hypothesis (given that the value of the test statistic are still unchanged).

This outcome will remain unchanged no matter whether the test is right-, left-, or two-tailed, since the negative critical t-values will be lesser when compared to the negative critical z-values and the positive critical t-values will be higher than the positive critical z-values.

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Answer:

H0 is rejected.There difference in the means for the three assembly methods.

Step-by-step explanation:

Calculating gives the following results.

ANOVA Table

Source        DF                  Sum                Mean            F Statistic      

                                       of Square         Square

Treatments       2               4560               2280             9.865

(between groups)

 

<u>Error            27                  6240                231.111                                   </u>

Total         29                   10,800  

Error SS= SST -SSTR = -10,800-4560=6240    

d.f for SSTr = r-1= 3-1

D.f for SST= n-r= 30-3= 27

Total D.f= d.f for SSTr+D.f for SST= 27+29

MSTR= SSTR/d.f= 4560/2= 2280

MSError= Error SS/ d.f= 6240/27=231.111

F test= MSTR/ MsError= 2280/231.11= 9.865

The P- value < 0.01 which  is less than 0.05 therefore H0 is rejected.

There is sufficient evidence to reject H0.

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3 years ago
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Fred, Norman, and Dave own a total of 128 comic books. If Dave owns 44 of them, what is the average
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Answer:

Mean = 42

Step-by-step explanation:

Given

Represent Fred with F; Norman with N and Dave with D.

Total = 128

D = 44

Required

Determine the arithmetic mean of F and N

First, we have that:

F + N + D = 128

Substitute 44 for D

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Subtract 44 from both sides

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Arithmetic mean of F and N is calculated as follows:

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Mean = \frac{84}{2}

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