Question

What are the conventional levels of significance set by most researchers? What do these values mean...

What are the conventional levels of significance set by most researchers? What do these values mean (i.e., p values of .05 and .01)?
Math   Statistics-And-Probability   
0 0
Add a commentTranscribed image text
Next >

Homework Answers

Answer #1

A significance level of '0.05' is conventionally set by most researchers, although probabilities as high as '0.10' as well as .01 as lower probabilities may also be used. Probabilities greater than '0.10' are rarely used.

A significance level of '0.05' indicates that the result is extreme enough as to have a 95% probability of appearing if the null hypothesis were true. We can also say that the  significance level is the probability of rejecting the null hypothesis when it is true. A significance level of 0.05 indicates that we have 5% chance of error of concluding that a difference exists when there is no actual difference.

Please revert in case of any doubt.

Please upvote. Thanks in advance

0 0
Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
what do significance levels and P values mean in hypothesis tests? what is statistical significance anyway?
what do significance levels and P values mean in hypothesis tests? what is statistical significance anyway?
An alpha (significance) level is set at .05 for a research study, What does this mean?...
An alpha (significance) level is set at .05 for a research study, What does this mean? A. A researcher is willing to reject the null hypothesis if the probability of the null hypothesis being true is lower than .05. B. A researcher is willing to reject the null hypothesis if the probability of the alternative hypothesis being true is lower than .05. C. A researcher is willing to reject the null hypothesis if the probability of the null hypothesis being...
Recall that a conflict of interest scenario was presented to a sample of 200 marketing researchers...
Recall that a conflict of interest scenario was presented to a sample of 200 marketing researchers and that 109 of these researchers disapproved of the actions taken. (a) Let p be the proportion of all marketing researchers who disapprove of the actions taken in the conflict of interest scenario. Set up the null and alternative hypotheses needed to attempt to provide evidence supporting the claim that a majority (more than 50 percent) of all marketing researchers disapprove of the actions...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three replications resulted in the following data: SST=286, SSA=24, SSB=22, SSAB=185.. Set up the ANOVA table and test for significance using a=.05. Show entries to 2 decimals, if necessary. If the answer is zero enter “0”. Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Factor A Factor B Interaction Error Total The -value for Factor A is...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three replications resulted in the following data: ,STT = 261, SSA=21, SSB=22, SSAB=165 Set up the ANOVA table and test for significance using a=.05 . Show entries to 2 decimals, if necessary. If the answer is zero enter “0”. Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Factor A Factor B Interaction Error Total The -value for...
Pairs of P-values and significance levels, α, are given. For each pair, state whether the observed...
Pairs of P-values and significance levels, α, are given. For each pair, state whether the observed P-value would lead to rejection of H0 at the given significance level. (a)    P-value = 0.062, α = 0.05 reject H0 do not reject H0     (b)    P-value = 0.002, α = 0.001 reject H0 do not reject H0     (c)    P-value = 0.492, α = 0.05 reject H0 do not reject H0     (d)    P-value = 0.062, α = 0.10 reject H0 do not reject H0     (e)    P-value = 0.041, α...
In a two-tailed hypothesis test of the mean using a 0.05 level of significance, researchers calculated...
In a two-tailed hypothesis test of the mean using a 0.05 level of significance, researchers calculated a p-value of 0.03. What conclusion can be drawn? The alternative hypothesis should be rejected because the p-value is so small. The null hypothesis is true because the p-value is less than the level of significance. The alternative hypothesis is 3% likely to be true. The null hypothesis should be rejected because the p-value is less than the level of significance. 1.The alternative hypothesis...
In an experiment designed to test the output levels of three different treatments, the following results...
In an experiment designed to test the output levels of three different treatments, the following results were obtained: SST= 420 SSTR=150 NT=19. Set up the ANOVA table and test for any significant difference between the mean output levels of the three treatments. Use A=.05. Source of Variation Sum of Squares Degrees of Freedom Mean Square (to 2 decimals) (to 2 decimals) -value (to 4 decimals) Treatments 150 Error Total 420 The P-value is - Select your answer -less than .01between...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three replications resulted in the following data: SST= 248, SSA= 22, SSB= 21, SSAB= 155. Set up the ANOVA table and test for significance using alpha= .05 . Show entries to 2 decimals, if necessary. If the answer is zero enter “0”. Source of Variation Sum of Squares Degrees of Freedom Mean Square F p-value Factor A Factor B Interaction Error Total
The calculations for a factorial experiment involving four levels of factor A, three levels of factor...
The calculations for a factorial experiment involving four levels of factor A, three levels of factor B, and three replications resulted in the following data: SST = 291, SSA = 20, SSB = 24, SSAB = 195. Set up the ANOVA table and test for significance using  = .05. Show entries to 2 decimals, if necessary. Round p-value to four decimal places. If your answer is zero enter "0". Source of Variation Sum of Squares Degrees of Freedom Mean Square F...
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT