1. The teacher in your class has assured you that a particular test can be compl
ID: 3320361 • Letter: 1
Question
1. The teacher in your class has assured you that a particular test can be completed in a mean time of 40 minutes but you are not convinced. You actually think the length of the test is different from 40 minutes, and you hypothesize that it should take a mean time of 45 minutes actually. You decide to get data from your friends and do a proper statistical test. (a) Write down the appropriate hypotheses for the test we want to perform using appropriate symbols that are well-defined. Note that at this stage you are only hypothesizing that the test time is different. (b) Write down the four factors that affect the statistical power of a test. (c) Suppose you ask 10 friends and you decide to conduct a test at the -0.05 significance level. Also assume that the population standard deviation is known to be 10 minutes. Compute the effect size that we are interested in and the statistical power of the test. (d) You have been told that doing the most appropriate test will give you higher statistical power. This means that instead of just testing for difference, if you tested that the actual population mean is greater than the null hypothesis value, you should get higher statistical power. Compute the power making this change (e) If you want higher statistical power, you can change any of the four factors in part b. Suppose you do the sensible thing of increasing sample size and ask 40 friends instead of just 10, and keep the two-sided test of part, compute the statistical power (f) You can increase this further by changing the hypothesis to greater than 40, while increasing sample size to 40. Compute the statistical power in this case.Explanation / Answer
a)Here you thinnk that the lenght of the test is different from 40 mins and so we have hypothised the same for 45 mins and so the hypothesis here would be written as
Ho: Mean test time is 45 mis. i.e. = 45 mins
Ha: Mean test time is not equal to 45 mins i.e. =/= 45mins
b) four factors that affect the statistical power of the test
c)now significance level of 5% is given with sd of 10 and n=10 number of friends
so looking at the z value from the teble for the 2 tail test we get z values as +/-1.96 &
= 45 mins is what we are testing for
So., solving the equation
Z= (X'-)/(sd / sqrt(n) we get
X' =(LCL) = -1.96 * 10/sqrt(10) + 45 =38.80194
X'=UCL = 1.96 * 10/sqrt(10) + 45 =51.19806
and so the range falls between 38.80194 & 51.19806 and so with 95% conf level or 5% significance level we can say that the null hypothesis that = 45 mins is correct
d) now in this case we are concerned for the actual population mean is greater than the null hypothesis value i.e. the test under observation and so
Ho: = 40 mins
H1: > 40 mins and now you notice that it has become a 1 tail test insted of the two tail in the previous case and so now the checking the same
Z= (X'-)/(sd / sqrt(n) we get
Z value as 1.64 for the 5% region towards right
1.64=(X'-40)/(10/sqrt(10))
we get X'=45.18614
and so we can say that if the mean lines between 45.18614 to infinite then only we can reject the null hypothesis while here the sample mean is 45 and so we can say that we cannot reject the null hypothesis and so here we say that the mean of the distribution is 40
rest of the questions can be answered just by changing n from 10 to 40 the procedure is one and the same.
As per the Chegg policy I have provided the answer of the first 4 sub questions in depth. I would love to answer the rest of the question as well but the time and policy are not in favour. Hope you understand. If you want the answer for the subsequent questions as well pls post it seperately. Hope you understand rest you can do them by yourself as i have provided the changes how to solve the rest now.
Hope the above explaination has helped you in understanding the problem Pls upvote the ans if it has really helped you. Good Luck!!
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