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Question 540:



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(a)    This is a one-tailed test. We can tell because the Null Hypothesis Ho, is pointing in only one direction (greater than or equal to 20). If it were two tailed, we'd see the not equal to sign.
(b)   The decision rule would be we'd reject the null hypothesis if the p-value is less than .05, meaning there is sufficient evidence the mean is LESS than 20. Another way to state this is that the decision rule is to reject the null hypothesis if the test statistics is less than the t-critical value with 35 degrees of freedom. Using the excel formula =TINV(.05,35) gets us 2.03.
(c)    We'd use the t-statistic as our test statistic. This would be a 1-sample t-test and we're testing it against the mean of 20. We want to know if there is enough evidence above chance levels that the observed mean of 21, from 36 observations is greater than 20.  To compute the test statistics we use these steps:
1. Subtract the observed mean from the test mean 21-20 = 1 and this becomes the numerator for our test statistic.
2. Find the standard error of the sample mean (SEM) which is the standard deviation divided by the square root of the sample size. 5/SQRT(36) = 5/6 = .833. This is the denominator for the test statistic.
3. Dividing the fraction we get 1/.833 = 1.20, which is our test t-statistic.
4. Now we need to find the 1-sided probability associated with this statistic. To find that we can use the excel function =TDIST(1.20,35,1) and we get the p-value .1191. Where the parameters are the test statistic, the degrees of freedom (36-1) and 1 tail probability. We can also use the calculator available here:
(d)   Our decision is NOT to reject the null hypothesis since this is above .05
(e)    P-value is .1191
(f)     There is not enough evidence to conclude the observed mean of 21 is comes from a population with mean less than 20, given the sample standard deviation of 5 from 36 samples.

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