Question 574:

Asked on January 14, 2009

Tags: 1-sample t-test , 1-sample Z-test

1

Answer:

No answer provided yet.

The sample size is pretty large, so the sample standard deviation will be a good estimate of the unknown population standard deviation, suggesting we could use the 1-sample Z test or the 1-sample t-test. The results will be very similar and I prefer to be on the slightly more conservative side and suggest we use the 1-sample t-test then see the results for the z-test as well.

 

We want to see if the sample mean of 218 is sufficient evidence for us to reject the null hypothesis, which is the mean is NOT equal to 200 (either greater than or less than 200).

1. The test statistic is derived by subtracting the test mean from the observed mean (218-200) =18 and dividing this result by the standard error of the mean.
2. The standard error of the mean is the standard deviation divided by the square root of the sample size. So the SEM is 30/SQRT(64) 30/8  = 40
3. So the t-statistic is 18/3.75 = 4.8
4. Now we just look up 4.8 using a t-table and we're looking for 2-sided area. We just need the degrees of freedom, which are n-1 = 64-1 = 63. If you don't like fumbling through a t-table, we can just use the percentiles from the t-distribution calculator and enter 4.8 and 63 for degrees of freedom. We get the p-value of less than .0001
5. Since the p-value is well less than the .05 cut-off level we would reject the null hypothesis and say that the observed mean of 218 is significantly different than 200 (p < .05).
6. If we were to use the 1-sample Z-test, we'd enter 4.8 in the z-score to percentile calculator and also get less than .0001 leading us to the same conclusion as the t-test.

For the second question the sample size is pretty large, so the sample standard deviation will be a good estimate of the unknown population standard deviation, suggesting we could use the 1-sample Z test or the 1-sample t-test. The results will be very similar and I prefer to be on the slightly more conservative side and suggest we use the 1-sample t-test then see the results for the z-test as well.

 

We want to see if the sample mean of 2300 is sufficient evidence for us to reject the null hypothesis, which is the mean is NOT equal to 2450 (either greater than or less than 2450).

1. The test statistic is derived by subtracting the test mean from the observed mean (2300-2450) =-150 and dividing this result by the standard error of the mean.
2. The standard error of the mean is the standard deviation divided by the square root of the sample size. So the SEM is 400/SQRT(100) 400/10  = 40
3. So the t-statistic is -150/4 = -3.75
4. Now we just look up -3.75 using a t-table and we're looking for 2-sided area. We just need the degrees of freedom, which are n-1 = 100-1 = 99. If you don't like fumbling through a t-table, we can just use the  percentiles from the t-distribution calculator and enter 3.75 (drop the negative sign) and 99 for degrees of freedom. We get the p-value of .003.
5. Since the p-value of .003 is less than the .01 cut-off level we would reject the null hypothesis and say that the observed mean of 2300is significantly different than 2450 (p < .01).
6. If we were to use the 1-sample Z-test, we'd enter 3.75 in the z-score to percentile calculator and also get less than .0017 or about .002 leading us to the same conclusion as the t-test.

For both questions you can use conduct the 1-sample t test using the online calculator at usablestats.com  and entering in the mean, standard deviation and sample size against the test mean.