Probability

The assumption for a single sample t test can be listed as follows:

1. The sample used should be random.
2. The random sample should be taken from independent observations.
3. The population distribution must be nearly normal or the size of sample should be large.

The purpose of this test is to state wherever there is significant difference between the hypothesized mean and the observed sample mean. If the difference is large the hypothesized population mean is rejected. If the difference is small we fail to reject the hypothesized population mean.

The inputs to the test can be summarized as follows:

• Sample population= 1,2,3,4,5,6
• The null Hypotheses (Ho)=0
• The alternative Hypotheses (Ha)≠0
• The level of Significance =0.05
• Sample size (n) = 6

From this inputs the console also calculates the sample mean(x̅ )= 3.5, the standard deviation (σ)

which is 1.870828693387. It can aslo be determined that the critical values are tcrit= ± 2.571 when relating to the T table for a 0.05 level of Significance and a degree of freedom (n-1) of 5.

T values, t = 4.5826, is a standardized value which represents the difference between the samplle mean and the hypothesized mean. It is calculated by the difference between the sample mean and the hypothesized mean divided by the standard error (standard error =standard deviation divide by square root of sample size).  As shown below.

[pic 1]

Hence [pic 2]

The next element of the output df is the degree of freedom. It represents the number of values in the calculation that can vary without affecting the sample mean. It is calculated using 1-n which in this case is 5.

Then we have the p value which represents the probability that the null hypothesis is true. The large the p-value the higher the probability that the hypothesized sample mean (0 in this example) is true.

The alternative Hypotheses represent the outcome if the null hypothesis is rejected. In this case the alternative hypothesis is Ha ≠0. margin of error

The confidence interval represent the range of values that ma contain the true mean value of the sample. The confidence interval is calculated using the formula x̅ ± margin of error (where margin of error=   tcrit * Standard Error).

In this example tcrit= ± 2.571 and  the margin of error is approximately 1.96363. Hence giving a confidence interval ( x̅ ± margin of error) of approximately (1.536686,5.463314 ).[pic 3]

The sample mean is the mean calculated using the given sample.

 From the result obtain we can conclude that the null hypothesis (Ho) was rejected. That is true mean of the sample is not 0.

The p value aslo tell use that the probability that the mean is 0 is about 0.005934  which is very same and hence intensify the conclusion that the null hypothesis should be rejected

In addition from the confidence interval we can be sure at 95% that the true mean is between 1.536686 and 5.463314

No I do not think that the data is meaningful to this test since the result just tell us that the mean is not 0. The test would be more relevant if an actual hypothesis was tested. For example, someone claim that the average weight of first year student is 50kg. If we take a random sample of weight of first year student from a population, we could calculated the a sample mean and compare it to the hypothesis using the t test to find out if wherever this hypothesis true.

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