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One-Sample T-Tests, Statistics Problem Example

Pages: 2

Words: 504

Statistics problem

The table below represents the numbers between 1 and 12 picked by my three friends from the neighborhood. In this scenario, the calculated sample mean serves as the average number of news programs of the television network watched by my neighborhood.

Number chosen by first friend Number chosen by second friend Number chosen by third friend Mean
3 10 5 (3+10+5)/3=9

Question 1

I think the sample mean would be equal to the national team. The sample from the neighborhood was represented in the national survey that the television network had conducted previously. Therefore, I expect the number of news programs watched by the neighborhood to be the same as that of the national population. This results in a null hypothesis that the sample mean would be equal to the population mean, represented mathematically as.

Question 2

Based on the null hypothesis, I will use a two-tailed t-test in the statistical analysis. This analysis will be the most appropriate one because my interest will be to identify whether the sample mean is greater or lesser than the population mean. This will assist in determining whether my neighborhood watches more or fewer number of news programs than the eight programs viewed nationally.

Question 3

If possible, I could have sampled 40 friends from my neighborhood. This would have made the sample size have higher statistical power than the sample of the sample of the three friends. This would make it possible for the conclusions made from this sample to be representative of the neighborhood population. Therefore, in my analysis, I will consider 40 respondents as my ideal sample size, N.

Question 4

If N=40, then the degrees of freedom (df) will be 39 (N-1). From the statistics table, the critical value of t at df of 39 (for the two-tailed test and a probability or alpha level of 0.05) is 2.0227.

Question 5

If my calculated or obtained value of t would be larger than my critical value of t, then I would reject the null hypothesis. This would show that the mean of neighborhood sample is significantly different from the mean of the national survey given the null hypothesis. Therefore, the neighborhood watches different number of programs than those viewed by the national population of the television network. In this case, the difference between the two values would have two interpretations. First, if obtained t value would be negative and greater than critical t value, then it would imply that the neighborhood watches a fewer news programs than the average programs watched nationally. Secondly, if the calculated value would be greater than tabled value of t and positive, then the neighborhood watches more programs than the national population.

If my obtained value would be smaller than the tabled t value, then I would accept the null hypothesis. It would indicate that the sample mean is not significantly (df = 39, p=0.05) different from the mean of the population obtained in the national survey. Therefore, there would be a high likelihood the number of news shows watched by the neighborhood is similar to that viewed nationally.

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