Hugh B. answered • 06/12/15
Tutor
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(36)
Statistical applications in Stata
Hi Mayra,
To answer this question, take steps as follow:
1. Compute the t statistic for the null hypothesis that the population mean is equal to 700 versus alternative hypothesis that the population mean is not equal to 700. The t statistic is given by
(sample mean -700) / [standard deviation /(square root of sample size)].
Given the numbers we have here, this is
(691 – 700) / [81/square root of 72].
I used Excel to compute this, and you can check my math, but I get
(-9)/(81/8.485281) = -0.94281. So the value of the t statistic is
t = -0.94281.
2. Make the t statistic that was computed in #1 the critical value (that is, the cutoff value)for deciding whether to accept or reject the null hypothesis. That is, treat it as though you will reject the null hypothesis if either t is greater than or equal to 0.94281 or t is less than or equal to -0.94281.
3. For this critical value of t, find the significance level of the hypothesis test. If you are using a text book with a t table in the appendices, you could use the t table in the appendices, but how you use the table depends on how it is set up, which varies across different textbooks. For that reason, it is easier for me to tell you what to do in Excel. In Excel, use the T.DIST.2T function – just paste the following into the formula bar "=T.DIST.2T(0.94281, 71)". The number that comes back, which is 0.348975, is the total probability of obtaining a value of the t statistic that is either greater than or equal to 0.94281 or less than -0.94281 IF THE NULL HYPOTHESIS IS TRUE. Therefore you can reject the null hypothesis at any significance level that is numerically larger than 0.348975 and you would accept the null hypothesis at any significance level that is numerically smaller than 0.348975. For example, for part (a) reject null hypothesis at significance level = 0.40 or significance level = 0.50, and for part (b) do not reject null hypothesis at significance level = 0.30 or significance level =0.25. Clearly you would also not reject the null hypothesis at more conventional significance levels like 0.05 and 0.01 and you should conclude that there is no evidence in this sample that the population mean weight is different than 700 pounds.
Hope this helps. Feel free to ask questions if not.
Kind regards,
Hugh
To answer this question, take steps as follow:
1. Compute the t statistic for the null hypothesis that the population mean is equal to 700 versus alternative hypothesis that the population mean is not equal to 700. The t statistic is given by
(sample mean -700) / [standard deviation /(square root of sample size)].
Given the numbers we have here, this is
(691 – 700) / [81/square root of 72].
I used Excel to compute this, and you can check my math, but I get
(-9)/(81/8.485281) = -0.94281. So the value of the t statistic is
t = -0.94281.
2. Make the t statistic that was computed in #1 the critical value (that is, the cutoff value)for deciding whether to accept or reject the null hypothesis. That is, treat it as though you will reject the null hypothesis if either t is greater than or equal to 0.94281 or t is less than or equal to -0.94281.
3. For this critical value of t, find the significance level of the hypothesis test. If you are using a text book with a t table in the appendices, you could use the t table in the appendices, but how you use the table depends on how it is set up, which varies across different textbooks. For that reason, it is easier for me to tell you what to do in Excel. In Excel, use the T.DIST.2T function – just paste the following into the formula bar "=T.DIST.2T(0.94281, 71)". The number that comes back, which is 0.348975, is the total probability of obtaining a value of the t statistic that is either greater than or equal to 0.94281 or less than -0.94281 IF THE NULL HYPOTHESIS IS TRUE. Therefore you can reject the null hypothesis at any significance level that is numerically larger than 0.348975 and you would accept the null hypothesis at any significance level that is numerically smaller than 0.348975. For example, for part (a) reject null hypothesis at significance level = 0.40 or significance level = 0.50, and for part (b) do not reject null hypothesis at significance level = 0.30 or significance level =0.25. Clearly you would also not reject the null hypothesis at more conventional significance levels like 0.05 and 0.01 and you should conclude that there is no evidence in this sample that the population mean weight is different than 700 pounds.
Hope this helps. Feel free to ask questions if not.
Kind regards,
Hugh
Hugh B.
Hi Mayra,
I am guessing that you are asking about the 71 in the T.DIST.2T function. Here is the explanation: Just like the normal distribution is really a family of bell-shaped distributions and we have to choose a specific mean and a specific variance to pick out a single normal distribution from this family of distributions, the Student's t distribution is also a family of bell-shaped distributions. All of the Student's t distributions have a mean of zero, and to pick out a single Student's t distribution, we need to choose a parameter that is called the "degrees of freedom" of the distribution. Degrees of freedom can be any number greater than 0, and the larger the degrees of freedom, the more closely the Student's t distribution resembles the standard normal distribution. The degrees of freedom for the t distribution determines is the variance of the distribution, or equivalently, how "fat" the tails of the distribution are.
In this problem then, the 71 is the degrees of freedom that we need to pick out a specific Student's t distribution. Next question: Why 71 as opposed to some other number? In general, we will use a Student's t distribution whenever we are testing a hypothesis about a sample mean or sample means and we have to estimate a population variance (or population variances) from a sample variance. In this particular type of problem, where we are only estimating 1 population variance, the degrees of freedom will always be the sample size minus one, or 72 - 1 = 71.
A little more explanation: Let N be the sample size, and what we have said is that degrees of freedom for this problem is N-1. The reason for this is that the term "degrees of freedom" refers to the number of independent observations that are used in the calculation of the sample variance. Since the sample variance is computed by summing the squares of deviations from the sample mean and dividing that sum by N-1, there are only N-1
independent observations used. The reason that there are only N-1 independent observations is because the sample mean is involved in the calculation, the sample mean and N-1 of the observations can be used to determine what the N-th observation is (that is, the "last observation has to be the value that generates the value of the sample mean that we have computed along with the other N-1 observations). In other words, in the sum that is used to compute the sample variance, only N-1 of the terms are independent because the value of the last observation in the sample can be computed from the sample mean and the other N-1 observations. This is often stated as "we lose a degree of freedom when we calculate a sample variance."
Later on, when you look at comparing two means from two different populations, you will see problems where you have to estimate two population variances, and in those problems, the degrees of freedom will be N-2.
Hope this helps.
Kind regards,
Hugh
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06/13/15
Mayra J.
Yes awesome ! i figure it was the degree of freedom!! thanks
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06/13/15
Hugh B.
Glad to help out, I will answer some more later. Best, Hugh
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06/13/15
Mayra J.
06/13/15