What test statistic will you use if the sample size is less than 30?
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Statistical Analysis The idea behind parametric tests is to provide the researcher with a statistical inference about the population by conducting statistically significant tests (like t-test) on the sample drawn from the population. The parametric test called t-test is based on a student’s t statistic. This statistic assumes that variables are drawn from the normal population. The mean of the population in this statistic of t-test has been assumed to be known. The distribution, called t-distribution, has a similar shape to that of a normal distribution, i.e. a bell shaped appearance. Statistics Solutions is the country’s leader in statistical consulting and t-test analysis. Use the calendar below to schedule a free 30-minute consultation.  Discover How We Assist to Edit Your Dissertation ChaptersAligning theoretical framework, gathering articles, synthesizing gaps, articulating a clear methodology and data plan, and writing about the theoretical and practical implications of your research are part of our comprehensive dissertation editing services.
The parametric test called t-test is useful for testing those samples whose size is less than 30. The reason behind this is that if the size of the sample is more than 30, then the distribution of the t-test and the normal distribution will not be distinguishable. The parametric test is used for conducting statistically significant tests in the testing of hypotheses. There are basically three types of t-tests: one sample t-test, two independent sample t-test and paired t-test. In the case of a one sample t-test, if a researcher in the field of psychology is working on a study where he wants to make sure that at least 65% of students will pass the IQ test, he can use the t-test. So, one sample t-test will be used after the hypothesis has been formulated in this particular case. The parametric test is then calculated by selecting an appropriate formula of t-test. In this case, the appropriate formula will be a t-test for a single mean. A selection of the level of significance is conducted to check the t-test of the null hypothesis. Usually, the researcher takes 0.05 as the appropriate level of significance while conducting the t-test. The level of significance refers to the minimum probability that there will be a false rejection of the null hypothesis. Now, if the value calculated from the t-test is more than the tabulated value, then the null hypothesis gets rejected at a particular level of significance. Similarly, if the value calculated from the t-test is less than the tabulated value, then the null hypothesis gets accepted at a particular level of significance. In two independent sample t-tests, two samples that are not at all related to each other are tested. The main idea behind two independent sample t-tests is to draw out a statistical inference about the comparison of two independent samples of data. For example, in the field of psychology, if the researcher wants to compare the IQ level of students living in region A and region B, then a two independent sample t-test is useful. The region A and the region B are not at all related to each other, i.e. they are independent of each other. The procedure for conducting this t-test is the same, except that now the sample number is double instead of single. Also, in the case of t-test for single mean and two independent sample t-tests, there are different formulas for the degree of freedom. The degree of freedom is referred to as the restriction that a researcher puts forward while conducting parametric tests, like t-test in this case. The paired sample t-test refers to that type of sample in which the variables form paired categories. For example, if a researcher wants to compare male and female smokers, paired sample t-test comes into play if the variable is in following form: male chain smoker and female chain smoker, male occasional smoker and female occasional smoker, etc. Reference: Introduction to the theory of statistics: Mood A.M., Graybill F.A., Boes D.C 8.2.3.1 - One Sample Mean t Test, Formulas8.2.3.1 - One Sample Mean t Test, FormulasFive Step Hypothesis Testing Procedure1. Check assumptions and write hypotheses Data must be quantitative. In order to use the t distribution to approximate the sampling distribution either the sample size must be large (\(\ge\ 30\)) or the population must be known to be normally distributed. The possible combinations of null and alternative hypotheses are:
where \( \mu_{0} \) is the hypothesized population mean. 2. Calculate the test statistic For the test of one group mean we will be using a \(t\) test statistic: Test Statistic: One Group Mean \(t=\frac{\overline{x}-\mu_0}{\frac{s}{\sqrt{n}}}\) \(\overline{x}\) = sample mean Note that structure of this formula is similar to the general formula for a test statistic: \(\frac{sample\;statistic-null\;value}{standard\;error}\) 3. Determine the p-value When testing hypotheses about a mean or mean difference, a \(t\) distribution is used to find the \(p\)-value. These \(t\) distributions are indexed by a quantity called degrees of freedom, calculated as \(df = n – 1\) for the situation involving a test of one mean or test of mean difference. The \(p\)-value can be found using Minitab. 4. Make a decision If \(p \leq \alpha\) reject the null hypothesis. If \(p>\alpha\) fail to reject the null hypothesis. 5. State a "real world" conclusion Based on your decision in Step 4, write a conclusion in terms of the original research question. The new few pages will walk you through examples before giving you the opportunity to do two on your own. 8.2.3.1.1 - Video Example: Book Costs8.2.3.1.1 - Video Example: Book CostsResearch question: Does the average Penn State student spend more than \$300 each semester on textbooks? In a sample of 226 Penn State students, the mean cost of a student’s textbooks was \$344 with a standard deviation of \$106. 8.2.3.1.2 : Example: Pulse Rate8.2.3.1.2 : Example: Pulse RateA research study measured the pulse rates of 57 college men and found a mean pulse rate of 70.4211 beats per minute with a standard deviation of 9.9480 beats per minute. Researchers want to know if the mean pulse rate for all college men is different from the current standard of 72 beats per minute. 1. Check assumptions and write hypotheses Pulse rates are quantitative. The sampling distribution will be approximately normally distributed because \(n \ge 30\). This is a two-tailed test because we want to know if the mean pulse rate is different from 72. \(H_{0}:\mu=72
\) 2. Calculate the test statistic Test Statistic: One Group Mean \(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}\) \(\overline{x}\) = sample mean \(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{70.4211-72}{\dfrac{9.9480}{\sqrt{57}}}=-1.198\) Our \(t\) test statistic is -1.198 3. Determine the p-value \(df=n-1=57-1=56\) \(p=0.117981+0.117981=0.235962\) Given that the null hypothesis is true and \(\mu=72\), the probability of taking a random sample of \(n=57\) and finding a sample mean this or more extremely different is 0.235962. This is our p-value. 4. Make a decision \(p>.05\), therefore we fail to reject the null hypothesis. 5. State a "real world" conclusion There is not sufficient evidence to state that the mean pulse of college men is different from 72. 8.2.3.1.3 - Example: Coffee8.2.3.1.3 - Example: CoffeeIn the population of Americans who drink coffee, the average daily consumption is 3 cups per day. A university wants to know if their students tend to drink more coffee than the national average. They ask a random sample of 50 students how many cups of coffee they drink each day and found \(\overline{x}=3.8\) and \(s=1.5\). Do they have evidence that their students drink more than the national average? 1. Check assumption and write hypotheses Amount of coffee consumed is a quantitative variable. We are given that random sampling methods were employed. Because \(n \ge 30\), we can approximate the sampling distribution using a t distribution. This is a right-tailed test because we want to know if the mean in the sample is greater than the national average. \(H_{0}:\mu= 3\) 2. Calculate the test statistic Test Statistic: One Group Mean \(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}\) \(\overline{x}\) = sample mean \(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{3.8-3}{\dfrac{1.5}{\sqrt{50}}}=3.771\) Our \(t\) test statistic is 3.771 3. Determine the p-value \(df=n-1=50-1=49\) Using Minitab, we can find that \(P(t > 3.771) =0.0002191\) p-value = 0.0002191 If \(\mu=3\), then the probability of taking a random sample of \(n=50\) and finding \(\overline{x} \geq 3\) is 0.0002191 4. Make a decision \(p\leq.05\), therefore we reject the null hypothesis. 5. State a "real world" conclusion There is evidence to state the mean number of coffees consumed in the population of all students at this university is greater than 3. 8.2.3.1.4 - Example: Transportation Costs8.2.3.1.4 - Example: Transportation CostsAccording to CNN, in 2011, the average American spent \$16,803 on housing. A suburban community wants to know if their residents spent less than this national average. In a survey of 30 randomly selected residents, they found that they spent an annual average of \$15,800 with a standard deviation of \$2,600. 1. Check assumptions and write hypotheses Housing costs are quantitative. Because \(n \ge 30\), the sampling distribution can be approximated using the \(t\) distribution. This is a left-tailed test because we want to know if residents of this community spent less than the national average. \(H_{0}:\mu=16803\) 2. Calculate the test statistic Test Statistic: One Group Mean \(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}\) \(\overline{x}\) = sample mean \(t=\dfrac{\overline{x}-\mu_0}{\dfrac{s}{\sqrt{n}}}=\dfrac{15800-16803}{\dfrac{2600}{\sqrt{30}}}=-2.113\) Our t test statistics is -2.113 3. Determine the p-value \(df=n-1=30-1=29\) This is a left-tailed test so we want to know the probability of \(t < -2.113\) Using Minitab we can find that \(p=0.0216634\) 4. Make a decision \(p\leq .05\), therefore we reject the null hypothesis. 5. State a "real world" conclusion There is evidence to state that on average residents of this community spent less than the national average on housing in 2011. Which test is used when sample size is less than 30?The parametric test called t-test is useful for testing those samples whose size is less than 30.
Which test can be used for sample size above 30?A t-test is necessary for small samples because their distributions are not normal. If the sample is large (n>=30) then statistical theory says that the sample mean is normally distributed and a z test for a single mean can be used. This is a result of a famous statistical theorem, the Central limit theorem.
Which test is meant for the sample below 30?So yes, you can use a t-test with a sample size which is smaller than 30.
What distribution is the best used when the sample is less than 30?Generally, the Student t distribution is used when the sample size is less than 30.
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