In a one-sample t-test, how is the alternative hypothesis formally expressed?

In a one-sample $$t$$-test, the null hypothesis posits that the true population mean ($$\mu$$) is identical to the hypothetical population mean ($$\mu_0$$), expressed formally as $$\mu = \mu_0$$. Conversely, the alternative hypothesis argues that the actual population mean significantly deviates from this expected standard of comparison, formalized as $$\mu 
eq \mu_0$$.

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The one-sample $$t$$-test is a statistical procedure used to compare a single sample mean ($$M$$) against a specific, hypothetical population mean ($$\mu_0$$) that serves as an interesting standard of comparison. By defining a null hypothesis ($$\mu = \mu_0$$) and an alternative hypothesis ($$\mu \neq \mu_0$$), researchers use this test to determine if the sample data provide enough statistical evidence to conclude that the true population mean differs from the hypothesized value.

One-Sample t-Test

A hypothetical population mean, symbolized mathematically as $$\mu_0$$, serves as a specific, interesting standard of comparison in statistical research. It represents an expected baseline value for an entire population, against which researchers can compare an actually observed sample mean ($$M$$) to draw statistical conclusions.

Hypothetical Population Mean

https://kpu.pressbooks.pub/psychmethods4e/

KPU Research Methods in Psychology - 4th American Edition

Statement that says there is no relationship between the two measured items

Null Hypothesis

Statement that says something is HAPPENING! New theory is true instead of an old one

Alternative Hypothesis

1. The dependent variable must be continuous (interval/ratio).
2. The observations are independent of one another.
3. The dependent variable should be approximately normally distributed.
4. The dependent variable should not contain any outliers.

Assumptions

To conduct a one-sample $$t$$-test, researchers compute the $$t$$ statistic using the exact formula: $$t = \frac{M - \mu_0}{\frac{SD}{\sqrt{N}}}$$. In this mathematical equation, $$M$$ represents the sample mean, $$\mu_0$$ is the hypothetical population mean serving as the standard of comparison, $$SD$$ is the standard deviation of the sample, and $$N$$ indicates the total sample size.

Formula for the One-Sample t-Test

The dependent-samples $$t$$-test, sometimes called the paired-samples $$t$$-test, is used to compare two means obtained from the same sample tested at two different times or under two distinct conditions. This statistical comparison is particularly appropriate for analyzing data from pretest-posttest designs or within-subjects experiments. Conceptually, it functions as a special case of the one-sample $$t$$-test.

Dependent-Samples t-Test

Null and Alternative Hypotheses for the One-Sample t-Test

When conducting a one-sample $$t$$-test, the degrees of freedom ($$df$$) are calculated by subtracting one from the total sample size ($$N - 1$$). This specific calculation dictates the exact shape of the $$t$$ distribution, enabling researchers to accurately identify the $$p$$ value associated with their test statistic.

Degrees of Freedom (One-Sample t-Test)

To demonstrate a one-sample $$t$$-test, imagine a health psychologist investigating whether university students accurately estimate the $$250$$ calories in a chocolate chip cookie. The null hypothesis ($$\mu = 250$$) assumes accurate estimation. The researcher asks a sample of $$10$$ students, resulting in a sample mean ($$M$$) of $$212.00$$ calories and a standard deviation ($$SD$$) of $$39.17$$. Plugging these statistics into the one-sample $$t$$-test formula yields a $$t$$ score of $$-3.07$$. For a two-tailed test with $$9$$ degrees of freedom, the critical value is $$\pm 2.262$$. Because the calculated $$t$$ score is more extreme than this critical value, the researcher rejects the null hypothesis and concludes that students significantly underestimate the cookie's calories. If the researcher had a strong theoretical reason to expect an underestimation, a one-tailed test with a less severe critical value of $$-1.833$$ could have been used, making it easier to reject the null hypothesis if the results trended in the expected direction.

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