# A tibble: 1 × 7
statistic t_df p_value alternative estimate lower_ci upper_ci
<dbl> <dbl> <dbl> <chr> <dbl> <dbl> <dbl>
1 1.99 12.8 0.0679 two.sided 4.90 -0.418 10.2
STAT 218 - Week 6, Lecture 3
The Wisconsin Fast Plant, Brassica campestris, has a very rapid growth cycle that makes it particularly well suited for the study of factors that affect plant growth.
In one such study, 7 plants were treated with the substance Ancymidol (ancy) and were compared to 8 control plants that were given ordinary water. Heights of all of the plants were measured, in cm, after 14 days of growth.
(\(df\) for this question is calculated as 12).
Let’s see an example for hypothesis testing.
The hypothesis that \(\mu_1\) and \(\mu_2\) are not equal is called an alternative hypothesis (or a research hypothesis)
\[ H_A: \mu_1 \neq \mu_2 \]
Its antithesis is the null hypothesis, \[ H_0: \mu_1 = \mu_2 \]
which asserts that \(\mu_1\) and \(\mu_2\) are equal. A researcher would usually express these hypotheses more informally and we can trace those hypotheses from the examples, problems and exercises in this course.
Alternatively we can express these hypotheses as following:
\[ \\H_0: \mu_1 - \mu_2 = 0 \\H_A: \mu_1 - \mu_2 \neq 0 \]
Remember our CI formula when we had one single quantitative variable. \[ \\95 \% \ CI = (\bar{y} \pm \ multiplier \ \times \ SE_{\bar{y}}) \]
Let’s revise our 95% confidence interval formula for comparing two means.
\[ \\95 \% \ CI = (\bar{y_1}-\bar{y_2}) \pm \ multiplier \times SE_{\bar{Y}_1 - \bar{Y}_2} \]
The formula that we have used so far is
\[ SE_{\bar{Y}} = \frac{s}{\sqrt{n}} \]
Naturally, we can say that taking the difference between two sample means is an estimate of the quantity (\(\mu_1\) - \(\mu_2\)).
However, the formula for the standard error of the difference (\(\bar{Y}_1\) - \(\bar{Y}_2\)) is a little different from what we initially thought.
\[ SE_{\bar{Y}_1 - \bar{Y}_2} = \sqrt{SE_1^2 + SE_2^2} \]
\[ SE_{\bar{Y}_1 - \bar{Y}_2} = \sqrt{ \frac {s_1^2}{n_1} + \frac {s_2^2}{n_2}} \]
\[ SE_{\bar{Y}_1 - \bar{Y}_2} = \sqrt{ \frac {4.8^2}{8} + \frac {4.7^2}{7}} = 2.46 \]
multiplier is 2.164 for this question.
\[ \\95 \% \ CI = (\bar{y_1}-\bar{y_2}) \pm \ multiplier \times SE_{\bar{Y}_1 - \bar{Y}_2} \]
\[ \\95 \% \ CI = (15.9 - 11) \pm \ 2.164 \times 2.46 \\ 4.9 \pm 5.32 \]
In our hypothetical example, we calculated 95% CI and we found it as (-0.42,10.22)
\[ test \ statistic = \frac{(\bar{y}_1 - \bar{y}_2) - (\mu_1 - \mu_2)}{SE(\bar{y}_1 - \bar{y}_2)} \]
\[ \frac{(15.9 - 11.0) - (0)}{2.46} = 1.99 \]
# A tibble: 1 × 7
statistic t_df p_value alternative estimate lower_ci upper_ci
<dbl> <dbl> <dbl> <chr> <dbl> <dbl> <dbl>
1 1.99 12.8 0.0679 two.sided 4.90 -0.418 10.2
Conclusion: We fail to reject the null hypothesis. That is, the data do not provide strong evidence that Ancymidol suppresses plant growth.
Ancymidol is known as a plant growth regulator that reduces plant growth by inhibiting gibberellin biosynthesis.
In our hypothetical example, we calculated 95% CI and we found it as (-0.4,10.2)
Is it reasonable to believe that the substance may affect plant growth? Discuss with your neighbor
Definition
\(\beta\) = Pr{lack of significant evidence for \(H_A\)} if \(H_A\) is true
OR
failing to reject the null hypothesis when the alternative is actually true.
If \(H_A\) is true, but we do not observe sufficient evidence to support \(H_A\), then we have made a Type II error.
Definition
\(\alpha\) = Pr{finding significant evidence for \(H_A\)} if \(H_0\) is true
OR
rejecting the null hypothesis when \(H_0\) is actually true.
From Essential Guide to Effect Sizes by Paul D. Ellis (2010)