Question

# In a similar study using the same design (sample of n = 9 participants), the individuals who wore the shirt produced an average estimate of M = 6.4 with SS = 162.

Study design

In a similar study using the same design (sample of $$n = 9$$ participants), the individuals who wore the shirt produced an average estimate of $$M = 6.4$$ with $$SS = 162$$.
The average number who said they noticed was (population mean) 3.1. Calculate a One Sample t-test using a two-tailed test with $$\displaystyle\alpha={.05}$$.

2020-11-09

Step 1
The null and alternative hypotheses are’
$$\displaystyle{H}_{{0}}:{m}={3.1}$$
$$\displaystyle{H}_{{1}}:{m}\ne{3.1}$$
The sample mean is 6.4 and the population mean is 3.1.
Here, $$SS=162$$ and $$n=9$$. The estimate of the standard deviation is
$$\displaystyle{s}=\sigma-\cap=\sqrt{{\frac{{{S}{S}}}{{{n}-{1}}}}}$$.
Therefore, the $$\displaystyle{s}=\sqrt{{\frac{{162}}{{8}}}}={4.5}$$.
Test statistic:
The test statistic for hypothesis test for mean is given by:
$$\displaystyle{t}=\frac{{{x}-\overline{-}\mu}}{{\frac{{s}}{\sqrt{{{n}}}}}}$$
$$\displaystyle=\frac{{{6.4}-{3.1}}}{{\frac{{4.5}}{\sqrt{{{9}}}}}}$$
$$= 2.2$$
The test statistic is 2.2.
Step 2
Critical value:
Here, sample size is 9.
Thus, the degrees of freedom for the test is $$8(=9-1)$$.
Therefore, the df is 15.
The critical value is $$\displaystyle\pm{2.306}$$ using the excel formula “=(T.INV.2T(0.05,8))”
Decision rule:
Denote t as test statistic value and $$\displaystyle{t}_{{\frac{\alpha}{{2}}}}$$ as the critical value.
Decision rule based on critical approach:
If $$\displaystyle{t}\le–{t}_{{\frac{\alpha}{{2}}}}{\left({\quad\text{or}\quad}\right)}{t}\ge{t}_{{\frac{\alpha}{{2}}}}$$, then reject the null hypothesis $$\displaystyle{H}_{{0}}$$.
If $$\displaystyle-{t}_{{\frac{\alpha}{{2}}}}{<}{t}{<}{t}_{{\frac{\alpha}{{2}}}}$$, then fail to reject the null hypothesis $$\displaystyle{H}_{{0}}$$.
Conclusion:
The test statistic value is given as 2.2 and critical value is $$\displaystyle\pm{2.306}$$.
Here, $$-2.306<2.2<2.306$$.
By the rejection rule fail to reject $$\displaystyle{H}_{{0}}$$.
The test statistic value is 2.2.
The critical/cut off values is $$\displaystyle\pm{2.131}$$