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
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asked 2020-11-08

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 \(\displaystyle\alpha={.05}\).

Answers (1)

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 \(\displaystyle\pm{2.131}\)

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