Question

Aa Aa Tips 4. Factors that influence a hypothesis test Suppose you conduct a hypothesis test about a population mean when the standard deviation is unknown by calculating at statistic and determining whether to reject the null hypothesis. Assume that your statistic is a positive value and that you will reject the null hypothesis when t is very large. For each of the potential changes to the components of your study design or the results listed below, determine the change to the t statistic, and then select "Increases," "Decreases," or "Stays the same" from the dropdown menu in the right-hand column. Tips Unlimited The t statistic red Change in Study Design or Results A decrease in the sample variance (82) A decrease in the obtained difference (M-1) An increase in the sample size (n) A decrease in the significance level (such as using a - .01 instead of a - .05) Options re er a FREE 7.

Answer 1

For the given problem,

Let $\mu$ denote the population mean, with unknown standard deviation. We need to compare this mean to a hypothesized value $\mu_{0}$ .

To test:
Hoi Ho
Vs say,
H . HO

The appropriate statistical test to test the above hypothesis would be a One sample t test, with test statistic given by,

$t=\frac{M-\mu_{0}}{s/\sqrt{n}} \sim t_{n-1}$

where, M = Sample mean, s² = Sample variation

1.Decrease in sample variance

From the test statistic, we find that the sample variance lies in the denominator of the test statistic and hence, is inversely proportional to t.Therefore,

Decrease in sample variance (s²) increases the t statistic.

2. Decrease in obtained difference

From the test statistic, we find that the difference $M-\mu$ lies in the numerator of the test statistic and hence, is directly proportional to t.Therefore,

Decrease in obtained difference decreases the t statistic.

3. Increase in sample size n:

From the t statistic,

$t=\frac{M-\mu}{\frac{s}{\sqrt{n}}}$

$=\frac{(M-\mu)\sqrt{n}}{s}$

We find that the difference n lies in the numerator of the test statistic and hence, is directly proportional to t.Therefore,

Increase in sample size n increases the t statistic.

4. Decrease in significance level alpha from 0.05 to 0.01.

Since, the observed / computed test statistic t has nothing to do with the significance level (it is the critical value of t that is obtained for a particular alpha, to which this observed t is compared to arrive at a decision of its significance), the t statistic remains the same.

5.

Given: $n=36,M=7.3,s=1.5$

Cohen's d can be computed using the formula:

$d=\frac{X-M}{s}$

For a student who scored 6.1,

$\dpi{80} d=\frac{6.1-7.3}{1.5}=-0.8$ and   $t=\frac{7.3-6.1}{1.5/\sqrt{36}}=4.8$

And

$r^{2}=\frac{t^{2}}{t^{2}+df}$

where, df for one sample t test = n - 1 = 36 - 1 = 35:

$\Rightarrow r^{2}=\frac{(4.80)^{2}}{(4.80)^{2}+35}$

= 0.397

Hence, using the guidelines for interpreting the effect sizes, for cohen's d:

Cohen's d	Effect size
0.2	Small
0.5	Medium
0.8	Large

The cohen's d obtained (d = -0.8) shows a large treatment effect.

r	Effect size
0.1	Small
0.3	Medium
0.5	Large

Here, $r=\sqrt{0.397}=0.63$ which suggests large effect.

Hence, using r², we may say that there is a large effect.

For n = 121, M = 6.1, s = 1.2:

Cohen's d can be computed as:

$d=\frac{X-M}{s}$

$=\frac{5.7-6.1}{1.2}$

= -0.33

Interpreting cohen's d, we may say that there is a small to medium treatment effect.

And t = 3.64, for 121 - 1 = 120 df

$r^{2}=\frac{(3.64)^{2}}{(3.64)^{2}+120}$

= 0.099

Here, $r=\sqrt{0.099}=0.32$ which suggests a medium effect

Interpreting r², we may say that there is a medium treatment effect.