Question

nd trustworthiness their experiment the rescarchers took photographs of B students (20 males with brown eyes, Why da we naturally tend to trust some strangers more than others? One group of researchers decided to study the relationship between eye color n r virkl eruitad 105 ntirinsntr to iudon ti-rtunkimere of nsch rtudant nhato Thi . n.int eale ba 1 mesnt u untauctuorthu snd 10 unne tauctworthu The RO rorne frnm ssk .diaie.n were then converted to z scores, and the average z score of each photo (across all 105 participants) was used for t analysis. Here is a summary of the results. Eve color n x s Brown 40 0.55 1.68 Blue 400.38 1.53 Can wa concluda from these data that bron-eyed students appaar mara trustworthy compared to thair blue-ayed counterparts? Test the hypothesis that the avarage scoras for the two groups are the same. State the null and alternative hypatheses Ho WBrow aue Ha rown Paue H Ho grown Halue H mwn niue Ha Blur Hai rown Report the test statistic, the degrees of freedom, and the Pvalue. (Round your test statistic to three decimal places, your degrees four decimal places.) freedom to the nearest whole number, and your P-value - value State your conclusion. (Use r0.05.) We do not reject Hn and can nat conclude that brown-eyed people seam more trustworthy according to this experiment. We reiect He and conclude that brown eved people seem more trustworthy according to this experiment.

Answer 1

SOLUTION:

From given data,

Eye color	n	$\bar{x}$
Brown	$n_{1}$ =40	$\bar{x}_{1}$=0.55	$s_{1}$ =1.68
Blue	$n_{2}$=40	$\bar{x}_{2}$-0.38	$s_{2}$=1.53

State the null and alternative hypotheses:

Null hypothesis :
Ho
:
L Brown
=
LBlue

Alternative hypothesis : $H_{a}$ :
L Brown
$\neq$
LBlue

2nd option is correct

Report the test statistic, the degree of freedom , and the P-value

The test statistic:

Standard error = SE = sqrt ($s_{1}^{2}$/$n_{1}$ +
Sa
/$n_{2}$) = sqrt (1.68²/40 + 1.53²/40​​​​​​​) = 0.3592805

test statistic = t = ($\bar{x}_{1}$ - $\bar{x}_{2}$) / SE

t = (0.55 - 0.38) / 0.3592805

t = 0.17 / 0.3592805

t = 0.4731

The degree of freedom :

degree of freedom = df = ($s_{1}^{2}$/$n_{1}$ +
Sa
/$n_{2}$​​​​​​​)² / [ ($s_{1}^{2}$/$n_{1}$)² / ($n_{1}$ -1) + (
Sa
/$n_{2}$)² / ($n_{2}$ -1)]

df = (1.68²/40 + 1.53²​​​​​​​/40)² / [ (1.68²/40)² / (40 -1) + (1.53²/40)² / (40​​​​​​​ -1)]

df = 0.01666229 / 0.0002154768

df = 77.327

Significance level = $\alpha$ = 0.05

The P-value :

P-value = P(t < 0.4731) = 0.637475

State the conclusion :

P-value =0.637475 > Significance level = $\alpha$ = 0.05 then

we reject and conclude that brown eyed people seem more trustworthy according to this experiment.

Ho

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