Question

In a studly involving depression in adokescents, you haws a ae trnie if your sample regreserts the krno a sample mean of R 8S w the population rman is μ-87"do.12 Since the caculled zscore-ano and Nr byond the crie " value of tL06, re ect lhe cacutaied 2.sere-G is nu tarytnd the c nic-value of ±し96,侧囟not ' . Since the calculated z-score is+ 0.83 and is the criical valu, 이 196, we do not reiecina tis sample mean is npresentative of the known popuain critical value tf t1隇, we de fut retect that this sample mean spresentative ot the known popiaa.

I need help.

Answer 1

$H_0: \mu = 87 \ \ H_1: \mu \neq 87$

$z = \frac{( \bar{x}-\mu) }{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{ (85-87) }{\frac{12}{\sqrt{25}}}$

= -0.8333

Table value of z at 0.05 level = 1.96

Rejection Region: Reject Ho if z < -1.96 or z > 1.96

Calculated z = -0.83 not in the rejection region

The null hypothesis is not rejected.

Option 2

Since the calculated z score -0.83 not beyond the critical value ±1.96, we do not reject this sample mean is representative of the known population.

Z Test of Hypothesis for the Mean
Data
Null Hypothesis                       m=	87
Level of Significance	0.05
Population Standard Deviation	12
Sample Size	25
Sample Mean	85
Intermediate Calculations
Standard Error of the Mean	2.4000
Z Test Statistic	-0.8333
Two-Tail Test
Lower Critical Value	-1.9600
Upper Critical Value	1.9600
p-Value	0.4047
Do not reject the null hypothesis