Question

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.05 significance level.

The null and alternative hypothesis would be:

H0:pO≥pCH0:pO≥pC
H1:pO<pCH1:pO<pC

H0:pO≤pCH0:pO≤pC
H1:pO>pCH1:pO>pC

H0:μO≤μCH0:μO≤μC
H1:μO>μCH1:μO>μC

H0:μO=μCH0:μO=μC
H1:μO≠μCH1:μO≠μC

H0:μO≥μCH0:μO≥μC
H1:μO<μCH1:μO<μC

H0:pO=pCH0:pO=pC
H1:pO≠pCH1:pO≠pC

The test is:

right-tailed

two-tailed

left-tailed

The sample consisted of 55 Orange Coast students, with a sample mean GPA of 2.69 and a standard deviation of 0.05, and 55 Coastline students, with a sample mean GPA of 2.71 and a standard deviation of 0.04.

The test statistic is:  (to 2 decimals)

The p-value is:  (to 2 decimals)

Based on this we:

• Fail to reject the null hypothesis
• Reject the null hypothesis

Answer 1

Given that, the sample consisted of 55 Orange Coast students, with a sample mean GPA of 2.69 and a standard deviation of 0.05, and 55 Coastline students, with a sample mean GPA of 2.71 and a standard deviation of 0.04.

Significance level $(\alpha)$ = 0.05

The null and alternative hypotheses are,

$H_0 : \mu_O \geq \mu_C$

$H_1: \mu_O < \mu_C$

Using TI-84 calculator we get,

Test statistic is -2.32

The p-value is 0.01

Here, p-value = 0.01 < significance level $(\alpha)$ = 0.05

So, we reject the null hypothesis

TI-84 Plus TEXAS INSTRUMENTS t--2 . 1 643 1 233 P= 0 1 1 25720 7 df= 1 . 035 1 873 1-2.3 x2 2.71 STAT PLOT F1 TBLSET F2 FORMAT F3 CALC F4 TABLE F INDOw 2ND A-LOCK ALPHA QUIT MODE LINK X,T,e,n INS DEL LIST STAT TEST A MATH MATRIX D x녔 ANGLE B DRAW C CLEAR APPS PRGM VARS CoS. F TAN G SIN COS TAN EE J e M 10* N LOG e* S LN RCL X u o w Q V P 8 L5 U 5 7 9 L4 T LG V 4 6 MEM " STO L1 Y L2 Z OFF ENTRY SOLVE 2 3 ON ENTER CATALOG ANS ? 0