Question

Test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students at the 0.005 significance level.
(a)The null and alternative hypothesis would be:_______
A. H 0 : μ O ≤ μ C
H 1 : μ O > μ C
B. H 0 : p O ≤ p C
H 1 : p O > p C
C. H 0 : p O ≥ p C
H 1 : p O < p C
D. H 0 : μ O = μ C
H 1 : μ O ≠ μ C
E. H 0 : p O = p C
H 1 : p O ≠ p C
F. H 0 : μ O ≥ μ C
H 1 : μ O < μ C
(b)The test is:_______
A. left-tailed
B. two-tailed
C. right-tailed
(c)The sample consisted of 60 Orange Coast students, with a sample mean GPA of 2.91 and a standard deviation of 0.05, and 60 Coastline students, with a sample mean GPA of 2.96 and a standard deviation of 0.03.
(d)The test statistic is:________ (to 2 decimals)
(e)The p-value is:________ (to 2 decimals)
(f)Based on this we:_______
A. Fail to reject the null hypothesis
B. Reject the null hypothesis

Answer 1

Answer:

a) F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

b) B. two-tailed

d) $t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$  

e) $p_v =P(t_{118}$  

f) B. Reject the null hypothesis

Step-by-step explanation:

Information provided

$\bar X_{O}=2.91$ represent the mean for the Orange Coast

$\bar X_{C}=2.96$ represent the mean for the Coastline

$s_{O}=0.05$ represent the sample standard deviation for Orange Coast

$s_{C}=0.03$ represent the sample standard deviation for Coastline

$n_{O}=60$ sample size for Orange Coast

$n_{C}=60$ sample size for Coastline

$\alpha=0.005$ Significance level provided

t would represent the statistic

Part a

For this case we want to test the claim that the mean GPA of Orange Coast students is smaller than the mean GPA of Coastline students

Null hypothesis:$\mu_{O} \geq \mu_{C}$  

Alternative hypothesis:$\mu_{O} < \mu_{C}$  

F. H 0 : μ O ≥ μ C

H 1 : μ O < μ C

Part b

For this case we need to conduct a left tailed test.

B. two-tailed

Part d

The statistic is given by:

$t=\frac{(\bar X_{O}-\bar X_{C})-\Delta}{\sqrt{\frac{s^2_{O}}{n_{O}}+\frac{s^2_{C}}{n_{C}}}}$ (1)  

And the degrees of freedom are given by $df=n_1 +n_2 -2=60+60-2=118$  

Replacing the info we got:

$t=\frac{(2.91-2.96)-0}{\sqrt{\frac{0.05^2}{60}+\frac{0.03^2}{60}}}}=-6.64$  

Part e

We can calculate the p value with this probability:

$p_v =P(t_{118}$  

Part f

Since the p value is a very low value compared to the significance level given of 0.005 we can reject the null hypothesis.

B. Reject the null hypothesis