Question

The following table gives results from two groups of students who took a nonproctored test. Use a 0.01 significance level to test the claim that the samples are from populations with the same mean.
Group 1: n=30 mean=70.29 s=22.09
Group 2: n=32 mean=74.26 s=18.15

Original claim:
Opposite claim:
Alternative and null hypotheses:
Significance level:
Test statistic:
P-value:
Reject or fail to reject:
Final Conclusion:

Answer 1

Answer:

Original claim is $\mu_{1} =\mu_{2}$

Opposite claim is $\mu_{1} \neq \mu_{2}$

Null and alternative hypotheses:

$H_{0}:\mu_{1} = \mu_{2}$

$H_{1} : \mu_{1} \neq \mu_{2}$

Significance level: 0.01

Test statistic:

We can use TI-84 calculator to find the test statistic and P-value. The steps are as follows:

Press STAT and the scroll right to TESTS

Scroll down to 2-SampTTest... and scroll to stats.

Enter below information.

$\bar{x_{1}}=70.29$

$Sx1=22.09$

$n_{1} = 30$

$\bar{x_{2}}=74.26$

$Sx2=18.15$

$n_{2} = 32$

$\mu_{1} \neq \mu_{2}$

Pooled: Yes

Calculate.

The output is in the attachment.

Therefore, the test statistic is:

$t=-0.78$

P-value: 0.4412

Reject or fail to reject: Fail to reject

Final Conclusion: Since the p-value is greater than the significance level, we, therefore, fail to reject the null hypothesis and conclude that the there is sufficient evidence to support the claim that the samples are from populations with the same mean.