Question

You wish to test the following claim (HaHa) at a significance level of α=0.002α=0.002.

      Ho:μ1=μ2
      Ha:μ1<μ2

You believe both populations are normally distributed, but you do not know the standard deviations for either. However, you also have no reason to believe the variances of the two populations are not equal. You obtain the following two samples of data.

Sample #1

Sample #2

87.5	100.8	78.3
62.9	108.3	76.4
100.8	73.4	61.8
84.5	83.5	77.2
84.2	87.1	71
81.7	66.9	67.5
98	53.5	81.3
72.5	63.8

100.1	90.2	95.7
108.8	88	97.1
97.4	104.1	85.7
89.8	99.2	92.5

What is the critical value for this test? (Report answer accurate to three decimal places.)
critical value =  

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
test statistic =  

The test statistic is...

• in the critical region
• not in the critical region

This test statistic leads to a decision to...

• reject the null
• accept the null
• fail to reject the null

As such, the final conclusion is that...

• There is sufficient evidence to warrant rejection of the claim that the first population mean is less than the second population mean.
• There is not sufficient evidence to warrant rejection of the claim that the first population mean is less than the second population mean.
• The sample data support the claim that the first population mean is less than the second population mean.
• There is not sufficient sample evidence to support the claim that the first population mean is less than the second population mean.

Please explain and use ti84

Answer 1

The null and alternative hypothesis is

Ho: μ1 = μ2

Ha: μ1 < μ2

Level of significance = α=0.002

n1 = 23

n2 = 12

However, you also have no reason to believe the variances of the two populations are not equal.

So we have to use here pooled variance.

Degrees of freedom = n1 + n2 - 2 = 33

Critical value = 3.094

By using TI-84 calculator we have to solve this question.

Enter values into TI-84 calculator.

Click on STAT ----------> Edit -----> Enter sample 1 data into L1 and Sample 2 data into L2.

Then click on STAT --------> TESTS --------> 2-SampTTest-------> Data ------>

List1: L1

List2: L2

Freq1: 1

Freq2: 1

Pooled: Yes

Calculate

Test statistic t = - 3.843

The test statistic is in the critical region.

This test statistic leads to a decision to reject the null.

Conclusion:

• There is not sufficient evidence to warrant rejection of the claim that the first population mean is less than the second population mean.

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