Question

You wish to test the claim that the first population mean is less than the second population meanat a significance level of α=0.001α=0.001.     

Ho:μ1=μ2Ho:μ1=μ2
Ha:μ1<μ2Ha:μ1<μ2

You obtain the following two samples of data.

Sample #1	Sample #2
43.3 56.2 64.2 49.7 95.4 61.0 85.0 49.1 50.9 49.1 61.0 63.2 72.7 81.7	41.4 55.8 54.5 83.6 93.5 79.2 74.9 77.8 49.8 48.8 43.5 90.5 90.5 76.2 47.7

1. What is the test statistic for this sample?
   
   test statistic =   Round to 3 decimal places.
   
2. What is the p-value for this sample?
   
   p-value =  Use Technology Round to 4 decimal places.
   
3. The p-value is...
   • less than (or equal to) αα
   • greater than αα
   
   
4. This test statistic leads to a decision to...
   • reject the null
   • accept the null
   • fail to reject the null
   
   
5. 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.

Answer 1

Ans:

1. test statistic =   Round to 3 decimal places. = -0.643

2. The p-value is  0.2627248

   3. greater than αα

4 . This test statistic leads to a decision to...

• fail to reject the null

As such, the final conclusion is that...

There is not sufficient sample evidence to support the claim that the first population mean is less than the second population mean.

############Explanation #######################

Here we want to test the claim that the first population mean is less than the second population.

Let be the mean of first population and be the mean of second population

Hence the null hypothesis is given as

vs the alternative hypothesis is given as

The given data is

Sample 1	Sample 2
43.3	41.4
56.2	55.8
64.2	54.5
49.7	83.6
95.4	93.5
61	79.2
85	74.9
49.1	77.8
50.9	49.8
49.1	48.8
61	43.5
63.2	90.5
72.7	90.5
81.7	76.2
	47.7

Calculating the sample mean

= 63.03571

= 68.57143

Obtaining the sample deviation

= 240.85016

Similarly

= 355.9446

Obtaining the pooled variance

= 300.528746

Therefore

s = 17.33576494

Rounding off

s = 17.336

Test statistic is given as

= -0.64331

Round to 3 decimal places

= -0.643

df= n1 +n2-2

= 27

Since this is one tailed hypothesis the p-value is given by

P[ t27  < -0.64331]=  0.2627248

At significance level of α=0.001

Since p-value > 0.001

We failed to reject the null hypothesis.

There is not sufficient sample evidence to support the claim that the first population mean is less than the second population mean.