Question

You wish to test the following claim ($H_a$) at a significance level of $\alpha =0.05$.

      $H_o:{\mu }_1={\mu }_2$
      $H_a:{\mu }_1<{\mu }_2$

You believe both populations are normally distributed, but you do not know the standard deviations for either. You should use a non-pooled test. You obtain a sample of size $n_1=16$ with a mean of $M_1=50.7$ and a standard deviation of $S{D}_1=7.4$ from the first population. You obtain a sample of size $n_2=12$ with a mean of $M_2=53.7$ and a standard deviation of $S{D}_2=18.2$ from the second population.

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

What is the p-value for this sample? For this calculation, use the conservative under-estimate for the degrees of freedom as mentioned in the textbook. (Report answer accurate to four decimal places.)
p-value = 

The p-value is...

• less than (or equal to) $\alpha$

• greater than $\alpha$

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.

Answer 1

Ans:

By using calculator,

test statistic = -2.185

Degrees of freedom =17

P value =0.0216

Yes we can use TI 84 calculator