Question

The following information was obtained from matched samples. Individual Method 1 Method 2 1 7 5 2 5 9 3 6 8 4 7 7 5 5 6 Refer to Exhibit 10-5. If the null hypothesis is tested at the 5% level, the null hypothesis _____.
a. should be rejected
b. should be revised
c. should not be rejected
d. None of these answers are correct

Answer 1

Answer:

$p_v =2*P(t_{(8)}$

c. should not be rejected

Step-by-step explanation:

1) Data given and notation

The data given is:

Method 1 : 7,5,6,7,5

Method 2: 5,9,8,7,6

$\bar X_{1}=6$ represent the mean for the method 1

$\bar X_{2}=7$ represent the mean for the method 2

$s_{1}=1$ represent the sample standard deviation for the method 1

$s_{2}=1.58$ represent the sample standard deviation for the method 2

$n_{1}=5$ sample size selected for the Consultant A

$n_{2}=5$ sample size selected for the Consultant B

$\alpha=0.05$ represent the significance level for the hypothesis test.

t would represent the statistic (variable of interest)

$p_v$ represent the p value for the test (variable of interest)

State the null and alternative hypotheses.

We need to conduct a hypothesis in order to check if the mean for the difference is equal to 0, the system of hypothesis would be:

Null hypothesis:$\mu_{1}- \mu_{2}=0$

Alternative hypothesis:$\mu_{1} - \mu_{2}\neq 0$

If we analyze the size for the samples both are less than 30 so for this case is better apply a t test to compare means, and the statistic is given by:

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

t-test: "Is used to compare group means. Is one of the most common tests and is used to determine whether the means of two groups are equal to each other".

Calculate the statistic

We can replace in formula (1) the info given like this:

$t=\frac{6-7}{\sqrt{\frac{1^2}{5}+\frac{1.58^2}{5}}}=-1.195$ (1)

P-value

The first step is calculate the degrees of freedom, on this case:

$df=n_{1}+n_{2}-2=5+5-2=8$

Since is a two sided test the p value would be:

$p_v =2*P(t_{(8)}$

Conclusion

If we compare the p value and the significance level given $\alpha=0.05$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis. So the best option would be:

c. should not be rejected