Questions: You wish to test the following claim (Ha) at a significance level of α=0.001. For the context of this problem, d= new - old where the first data set represents the rating for an old product and the second data set represents the scores for the new product. Ho: μd=0 Ha: μd<0 You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: old new --- --- 50.6 0.3 49 40.6 59.5 25.2 59.1 39.5 57 33.3 47.7 23 What is the test statistic for this sample? (Report answer accurate to three decimal places.) t= The p -value is ... This p -value leads to a decision to... - reject the null - accept the null - fail to reject the null

You wish to test the following claim (Ha) at a significance level of α=0.001. For the context of this problem, d= new - old where the first data set represents the rating for an old product and the second data set represents the scores for the new product.

Ho: μd=0
Ha: μd<0

You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of data:

old new
--- ---
50.6 0.3
49 40.6
59.5 25.2
59.1 39.5
57 33.3
47.7 23

What is the test statistic for this sample? (Report answer accurate to three decimal places.)
t=
The p -value is ...
This p -value leads to a decision to...
- reject the null
- accept the null
- fail to reject the null

Transcript text: You wish to test the following claim $\left(H_{a}\right)$ at a significance level of $\alpha=0.001$. For the context of this problem, $d=$ new - old where the first data set represents the rating for an old product and the second data set represents the scores for the new product. \[ \begin{array}{l} H_{o}: \mu_{d}=0 \\ H_{a}: \mu_{d}<0 \end{array} \] You believe the population of difference scores is normally distributed, but you do not know the standard deviation. You obtain the following sample of data: \begin{tabular}{|r|r|} \hline old & new \\ \hline 50.6 & 0.3 \\ \hline 49 & 40.6 \\ \hline 59.5 & 25.2 \\ \hline 59.1 & 39.5 \\ \hline 57 & 33.3 \\ \hline 47.7 & 23 \\ \hline \end{tabular} What is the test statistic for this sample? (Report answer accurate to three decimal places.) $t=$ $\square$ The p -value is ... $\square$ This p -value leads to a decision to... reject the null accept the null fail to reject the null

Solution

Solution Steps

Step 1: Calculate the Mean of Differences

The mean of the differences $ \mu $ is calculated as follows: \[ \mu = \frac{\sum_{i=1}^N x_i}{N} = \frac{-161.0}{6} = -26.833 \] Thus, the mean of differences is $ -26.833 $.

Step 2: Calculate the Standard Deviation

The variance $ \sigma^2 $ is calculated using the formula: \[ \sigma^2 = \frac{\sum (x_i - \mu)^2}{n-1} = 202.583 \] The standard deviation $ \sigma $ is then: \[ \sigma = \sqrt{202.583} = 14.233 \] Thus, the standard deviation of differences is $ 14.233 $.

Step 3: Calculate the Standard Error

The standard error $ SE $ is calculated as: \[ SE = \frac{\sigma}{\sqrt{n}} = \frac{14.233}{\sqrt{6}} = 5.811 \]

Step 4: Calculate the Test Statistic

The test statistic $ t $ is calculated using the formula: \[ t = \frac{\bar{x} - \mu_0}{SE} = \frac{-26.833 - 0}{5.811} = -4.618 \]

Step 5: Calculate the p-value

For a left-tailed test, the p-value is: \[ P = T(z) = 0.003 \]

Step 6: Decision Based on p-value

Since the p-value $ 0.003 $ is less than the significance level $ \alpha = 0.001 $, we fail to reject the null hypothesis.