Answer to Question #194005 in Statistics and Probability for Alice

Question

Comparing dexterity-test scores of workers on the day shift versus those on the night shift, the production manager of a large electronics plant finds that a sample of 37 workers from the day shift have an average score of 73.1, with a standard deviation of 12.3. For 42 workers from the night shift, the average score was 77.3, with a standard deviation of 8.4. Assuming the population standard deviations are equal, use the 0.05 level of significance in comparing the average scores for the two shifts. Using the appropriate statistical table, what is the most accurate statement we can make about the p-value for this test? Construct and interpret the 95% confidence interval for the difference between the population means.

Answer 1

Null Hypothesis, $H_o:\mu_1=\mu_2$

There is no significant difference between the average score of the two shifts.

Alternative Hypothesis:

$H_1:\mu_1\neq \mu_2$

There is a significant difference between the average scores pf the two shifts.

Level of significance $\alpha=0.05$

Here,

$\bar{x_1}=73.1,\bar{x_2}=77.3,s_1=12.3,s_2=8.4,n_1=37,n_2=42$

First we calculate the pooled variance-

$s_p^2=\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$

   $=\dfrac{(37-1)(12.30)^2+(42-1)(8.40)^2}{37+42-2}=108.30$

The test statistics is-

$t=\dfrac{(\bar{x_1}-\bar{x_2})-(\mu_1-\mu_2)_0}{\sqrt{s_p^2(\dfrac{1}{n_1}+\dfrac{1}{n_2}})}$

 

 

  $=\dfrac{73.10-77.30}{\sqrt{108.30(\dfrac{1}{37}+\dfrac{1}{42})}}$

 =-1.79

Degree of freedom-

$df=n_1+n_2-2=37+42-2=77$

Using the test statistics and 77 degrees of freedom, we can look up an appropriate t value using the t-table. As we can see, this p-value is between 0.05 and 0.10.

Because this p-value is greater than 0.05, we are fail to reject null hypothesis.

and conclude there is no significant difference between the average scores of the two shifts.

Confidence interval-

$\alpha=0.05,t_{\frac{\alpha}{2}}=1.991$

95% confidence interval for the difference between the population means is-

 $=(\bar{x_1}-\bar{x_2})\pm t_{\frac{\alpha}{2}} \sqrt{s_p^2(\dfrac{1}{n_1}+\dfrac{1}{n_2})}$

$=(73.10-77.30) \pm 1.991 \sqrt{108.3(\dfrac{1}{37}+\dfrac{1}{42})}$

 

$=-4.20\pm 4.67\\[9pt]=(-8.87,0.47)$

Because our confidence interval above includes 0, we can determine that the means above are equal.