Questions: Mandrake Falls High School is experimenting with a weekend course in laboratory techniques. Of the 200 students enrolled in lab classes at Mandrake, only 73 have been able to take the techniques course. Mandrake is interested in evaluating the course's effectiveness in propagating safety in the laboratories. During regular lab classes, lab instructors have recorded harmful lab incidents: accidents, misuse of lab equipment, etc. The school is looking at the data and examining two variables: laboratory performance ("involved in no incident", "involved in exactly one incident", or "involved in 2+ incidents") and status regarding lab techniques course ("took the techniques course" or "didn't take the techniques course"). The contingency table below gives a summary of the data that have been gathered so far. In each of the 6 cells of the table are three numbers: the first number is the observed cell frequency (f0); the second number is the expected cell frequency ( fk ) under the assumption that there is no relationship between students' laboratory performances and whether or not they took the techniques course; and the third number is the following value. (Observed cell frequency - Expected cell frequency )^2 / Expected cell frequency The numbers labeled "Total" are totals for observed frequency. - Laboratory performance: Involved in no incident, Involved in exactly one incident, Involved in 2+ incidents - Status regarding lab techniques course: Took the techniques course, Didn't take the techniques course Laboratory Performance Involved in no incident Involved in exactly one incident Involved in 2+ incidents Total ------------------------------------------------------------------------------------------------------------------- Took the techniques course 31 26 16, 15.70, 0.006 73 Didn't take the techniques course 66 34 27, 27.31, 0.004 127 Total 97 60 43 200 (a) Determine the type of test statistic to use. Type of test statistic: (b) Find the value of the test statistic. (Round to two or more decimal places.)

Mandrake Falls High School is experimenting with a weekend course in laboratory techniques. Of the 200 students enrolled in lab classes at Mandrake, only 73 have been able to take the techniques course. Mandrake is interested in evaluating the course's effectiveness in propagating safety in the laboratories. During regular lab classes, lab instructors have recorded harmful lab incidents: accidents, misuse of lab equipment, etc. The school is looking at the data and examining two variables: laboratory performance ("involved in no incident", "involved in exactly one incident", or "involved in 2+ incidents") and status regarding lab techniques course ("took the techniques course" or "didn't take the techniques course"). The contingency table below gives a summary of the data that have been gathered so far. In each of the 6 cells of the table are three numbers: the first number is the observed cell frequency (f0); the second number is the expected cell frequency ( fk ) under the assumption that there is no relationship between students' laboratory performances and whether or not they took the techniques course; and the third number is the following value. (Observed cell frequency - Expected cell frequency )^2 / Expected cell frequency The numbers labeled "Total" are totals for observed frequency. - Laboratory performance: Involved in no incident, Involved in exactly one incident, Involved in 2+ incidents - Status regarding lab techniques course: Took the techniques course, Didn't take the techniques course Laboratory Performance Involved in no incident Involved in exactly one incident Involved in 2+ incidents Total ------------------------------------------------------------------------------------------------------------------- Took the techniques course 31 26 16, 15.70, 0.006 73 Didn't take the techniques course 66 34 27, 27.31, 0.004 127 Total 97 60 43 200 (a) Determine the type of test statistic to use. Type of test statistic: (b) Find the value of the test statistic. (Round to two or more decimal places.)
Transcript text: Mandrake Falls High School is experimenting with a weekend course in laboratory techniques. Of the 200 students enrolled in lab classes at Mandrake, only 73 have been able to take the techniques course. Mandrake is interested in evaluating the course's effectiveness in propagating safety in the laboratories. During regular lab classes, lab instructors have recorded harmful lab incidents: accidents, misuse of lab equipment, etc. The school is looking at the data and examining two variables: laboratory performance ("involved in no incident", "involved in exactly one incident", or "involved in $2+$ incidents") and status regarding lab techniques course ("took the techniques course" or "didn't take the techniques course"). The contingency table below gives a summary of the data that have been gathered so far. In each of the 6 cells of the table are three numbers: the first number is the observed cell frequency $\left(f_{0}\right)$; the second number is the expected cell frequency ( $f_{k}$ ) under the assumption that there is no relationship between students' laboratory performances and whether or not they took the techniques course; and the third number is the following value. \[ \frac{\left(f_{0}-f_{5}\right)^{2}}{f_{\Sigma}}=\frac{\text { (Observed cell frequency - Expected cell frequency })^{2}}{\text { Expected cell frequency }} \] The numbers labeled "Total" are totals for observed frequency. \begin{tabular}{|c|c|c|c|c|c|} \hline \multicolumn{2}{|l|}{\multirow[t]{2}{*}{}} & \multicolumn{4}{|c|}{Laboratory performance} \\ \hline & & Involved in no incident & Involved in exactly one incident & Involved in $2+$ incidents & Total \\ \hline Status regarding & Took the techniques course & \[ 31 \] & \begin{tabular}{l} 26 \\ \end{tabular} & \begin{tabular}{l} 16 \\ 15.70 \\ 0.006 \end{tabular} & 73 \\ \hline lab techniques course & Didn't take the techniques course & \begin{tabular}{l} 66 \\ \end{tabular} & \begin{tabular}{l} 34 \\ \end{tabular} & \begin{tabular}{l} 27 \\ 27.31 \\ 0.004 \end{tabular} & 127 \\ \hline & Total & 97 & 60 & 43 & 200 \\ \hline \end{tabular} (a) Determine the type of test statistic to use. Type of test statistic: $\square$ (b) Find the value of the test statistic. (Round to two or more decimal places.) $\square$
failed

Solution

failed
failed

Solution Steps

Step 1: Calculate Expected Frequencies

To determine the expected frequencies under the assumption of independence between the two categorical variables, we use the formula:

\[ E = \frac{R_i \times C_j}{N} \]

where \( R_i \) is the total for row \( i \), \( C_j \) is the total for column \( j \), and \( N \) is the total number of observations.

The expected frequencies for each cell are calculated as follows:

  • For cell (1, 1): \[ E = \frac{73 \times 97}{200} = 35.405 \]

  • For cell (1, 2): \[ E = \frac{73 \times 60}{200} = 21.9 \]

  • For cell (1, 3): \[ E = \frac{73 \times 43}{200} = 15.695 \]

  • For cell (2, 1): \[ E = \frac{127 \times 97}{200} = 61.595 \]

  • For cell (2, 2): \[ E = \frac{127 \times 60}{200} = 38.1 \]

  • For cell (2, 3): \[ E = \frac{127 \times 43}{200} = 27.305 \]

Thus, the expected frequencies are: \[ \begin{bmatrix} 35.405 & 21.9 & 15.695 \\ 61.595 & 38.1 & 27.305 \end{bmatrix} \]

Step 2: Calculate Chi-Square Components

Next, we calculate the Chi-Square components for each cell using the formula:

\[ \frac{(O - E)^2}{E} \]

where \( O \) is the observed frequency and \( E \) is the expected frequency.

The calculations for each cell are as follows:

  • For cell (1, 1): \[ \frac{(31 - 35.405)^2}{35.405} = 0.5481 \]

  • For cell (1, 2): \[ \frac{(26 - 21.9)^2}{21.9} = 0.7676 \]

  • For cell (1, 3): \[ \frac{(16 - 15.695)^2}{15.695} = 0.0059 \]

  • For cell (2, 1): \[ \frac{(66 - 61.595)^2}{61.595} = 0.315 \]

  • For cell (2, 2): \[ \frac{(34 - 38.1)^2}{38.1} = 0.4412 \]

  • For cell (2, 3): \[ \frac{(27 - 27.305)^2}{27.305} = 0.0034 \]

The sum of the Chi-Square components is: \[ \chi^2 = 0.5481 + 0.7676 + 0.0059 + 0.315 + 0.4412 + 0.0034 = 2.0812 \]

Step 3: Determine the Test Statistic and Critical Value

The Chi-Square test statistic is given by: \[ \chi^2 = 2.0812 \]

To determine the critical value for the Chi-Square distribution at a significance level of \( \alpha = 0.1 \) with \( df = 2 \), we find: \[ \chi^2_{\alpha, df} = \chi^2_{(0.1, 2)} = 4.6052 \]

Step 4: Calculate the P-Value

The p-value associated with the Chi-Square statistic is calculated as: \[ P = P(\chi^2 > 2.0812) = 0.3532 \]

Step 5: Conclusion

Based on the calculated Chi-Square statistic, critical value, and p-value, we can summarize the results of the hypothesis test regarding the independence of laboratory performance and the status of taking the techniques course.

Final Answer

Part 1:

  • Expected frequencies:
    • Took the techniques course:
      • Involved in no incident: \( \approx 35.41 \)
      • Involved in exactly one incident: \( \approx 21.90 \)
      • Involved in \( 2+ \) incidents: \( \approx 15.70 \)
    • Didn't take the techniques course:
      • Involved in no incident: \( \approx 61.60 \)
      • Involved in exactly one incident: \( \approx 38.10 \)
      • Involved in \( 2+ \) incidents: \( \approx 27.31 \)

Part 2: (a) Type of test statistic: Chi-Square

(b) Value of the test statistic: \( \chi^2 \approx 2.08 \)

Thus, the final answers are:

  • Part 1: Expected frequencies filled in as described.
  • Part 2(a): Chi-Square
  • Part 2(b): \( \boxed{2.08} \)
Was this solution helpful?
failed
Unhelpful
failed
Helpful

Questions asked by other users

failedAnswered
It takes 29 days for the Moon to go through its cycle of full moon, half moon, new moon, half moon, and then back to full moon. The function P models the percentage of the Moon that is visible t days after the cycle begins, where t=0 corresponds to a full moon on the night of June 5, 2020. Based on the model: which of the following best describes how the percentage of the Moon that is visible is changing on the night of June 25, 2020? It is increasing by 8.3% per day. It is increasing by 10.1% per day. It is increasing by 31.5% per day. It is decreasing by 8.3% per day. It is decreasing by 10.1% per day.
failedAnswered
A spring stretches by 0.0154 m when a 4.91-kg object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is f=5.52 Hz?
failedAnswered
Which type of DNA mutation occurs when a base is removed? deletion substitution insertion none of these
failedAnswered
In the knee joint, the femur articulates with the □ and the patella.
failedAnswered
19 ft = yd