At an assembly plant for light trucks, routine monitoring of the quality of welds yields the following data: Number of Welds
High Moderate Low
Quality Quality Quality
Day Shift
Evening Shift
Night Shift 467 191 42
445 171 34
254 129 17
Can you conclude that the quality varies among shifts?
a. State the appropriate null hypothesis.
b. Compute the expected values under the null hypothesis.
c. Compute the value of the chi-square statistic.
d. Find the P-value. What do you conclude?

Kindly check explanation

Step-by-step explanation:

H0 : quality does not vary among shift

H1 : quality varies among shift

The expected values :

(Row * column) / grand total

Given the data:

Observed values :

total

467 191 42 __700

445 171 34 __650

254 129 17 _ 400

Total __ 1166 491 93 _ 1750

Expected value count using the formula :.

Expected Values:

466.4 196.4 37.2

433.086 _182.371___ 34.5429

266.514_ 112.229 21.2571

The Chisquare statistic (χ²) :

χ² = (observed - Expected)²/ observed

χ² = 5.76045

The degree of freedom = (row - 1) * (column - 1)

Degree of freedom = (3-1)*(3-1) = 2*2 = 4

Pvalue = 0.2178

Pvalue > α ; We foal to reject H0 ; Hence, we conclude that quality does not vary among shift

amount of oatypop that contains the prize = 40%

amount   of oatypop which doesn't contain the prize = (100-40)=60%

probability that hannah will win a prize

    = s +f s +f f s+ff   f s+ff f f s+ infinity.where f= failure,s= success

=   or 60%

probability that hannah only has to buy 3 or less boxes before getting a prize= (s) and ( f s) and (ff s) and (f f f s)

        = = 87.04 %  

step-by-step explanation:

step-by-step explanation:

mean or average is 40.6