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3 years ago

All analysis of variance procedures require that the compared populations have equal variances.

Business

1 answer:

nikdorinn [45]3 years ago

7 0

Answer: b. False

All analysis of variance procedures <em><u>assume</u></em> that the compared populations have equal variances.

In all analysis of variance procedures, tests like F-test, Bartlett’s test, Levene’s test and Brown-Forsythe test are used to verify or test the assumption if k samples are from populations with equal variances.

When two or more populations have equal variances, we say that homoscedasticity or homogeneity of variances exist.

The F-test and Bartlett’s test yield best results only if the population is normally distributed. However, Levene’s test and Brown-Forsythe are known to yield good results for data that is not normally distributed.

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A produce distributor uses 776 packing crates a month, which it purchases at a cost of $9 each. The manager has assigned an annu

Westkost [7]

Answer:

$261.42

Explanation:

economic order quantity (EOQ) = √(2SD/H)

S = cost per order = $31

D = annual demand = 776 x 12 = 9,312

H = holding cost = $9 x 36% = $3.24

EOQ = √[(2 x $31 x 9,312) / $3.24] = √178,192.59 = 422.13 ≈ 422

total ordering and holding costs considering EOQ:

ordering costs = (9,312 / 422) x $31 = $684.06

holding costs = $3.24 x (422/2) = $683.64

total = $1,367.70

current costs:

ordering costs = $31 x 12 = $372

holding costs = $3.24 x (776/2) = $1,257.12

total = $1,629.12

annual savings = $1,629.12 - $1,367.70 = $261.42

3 0

3 years ago

Data for the two departments of Kimble & Pierce Company for June of the current fiscal year are as follows: Drawing Departme

Vitek1552 [10]

Answer:

Direct material equivalent units = 60,000

Conversion equivalent units = 61,200

Explanation:

The direct materials and conversion equivalent units of production is shown below:-

Winding Department

Whole Units Direct Materials Conversion

Equivalent unit Equivalent units

Starting Inventory 4000 0 2,600

in process (a) (4,000 × 65%)

June Started and 56,500 56,500 56,500

completed (b) (60,500-4,000)

Transferred to 60,500 56,500 59,100

finished goods (a+b)

Ending Inventory 3,500 3,500 2,100

in process (3,500 × 60%)

Total 64,000 60,000 61,200

8 0

3 years ago

On December 31, 2011, Colonial Corporation had the following account balances related to credit sales and receivables prior to r

kow [346]

Answer

Colonial corporation

Journal entries

A. $2,150 uncollectible of the year end Accounts receivable balance.

Debit Uncollectible Account with $2,150

Credit Account Receivable Account with $2,150

(Being aged debt that is uncollectible at year end)

Debit Allowance for doubtful debt with $200

Credit Uncollectible debt Account with $200

(Being reversal of allowance for doubtful debt no longer required after ageing review of debt)

B. 1% of credit sales is uncollectible

1% of $400,000 = $4,000

Debit Uncollectible Account with $4,000

Credit Account Receivable Account with $4,000

(Being 1% of credit sales that is uncollectible at year end)

Debit Allowance for doubtful debt with $200

Credit Uncollectible debt Account with $200

(Being reversal of allowance for doubtful debt no longer required after ageing review of debt)

C. (assuming a above but allowance for doubtful debt is a debit balance of $200)

i. $2,150 uncollectible of the year end Accounts receivable balance.

Debit Uncollectible Account with $2,150

Credit Account Receivable Account with $2,150

(Being aged debt that is uncollectible at year end)

ii. 1% of credit sales is uncollectible

1% of $400,000 = $4,000

Debit Uncollectible Account with $4,000

Credit Account Receivable Account with $4,000

(Being 1% of credit sales that is uncollectible at year end)

The debit receivable balance could relate to debts once written off but now being paid.

5 0

3 years ago

Mediation is useful when it puts the solution to a dispute in the hands of the

vfiekz [6]

Answer:

Jssj

Explanation:

3 0

3 years ago

Printed circuit cards are placed in a functional test after being populated with semiconductor chips. A lot contains 140 cards,

suter [353]

Answer:

a) 0.964

b) 0.543

Explanation:

Given data:

Total number of card is 140

A) if 20 chips are defective then

we have to find probability that at least one chip in sample is defective

120 chips are non defective

Defective chip denoted by D

hence Probability of at least one chip defective is P(D>1)

as P(D>1) = 1 - P(D=0)

$P(D=0) = \frac{\binom{20}{0}\binom{120}{20}}{\binom{140}{20}}$

P(D)=0.036

so P(D>1) = 1 - P(D=0) = 1-0.036 = 0.964

B) if 5 chips are defective then

we have to find probability that at least one chip in sample is defective

135 chips are non defective

we have to find P(D>1)

so, P(D>1) =1 - P(D=0)

$P(D=0) = \frac{\binom{5}{0}\binom{135}{20}}{\binom{140}{20}}$

P(D=0) = 0.457

so, $P(D\leq 1) = 1 - 0.457 = 0.543$

5 0

3 years ago