Pryce Company owns equipment that cost $65,000 when purchased on January 1, 2012. It has been depreciated using the straight-lin
e method based on an estimated salvage value of $5,000 and an estimated useful life of 5 years. InstructionsPrepare Pryce Company's journal entries to record the sale of the equipment in these four independent situations. (a) Sold for $31,000 on January 1, 2015. (b) Sold for $31,000 on May 1, 2015. (c) Sold for $11,000 on January 1, 2015. (d) Sold for $11,000 on October 1, 2015
Monthly depreciation is calculated as follows: 65,000 (Cost) – 5,000 (Salvage Value) ÷ 60 (5 years X 12 months per year) = 1,000 in depreciation per month.
a) Accumulated depreciation from 1/1/12 to 1/1/15 is 36,000 (12 months for 2012, 2013, & 2014). Assuming that sale was a Cash sale, the journal entry would look like this: 1/1/15 Cash (DR) 31,000 Accumulated Depreciation (DR) 36,000 Equipment (CR) 65,000 Gain on Sale of Equipment (CR) 2,000
b) Accumulated depreciation from 1/1/12 to 5/1/15 is 40,000 (12 months for 2012, 2013, 2014, & 4 months for 2015). Assuming that sale was a Cash sale, the journal entry would look like this: 5/1/15 Cash (DR) 31,000 Accumulated Depreciation (DR) 40,000 Equipment (CR) 65,000 Gain on Sale of Equipment (CR) 6,000
c) Accumulated depreciation from 1/1/12 to 1/1/15 is 36,000 (12 months for 2012, 2013, & 2014). Assuming that sale was a Cash sale, the journal entry would look like this: 1/1/15 Cash (DR) 11,000 Accumulated Depreciation (DR) 36,000 Loss on Sale of Equipment (DR) 18,000 Equipment (CR) 65,000
d) Accumulated depreciation from 1/1/12 to 10/1/15 is 45,000 (12 months for 2012, 2013, 2014, & 9 months for 2015). Assuming that sale was a Cash sale, the journal entry would look like this: 10/1/15 Cash (DR) 11,000 Accumulated Depreciation (DR) 45,000 Loss on Sale of Equipment (DR) 9,000 Equipment (CR) 65,000
Based on the one-sample t-test that Mark is using, the two true statements are:
c.)The value for the degrees of freedom for Mark's sample population is five.
d.)The t-distribution that Mark uses has thicker tails than a standard normal distribution.
<h3>What are the degrees of freedom?</h3><h3 />
The number of subjects in the data given by Mark is 6 subjects.
The degrees of freedom can be found as:
= n - 1
= 6 - 1
= 5
This is a low degrees of freedom and one characteristic of low degrees of freedom is that their tails are shorter and thicker when compared to standard normal distributions.
Options for this question are:
a.)The t-distribution that Mark uses has thinner tails than a standard distribution.
b.)Mark would use the population standard deviation to calculate a t-distribution.
c.)The value for the degrees of freedom for Mark's sample population is five.
d.)The t-distribution that Mark uses has thicker tails than a standard normal distribution.
e.)The value for the degrees of freedom for Mark's sample population is six.
