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rosijanka [135]
3 years ago
13

You interview a random sample of 50 adults. the results of the survey show that 46​% of the adults said they were more likely to

buy a product when there are free samples. at alphaequals0.05​, can you reject the claim that at least 51​% of the adults are more likely to buy a product when there are free​ samples?
Business
1 answer:
photoshop1234 [79]3 years ago
7 0

Answer:

z=\frac{0.46 -0.51}{\sqrt{\frac{0.51(1-0.51)}{50}}}=-0.707  

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who said they were more likely to buy a product when there are free samples is higher or equal than 0.51

Explanation:

Data given and notation

n=50 represent the random sample taken

\hat p==0.46 estimated proportion of adults who said they were more likely to buy a product when there are free samples

p_o=0.51 is the value that we want to test

\alpha=0.05 represent the significance level

Confidence=95% or 0.95

z would represent the statistic (variable of interest)

p_v represent the p value (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to test the claim that the true proportion is at least 0.51 or no.:  

Null hypothesis:p \geq 0.51  

Alternative hypothesis:p < 0.51  

When we conduct a proportion test we need to use the z statistic, and the is given by:  

z=\frac{\hat p -p_o}{\sqrt{\frac{p_o (1-p_o)}{n}}} (1)  

The One-Sample Proportion Test is used to assess whether a population proportion \hat p is significantly different from a hypothesized value p_o.

Calculate the statistic  

Since we have all the info requires we can replace in formula (1) like this:  

z=\frac{0.46 -0.51}{\sqrt{\frac{0.51(1-0.51)}{50}}}=-0.707  

Statistical decision  

It's important to refresh the p value method or p value approach . "This method is about determining "likely" or "unlikely" by determining the probability assuming the null hypothesis were true of observing a more extreme test statistic in the direction of the alternative hypothesis than the one observed". Or in other words is just a method to have an statistical decision to fail to reject or reject the null hypothesis.  

The significance level provided \alpha=0.05. The next step would be calculate the p value for this test.  

Since is left tailed the p value would be:  

p_v =P(z  

So the p value obtained was a very high value and using the significance level given \alpha=0.05 we have p_v>\alpha so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can said that at 5% of significance the proportion of adults who said they were more likely to buy a product when there are free samples is higher or equal than 0.51

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You deposited​ ($1,000) in a savings account that pays 8 percent​ interest, compounded​ quarterly, planning to use it to finish
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Answer:

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Answer:

a.

                                                                       Debit   Credit

December 31, 2017

Lease Equipment Under Capital Leases    $166,794  

                                                      Lease Liability    $166,794

December 31, 2017/January 1, 2018

Lease Liability                                        $40,000  

                                                         Cash             $40,000

b.                                           Debit               Credit

December 31, 2018

Depreciation Expense  $23,828  

          Accumulated Depreciation      $23,828

December 31, 2018/January 1, 2019

Interest Expense           $12,679  

Lease Liability          $27,321  

                           Cash                     $40,000

c.                                             Debit     Credit

December 31, 2019

Depreciation Expense        $23,828  

  Accumulated Depreciation  $23,828

December 31, 2019/January 1, 2020

Interest Expense                    $9,947  

Lease Liability                 $30,053  

                Cash                         $40,000

d. Balance Sheet

December 31,2019

Property Plant and Equipment                             Current Liabilities  

Leased Equipment Under Capital Leases $166,794 Lease Liability $33,058

Less Accumulated Depreciation $47,656  

                                                        $119,138                Long Term  

                                                                                      Lease Liability $36,362

Explanation:

a. The journal entries, that should be recorded on January 1, and December 31, 2017, by Steel would be as follows:

                                                                       Debit   Credit

December 31, 2017

Lease Equipment Under Capital Leases    $166,794  

                                                      Lease Liability    $166,794

December 31, 2017/January 1, 2018

Lease Liability                                        $40,000  

                                                         Cash             $40,000

Lease Equipment Under Capital Leases=(40,000*PVIFA(10%,Years = 40,000*4.16986))= $166,794  

b. The journal entries, that should be recorded on January 1 and December 31, 2018, by Steel would be as follows:

                                          Debit               Credit

December 31, 2018

Depreciation Expense  $23,828  

          Accumulated Depreciation      $23,828

December 31, 2018/January 1, 2019

Interest Expense           $12,679  

Lease Liability          $27,321  

                           Cash                     $40,000

Depreciation Expense= (166,794/7)=$23,828

Interest Expense [(166,794 - 40,000)*10%]=$12,679  

Lease Liability=(40,000 - 12,679)=$27,321

c. The journal entries, that should be recorded on January 1, and December 31, 2019, by Steel would be as follows:

                                            Debit     Credit

December 31, 2019

Depreciation Expense        $23,828  

  Accumulated Depreciation  $23,828

December 31, 2019/January 1, 2020

Interest Expense                    $9,947  

Lease Liability                 $30,053  

                Cash                         $40,000

d. The amounts that would appear on Steel's December 31, 2019, balance sheet relative to the lease arrangement would be as follows:

Balance Sheet

December 31,2019

Property Plant and Equipment                             Current Liabilities  

Leased Equipment Under Capital Leases $166,794 Lease Liability $33,058

Less Accumulated Depreciation $47,656  

                                                        $119,138                Long Term  

                                                                                      Lease Liability $36,362

8 0
4 years ago
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