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enyata [817]
4 years ago
15

Erin has previously recorded all credit card activity manually using the Expense transaction screen and reconciled the account u

sing the Reconciliation Tool. After connecting her credit card in the Banking Center, she doesn’t see any matches for the transactions she previously entered and reconciled.
Mathematics
1 answer:
Ira Lisetskai [31]4 years ago
4 0

Answer:

The steps Erin has to take for the reconciliation of her account and activities is as follows: Select the reconciled transactions, Select Batch actions, and Modify the selected ones.

Step-by-step explanation:

Solution

Since Erin could not detect any matches for the transactions she has entered before and enrolled, she needs to take the following processes to reconcile back all her credit activities which is stated below:

Process 1 :Select the reconciled transactions

Process 2 :Batch Actions

Process 3: Modify Selected

From the process stated above Erin can first of all choose the reconciled transactions, after that she can select the batch actions and lastly modify the ones that was selected with the aim of putting or adding them back in the account reconciliation.

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We conclude that the mean waiting time is less than 10 minutes.

Step-by-step explanation:

We are given that a public bus company official claims that the mean waiting time for bus number 14 during peak hours is less than 10 minutes.

Karen took bus number 14 during peak hours on 18 different occasions. Her mean waiting time was 7.8 minutes with a standard deviation of 2.5 minutes.

Let \mu = <u><em>mean waiting time for bus number 14.</em></u>

So, Null Hypothesis, H_0 : \mu \geq 10 minutes      {means that the mean waiting time is more than or equal to 10 minutes}

Alternate Hypothesis, H_A : \mu < 10 minutes    {means that the mean waiting time is less than 10 minutes}

The test statistics that would be used here <u>One-sample t test statistics</u> as we don't know about the population standard deviation;

                       T.S. =  \frac{\bar X-\mu}{\frac{s}{\sqrt{n} } }  ~ t_n_-_1

where, \bar X = sample mean waiting time = 7.8 minutes

             s = sample standard deviation = 2.5 minutes

             n = sample of different occasions = 18

So, <u><em>test statistics</em></u> =  \frac{7.8-10}{\frac{2.5}{\sqrt{18} } }  ~ t_1_7

                              =  -3.734

The value of t test statistics is -3.734.

Now, at 0.01 significance level the t table gives critical value of -2.567 for left-tailed test.

Since our test statistic is less than the critical value of t as -3.734 < -2.567, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which <u>we reject our null hypothesis</u>.

Therefore, we conclude that the mean waiting time is less than 10 minutes.

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4 years ago
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