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kirill115 [55]
3 years ago
8

The average local cell phone call length was reported to be 2.27 minutes. A random sample of 20 phone calls showed an average of

2.98 minutes in length with a standard deviation of 0.98 minutes.
At α = 0.05 can it be concluded that the average differs from the population average?
Report your answer to two decimal places.
Note: Find the test statistic ONLY to test this claim using an appropriate formula. You must show your work.
Mathematics
1 answer:
Afina-wow [57]3 years ago
6 0

Answer:

Yes, at α = 0.05 it can be concluded that the average differs from the population average.

Step-by-step explanation:

We are given that the average local cell phone call length was reported to be 2.27 minutes. A random sample of 20 phone calls showed an average of 2.98 minutes in length with a standard deviation of 0.98 minutes.

<em>Let </em>\mu<em> = population average local cell phone call length</em>

SO, Null Hypothesis, H_0 : \mu = 2.27 minutes  {means that the average is same as that of population average}

Alternate Hypothesis, H_a : \mu \neq 2.27 minutes  {means that the average differs from the population average}

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

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

where, \bar X = sample average call length = 2.98 minutes

            s = sample standard deviation = 0.98 minutes

            n = sample of phone calls = 20

So, <em><u>test statistics</u></em>  =  \frac{2.98-2.27}{\frac{0.98}{\sqrt{20} } }  ~ t_1_9

                               =  <u>3.24</u>

Now, at 0.05 level of significance the t table gives critical values between -2.093 and 2.093 at 19 degree of freedom for two-tailed test. Since our test statistics does not lie within these range of critical values so we have sufficient evidence to reject our null hypothesis as test statistics will fall in the rejection region.

Therefore, we conclude that the average differs from the population average.

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Account A and Account B both have a principal of $2,000 and an annual interest rate of 2%. No additional deposits or withdrawals
GuDViN [60]

Answer:

Account B earns more interest.

After 20 years, account B will have earned $171.89 more.

Step-by-step explanation:

Let's calculate the total for each account.

Account A:

Account A earns simple interest. We know that the principal value is $2000 and the interest rate is 2% or 0.02. We can use the simple interest formula:

A=P(1+rt)

Where A is the future value, P is the principal, r is the rate, and t is the time in years.

So, let's substitute 2000 for P, 0.02 for r, and 20 for t. This yields:

A=2000(1+0.02(20))

Multiply and add:

A=2000(1+0.4)=2000(1.4)

Multiply. So, the total amount of money in Account A after 20 years is:

A=\$2800

Since we initially deposited $2000 and our total is now $2800, this means that we earned an interest of 2800-2000=\$ 800

Account B:

Account B earns compound interest. Like Account A, Account B has a principal value of $2000 and the interest rate is 2% or 0.02. We also know that it's compounded annually, so once per year. We can use the compound interest formula:

B=P(1+\frac{r}{n}})^{nt}

Where B is the future value, P is the principal, r is the rate, n is the times compounded per year, and t is the time in years.

So, let's substitute 2000 for P, 0.02 for r, n for 1 (since it's compounded annually), and t for 20. This yields:

B=2000(1+\frac{0.02}{1})^{(1)(20)}

Simplify this to acquire:

B=2000(1.02)^{20}

Evaluate. Use a calculator. So, after 20 years, the amount of money in Account B is:

B\approx\$2971.89

Since our principal was $2000, this means that we earned an interest of approximately  2971.89-2000=\$ 971.89.

So, Account A earned an interest of $800 and Account B earned an approximate interest of $971.89.

So, Account B earned more interest.

And it earned 971.89-800=\$ 171.89 more than Account A.

And we're done!

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