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SIZIF [17.4K]
2 years ago
7

Help please

Mathematics
1 answer:
hoa [83]2 years ago
7 0
1.) One hundred twenty-five and forty-seven thousandths

2.) 91/100.

3.) 2769/1000

4.) 0.074

5.) 3.705

6.) 75.69

7.) 0.162 is greater than 0.07

8.) 8.049 is greater than 8.0094

9.) 6

10.) 5.678

11.) 5.7



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Use the standard norml distribution or the t-distribution to construct a 99% confidence interval for the population mean. Justif
dalvyx [7]

Answer:

E) we will use t- distribution because is un-known,n<30

the confidence interval is (0.0338,0.0392)

Step-by-step explanation:

<u>Step:-1</u>

Given sample size is n = 23<30 mortgage institutions

The mean interest rate 'x' = 0.0365

The standard deviation 'S' = 0.0046

the degree of freedom = n-1 = 23-1=22

99% of confidence intervals t_{0.01} =2.82  (from tabulated value).

The mean value = 0.0365

x±t_{0.01} \frac{S}{\sqrt{n-1} }

0.0365±2.82 \frac{0.0046}{\sqrt{23-1} }

0.0365±2.82 \frac{0.0046}{\sqrt{22} }

0.0365±2.82 \frac{0.0046}{4.690 }

using calculator

0.0365±0.00276

Confidence interval is

(0.0365-0.00276,0.0365+0.00276)

(0.0338,0.0392)

the mean value is lies between in this confidence interval

(0.0338,0.0392).

<u>Answer:-</u>

<u>using t- distribution because is unknown,n<30,and the interest rates are not normally distributed.</u>

4 0
3 years ago
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