Condense the expression into a

Mathematics

1 answer:

liberstina [14]1 year ago

7 0

$2(\log 18-\log 3)+\dfrac 12 \log \dfrac{1}{16}=2 \log \dfrac{18}{3}+\log \sqrt{\dfrac{1}{16}}\\=2 \log 6+\log \dfrac{1}{4}=\log 6^2+\log \dfrac{1}{4}=\log 36+\log \dfrac{1}{4}\\=\log \dfrac{36}{4}=\log 9$

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The National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time requir

Paha777 [63]

Answer:

95% confidence interval for the mean time required to earn a bachelor’s degree by all college students is [5.10 years , 5.20 years].

Step-by-step explanation:

We are given that the National Center for Education Statistics surveyed a random sample of 4400 college graduates about the lengths of time required to earn their bachelor’s degrees. The mean was 5.15 years and the standard deviation was 1.68 years respectively.

Firstly, the pivotal quantity for 95% confidence interval for the population mean is given by;

P.Q. = $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ ~ N(0,1)

where, $\bar X$ = sample mean time = 5.15 years

$\sigma$ = sample standard deviation = 1.68 years

n = sample of college graduates = 4400

$\mu$ = population mean time

Here for constructing 95% confidence interval we have used One-sample z test statistics although we are given sample standard deviation because the sample size is very large so at large sample values t distribution also follows normal.

So, 95% confidence interval for the population mean, $\mu$ is ;

P(-1.96 < N(0,1) < 1.96) = 0.95 {As the critical value of z at 2.5%

level of significance are -1.96 & 1.96}

P(-1.96 < $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$ < 1.96) = 0.95

P( $-1.96 \times {\frac{\sigma}{\sqrt{n} } }$ < ${\bar X-\mu}$ < $1.96 \times {\frac{\sigma}{\sqrt{n} } }$ ) = 0.95