Question

1) Two hundred randomly selected women in a certain city, including those who are working, were asked the time they spend in the beauty salon for each visit. The average time was 44 minutes with a standard deviation of 16 minutes. Construct an 80% confidence interval for the mean time spent in the salon by all the women in the city.

a.[43.05, 44.95]

b.[43.76, 44.24]

c.[42.55, 45.45]

d.[43.64, 44.36]

e.[42.45, 45.55]

2) In order to estimate the mean number of books purchased by customers, a bookstore samples 64 customers and obtains a sample mean of 6 books with a sample standard deviation of 2. Construct a 95% confidence interval for the mean number of books purchased by all the customers visiting the store.

a.[5.51, 6.49]

b.[5.71, 6.29]

c.[5.59, 6.41]

d.[5.65, 6.35]

e.[5.50, 6.50]

Answer 1

Solution : ( 1 )

$\mathrm{c.\quad\:}\left ( 42.55,\quad\:45.45 \right )$

$\mathrm{Explanation\::}$

$\mathrm{80\%\:confidence\:interval\:for\:the\:mean\:time\:spent\:in\:the\:salon\:by\:all\:the\:women\:in\:the\:city\::}$

$=\left ( 44-1.28* \frac{16}{\sqrt{200}},\quad\:44+1.28* \frac{16}{\sqrt{200}} \right )$

$=\left ( 44-1.28* \frac{2^4}{\sqrt{2^3*5^2}},\quad\:44+1.28* \frac{2^4}{\sqrt{2^3*5^2}} \right )$

$=\left ( 44-1.28* \frac{2^4}{\sqrt{2^3}\sqrt{5^2}},\quad\:44+1.28* \frac{2^4}{\sqrt{2^3}\sqrt{5^2}} \right )$

$=\left ( 44-1.28* \frac{2^4}{2^{\frac{3}{2}}*5},\quad\:44+1.28* \frac{2^4}{2^{\frac{3}{2}}*5} \right )$

$=\left ( 44-1.28* \frac{2^{\frac{5}{2}}}{5},\quad\:44+1.28* \frac{2^{\frac{5}{2}}}{5} \right )$

$=\left ( 44-1.28* \frac{2^2\sqrt{2}}{5},\quad\:44+1.28* \frac{2^2\sqrt{2}}{5} \right )$

$=\left ( 44-1.28* \frac{4\sqrt{2}}{5},\quad\:44+1.28* \frac{4\sqrt{2}}{5} \right )$

$=\left ( 44- \frac{1.28*4\sqrt{2}}{5},\quad\:44+ \frac{1.28*4\sqrt{2}}{5} \right )$

$=\left ( 44- \frac{5.12\sqrt{2}}{5},\quad\:44+ \frac{5.12\sqrt{2}}{5} \right )$

$=\left ( 44- \frac{5.12*1.414}{5},\quad\:44+ \frac{5.12*1.414}{5} \right )$

$=\left ( 44- \frac{7.24077}{5},\quad\:44+ \frac{7.24077}{5} \right )$

$=\left ( 44- 1.448154,\quad\:44+ 1.448154 \right )$

$=\left ( 42.551846,\quad\:45.448154 \right )$

$=\left ( 42.55,\quad\:45.45 \right )$​​​​​​​

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Solution : ( 2 )

$\mathrm{a.\quad\:}\left ( 5.51,\quad\:6.49 \right )$

$\mathrm{Explanation\::}$

$\mathrm{95\%\:confidence\:interval\:for\:the\:mean\:number\:of\:books\:purchased\:by\:all\:the\:customers\:visiting\:the\:store\::}$

$=\left ( 6-1.96* \frac{2}{\sqrt{64}},\quad\:6+1.96* \frac{2}{\sqrt{64}} \right )$

$=\left ( 6-1.96* \frac{2}{8},\quad\:6+1.96* \frac{2}{8} \right )$

$=\left ( 6-1.96* \frac{1}{4},\quad\:6+1.96* \frac{1}{4} \right )$

$=\left ( 6-\frac{1.96}{4},\quad\:6+\frac{1.96}{4} \right )$

$=\left ( 6-0.49,\quad\:6+0.49 \right )$

$=\left ( 5.51,\quad\:6.49 \right )$