Question

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3. A restaurant association collected information on the number of meals eaten outside the home per week by young married couples. A survey of 20 couples showed the sample mean number of meals eaten outside home was 2.76 meals per week, with a S.D. of 0.75 meals per week. Construct a 90% confidence interval for the population mean. Construct a 99% confidence interval for the population mean. If the population standard deviation is known to be 0.45 meals per week, construct the 90% confidence interval for the population mean. If the population standard deviation is known to be 0.45 meals per week, construct the 99% confidence interval for the population mean. a. b. c. d.

Answer 1

Ans(3) The data is given for the sample of 20 couples, and the sample mean $(\bar{x}=2.76)$ and the standard deviation $(\sigma=0.75)$ is as follows. Now using this sample data we need to construct confidence interval for the population mean.

(a) The formula to calculate the confidence interval for population mean is : $\bar{x}\pm t_{(\frac{\alpha}{2},n-1)}\frac{s}{\sqrt{n}}$ ,

sample mean, $\bar{x}=2.76$

t-value for 90% confidence interval is, $t_{\alpha/2,n-1}=t_{\frac{0.1}{2},19}=1.7291$

sample standard deviation, $s=0.75$

sample size, $n=20$

90% confidence interval for population mean$(\mu)$ :

Lower limit:

$\bar{x}-t_{(\frac{0.1}{2},19)}\frac{s}{\sqrt{n}}=2.76-1.7291\frac{0.75}{\sqrt{20}}=2.470021\approx 2.47$

Upper limit:      

$\bar{x}+t_{(\frac{0.1}{2},19)}\frac{s}{\sqrt{n}}=2.76+1.7291\frac{0.75}{\sqrt{20}}=3.049978\approx 3.05$

So the 90% confidence interval for the population mean $\mu$ (meals per week) is (2.47, 3.05), we can conclude that we are 90% confident that this interval will capture the true population mean, $\mu$

(b) Calculation for the 99% confidence interval for population mean $(\mu)$:

$\bar{x}\pm t_{(\frac{\alpha}{2},n-1)}\frac{s}{\sqrt{n}}$

t-value associated with the 99% confidence is, $t_{\alpha/2,n-1}=t_{\frac{0.01}{2},19}=2.8609$

Lower limit:  

$\bar{x}-t_{(\frac{0.01}{2},19)}\frac{s}{\sqrt{n}}=2.76-2.8609\frac{0.75}{\sqrt{20}}=2.280212\approx 2.28$

Upper limit:    

$\bar{x}+t_{(\frac{0.01}{2},19)}\frac{s}{\sqrt{n}}=2.76+2.8609\frac{0.75}{\sqrt{20}}=3.23978\approx 3.24$

So the 90% confidence interval for the population mean, $\mu$(meals per week) is (2.28, 3.24)

(c) Since we have given the value of population standard deviation$(\sigma=0.45)$, so the formula used for confidence interval is,   $\bar{x}\pm z\frac{\sigma}{\sqrt{n}}$

Calculation for the 90% confidence interval for population mean, when population standard deviation is given as $\sigma=0.45$.

$\bar{x}=2.76,\:\sigma=0.45,\:n=20$, z-value for 90% confidence is, $z=1.6449$

lower limit:

$\bar{x}-z\frac{\sigma}{\sqrt{n}}=2.76-1.6449\frac{0.45}{\sqrt{20}}=2.59448\approx 2.59$

Upper limit:

$\bar{x}+z\frac{\sigma}{\sqrt{n}}=2.76+1.6449\frac{0.45}{\sqrt{20}}=2.925514\approx 2.93$

So the 90% confidence interval for the population mean, $\mu$(meals per week) is (2.59, 2.93)

(d) Calculation for the 99% confidence interval for the population mean $(\mu)$:

z-value associated with 99% confidence is, $z=2.5758$

Lower limit:

$\bar{x}-z\frac{\sigma}{\sqrt{n}}=2.76-2.5758\frac{0.45}{\sqrt{20}}=2.500815\approx 2.50$

Upper limit:

$\bar{x}+z\frac{\sigma}{\sqrt{n}}=2.76+2.5758\frac{0.45}{\sqrt{20}}=3.01918\approx 3.02$

So the 90% confidence interval for the population mean, $\mu$ (meals per week) is (2.50, 3.02)