Chapter 8, Problem 101E

(a)

To determine

To find: The 95% confidence interval for question (a).

Answer to Problem 101E

Solution: The 95% confidence interval for question (a) is $\left(0.5278,0.5822\right)$.

Explanation of Solution

Calculation: The formula for confidence interval is defined as;

$\widehat{p}\pm m$

Where m is the margin of error which is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

$z^{*}$ is the value of the standard normal density curve and $S{E}_{\widehat{p}}$ is the standard error . The standard error is the estimate of the standard deviation of a statistic. The formula for standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ and sample size n is defined as:

$S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

The sample proportion is provided as $\widehat{p}=0.555$ and the sample size is $n=1280$ . Substitute the values in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{0.555\left(1-0.555\right)}{1280}}\\ =\sqrt{\frac{0.246975}{1280}}\\ =0.0139\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.96\times 0.0139\\ =0.027244\end{array}$

The 95% confidence interval for question (a) is obtained as;

$\begin{array}{c}0.555\pm 0.027244=\left(0.555-0.027244,0.555+0.027244\right)\\ =\left(0.5278,0.5822\right)\end{array}$

Interpretation: There is 95% confidence that between 52.78% and 58.22% of students feel burdened by their student loan payments.

(b)

To determine

To find: The 95% confidence interval for question (b).

Answer to Problem 101E

Solution: The 95% confidence interval for question (b) is $\left(0.5168,0.5712\right)$.

Explanation of Solution

Calculation: The formula for confidence interval is defined as;

$\widehat{p}\pm m$

Where m is the margin of error which is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

$z^{*}$ is the value of the standard normal density curve and $S{E}_{\widehat{p}}$ is the standard error . The standard error is the estimate of the standard deviation of a statistic. The formula for standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ and sample size n is defined as:

$S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

The sample proportion is provided as $\widehat{p}=0.544$ and the sample size is $n=1280$ . Substitute the values in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{0.544\left(1-0.544\right)}{1280}}\\ =\sqrt{\frac{0.248064}{1280}}\\ =0.0139\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.96\times 0.0139\\ =0.027244\end{array}$

The 95% confidence interval for question (b) is obtained as;

$\begin{array}{c}0.544\pm 0.027244=\left(0.544-0.027244,0.544+0.027244\right)\\ =\left(0.5168,0.5712\right)\end{array}$

Interpretation: There is 95% confidence that between 51.68% and 57.12% of students would like to borrow lesser loan if the loan begins again.

(c)

To determine

To find: The 95% confidence interval for question (c).

Answer to Problem 101E

Solution: The 95% confidence interval for question (c) is $\left(0.3169,0.3691\right)$.

Explanation of Solution

Calculation: The formula for confidence interval is defined as;

$\widehat{p}\pm m$

Where m is the margin of error which is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

$z^{*}$ is the value of the standard normal density curve and $S{E}_{\widehat{p}}$ is the standard error . The standard error is the estimate of the standard deviation of a statistic. The formula for standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ and sample size n is defined as:

$S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

The sample proportion is provided as $\widehat{p}=0.343$ and the sample size is $n=1280$ . Substitute the values in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{0.343\left(1-0.343\right)}{1280}}\\ =\sqrt{\frac{0.225351}{1280}}\\ =0.0133\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.96\times 0.0133\\ =0.026068\end{array}$

The 95% confidence interval for question (b) is obtained as;

$\begin{array}{c}0.343\pm 0.026068=\left(0.343-0.026068,0.343+0.026068\right)\\ =\left(0.3169,0.3691\right)\end{array}$

Interpretation: There is 95% confidence that between 31.69% and 36.19% of students disagreed that the education loans are not more financial hardship than expected at the time of taking loans.

(d)

To determine

To find: The 95% confidence interval for question (d).

Answer to Problem 101E

Solution: The 95% confidence interval for question (d) is $\left(0.5620,0.6160\right)$.

Explanation of Solution

Calculation: The formula for confidence interval is defined as;

$\widehat{p}\pm m$

Where m is the margin of error which is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

$z^{*}$ is the value of the standard normal density curve and $S{E}_{\widehat{p}}$ is the standard error . The standard error is the estimate of the standard deviation of a statistic. The formula for standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ and sample size n is defined as:

$S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

The sample proportion is provided as $\widehat{p}=0.589$ and the sample size is $n=1280$ . Substitute the values in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{0.589\left(1-0.589\right)}{1280}}\\ =\sqrt{\frac{0.242079}{1280}}\\ =0.0138\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.96\times 0.0138\\ =0.027048\end{array}$

The 95% confidence interval for question (b) is obtained as;

$\begin{array}{c}0.589\pm 0.027048=\left(0.589-0.027048,0.589+0.027048\right)\\ =\left(0.5620,0.6160\right)\end{array}$

Interpretation: There is 95% confidence that between 56.20% and 61.60% of students agreed that payment of loans is unpleasant even though the benefits of education loans are worthy.

(e)

To determine

To find: The 95% confidence interval for question (e).

Answer to Problem 101E

Solution: The 95% confidence interval for question (e) is $\left(0.5620,0.6160\right)$.

Explanation of Solution

Calculation: The formula for confidence interval is defined as;

$\widehat{p}\pm m$

Where m is the margin of error which is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

$z^{*}$ is the value of the standard normal density curve and $S{E}_{\widehat{p}}$ is the standard error . The standard error is the estimate of the standard deviation of a statistic. The formula for standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ and sample size n is defined as:

$S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

The sample proportion is provided as $\widehat{p}=0.589$ and the sample size is $n=1280$ . Substitute the values in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{0.589\left(1-0.589\right)}{1280}}\\ =\sqrt{\frac{0.242079}{1280}}\\ =0.0138\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.96\times 0.0138\\ =0.027048\end{array}$

The 95% confidence interval for question (b) is obtained as;

$\begin{array}{c}0.589\pm 0.027048=\left(0.589-0.027048,0.589+0.027048\right)\\ =\left(0.5620,0.6160\right)\end{array}$

Interpretation: There is 95% confidence that between 56.20% and 61.60% of students are satisfied with the investment for education loan which worth for career opportunities.

(f)

To determine

To find: The 95% confidence interval for question (f).

Answer to Problem 101E

Solution: The 95% confidence interval for question (f) is $\left(0.6903,0.7397\right)$.

Explanation of Solution

Calculation: The formula for confidence interval is defined as;

$\widehat{p}\pm m$

Where m is the margin of error which is defined as:

$m=z^{*}\times S{E}_{\widehat{p}}$

$z^{*}$ is the value of the standard normal density curve and $S{E}_{\widehat{p}}$ is the standard error . The standard error is the estimate of the standard deviation of a statistic. The formula for standard error $S{E}_{\widehat{p}}$ of sample proportion $\widehat{p}$ and sample size n is defined as:

$S{E}_{\widehat{p}}=\sqrt{\frac{\widehat{p}\left(1-\widehat{p}\right)}{n}}$

The sample proportion is provided as $\widehat{p}=0.715$ and the sample size is $n=1280$ . Substitute the values in the standard error formula. So,

$\begin{array}{c}S{E}_{\widehat{p}}=\sqrt{\frac{0.715\left(1-0.715\right)}{1280}}\\ =\sqrt{\frac{0.203775}{1280}}\\ =0.0126\end{array}$

The value of $z^{*}$ for 95% confidence level is $z^{*}=1.96$ from the standard normal table.

So, the margin of error is obtained as:

$\begin{array}{c}m=z^{*}\times S{E}_{\widehat{p}}\\ =1.96\times 0.0126\\ =0.024696\end{array}$

The 95% confidence interval for question (b) is obtained as;

$\begin{array}{c}0.715\pm 0.024696=\left(0.715-0.024696,0.715+0.024696\right)\\ =\left(0.6903,0.7397\right)\end{array}$

Interpretation: There is 95% confidence that between 69.03% and 73.97% of students are satisfied with the investment for student loan which worth for personal growth.