Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent" The ratings were obtained at one university in a state. Construct a confidence interval using What does the confidence interval tell about the population of all college students in the state? $$ 3.6,30,4.0,4.6,3.1,4.5,3.4,4.4,4.6,4.0,4.6,3.7,3.0,3.7,40 $$ ? What is the confidence interval for the population mean $$ \mu $$ ? $$ \square

Question

Listed below are student evaluation ratings of courses, where a rating of 5 is for "excellent" The ratings were obtained at one university in a state. Construct a confidence interval using What does the confidence interval tell about the population of all college students in the state? $$ 3.6,30,4.0,4.6,3.1,4.5,3.4,4.4,4.6,4.0,4.6,3.7,3.0,3.7,40 $$ ? What is the confidence interval for the population mean $$ \mu $$ ? $$ \square<\mu<\square $$ (Round to two decimal places as needed)

UpStudy AI · Accepted Answer

Given the (corrected) data for ratings

$$
3.6,\;3.0,\;4.0,\;4.6,\;3.1,\;4.5,\;3.4,\;4.4,\;4.6,\;4.0,\;4.6,\;3.7,\;3.0,\;3.7,\;4.0,
$$

we have a sample size of $$ n=15 $$. (Note that the entries “30” and “40” are interpreted as $$ 3.0 $$ and $$ 4.0 $$ respectively so that all ratings fall on the scale with $$ 5 $$ representing “excellent”.)

**Step 1. Compute the Sample Mean**

The sample mean is given by

$$
\bar{x}=\frac{1}{n}\sum_{i=1}^{n} x_i.
$$

Summing the numbers

$$
3.6+3.0+4.0+4.6+3.1+4.5+3.4+4.4+4.6+4.0+4.6+3.7+3.0+3.7+4.0=58.2,
$$

so

$$
\bar{x}=\frac{58.2}{15} \approx 3.88.
$$

**Step 2. Compute the Sample Standard Deviation**

We first compute each squared deviation, $$ (x_i-\bar{x})^2 $$:

$$
\begin{array}{rcl}
(3.6-3.88)^2 & = & (-0.28)^2=0.0784,\[1mm]
(3.0-3.88)^2 & = & (-0.88)^2=0.7744,\[1mm]
(4.0-3.88)^2 & = & (0.12)^2=0.0144,\[1mm]
(4.6-3.88)^2 & = & (0.72)^2=0.5184,\[1mm]
(3.1-3.88)^2 & = & (-0.78)^2=0.6084,\[1mm]
(4.5-3.88)^2 & = & 0.62^2=0.3844,\[1mm]
(3.4-3.88)^2 & = & (-0.48)^2=0.2304,\[1mm]
(4.4-3.88)^2 & = & 0.52^2=0.2704,\[1mm]
(4.6-3.88)^2 & = & 0.72^2=0.5184,\[1mm]
(4.0-3.88)^2 & = & 0.12^2=0.0144,\[1mm]
(4.6-3.88)^2 & = & 0.72^2=0.5184,\[1mm]
(3.7-3.88)^2 & = & (-0.18)^2=0.0324,\[1mm]
(3.0-3.88)^2 & = & (-0.88)^2=0.7744,\[1mm]
(3.7-3.88)^2 & = & (-0.18)^2=0.0324,\[1mm]
(4.0-3.88)^2 & = & 0.12^2=0.0144.
\end{array}
$$

Summing these squared deviations gives

$$
0.0784 + 0.7744+0.0144+0.5184+0.6084+0.3844+0.2304+0.2704+0.5184+0.0144+0.5184+0.0324+0.7744+0.0324+0.0144 \approx 4.7840.
$$

The sample variance is calculated using

$$
s^2=\frac{\sum (x_i-\bar{x})^2}{n-1}=\frac{4.7840}{14}\approx 0.3417,
$$

and therefore the sample standard deviation is

$$
s\approx\sqrt{0.3417}\approx 0.585.
$$

**Step 3. Construct the 95% Confidence Interval**

The confidence interval for the population mean $$ \mu $$ is given by

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

where for a 95% confidence level and $$ n-1=14 $$ degrees of freedom, the critical value is approximately

$$
t_{0.025,14}\approx2.145.
$$

The standard error is

$$
\frac{s}{\sqrt{n}}=\frac{0.585}{\sqrt{15}}\approx\frac{0.585}{3.873}\approx0.151.
$$

The margin of error is

$$
E=2.145\times0.151\approx0.324.
$$

Thus, the confidence interval is

$$
3.88-0.324<\mu<3.88+0.324,
$$

or

$$
3.56<\mu<4.20.
$$

**Interpretation**

This 95% confidence interval means that we are 95% confident that the true mean student rating for courses among the population of all college students in the state lies between $$ 3.56 $$ and $$ 4.20 $$.

The final answer for the confidence interval is:

$$
\boxed{3.56<\mu<4.20.}
$$
The confidence interval for the population mean $$ \mu $$ is $$ 3.56 < \mu < 4.20 $$.

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