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. Constr What does the confidence interval tell about the population of all college students in the state? $$ 4.0,3.1,4.3,4.9,3.1,4.0,3.3,4.4,4.4,4.1,4.4,3.7,3.1,4.1,3.9 $$ ? 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. Constr What does the confidence interval tell about the population of all college students in the state? $$ 4.0,3.1,4.3,4.9,3.1,4.0,3.3,4.4,4.4,4.1,4.4,3.7,3.1,4.1,3.9 $$ ? What is the confidence interval for the population mean $$ \mu $$ ? $$ \square<\mu<\square $$ (Round to two decimal places as needed.)

UpStudy AI · Accepted Answer

We begin by computing the sample mean and standard deviation for the provided ratings

$$
4.0,\; 3.1,\; 4.3,\; 4.9,\; 3.1,\; 4.0,\; 3.3,\; 4.4,\; 4.4,\; 4.1,\; 4.4,\; 3.7,\; 3.1,\; 4.1,\; 3.9.
$$

**Step 1. Compute the Sample Mean**

The sample mean is given by

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

There are $$ n=15 $$ observations. Their sum is

$$
4.0+3.1+4.3+4.9+3.1+4.0+3.3+4.4+4.4+4.1+4.4+3.7+3.1+4.1+3.9=58.8.
$$

Thus, the sample mean is

$$
\bar{x}=\frac{58.8}{15}\approx 3.92.
$$

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

The sample standard deviation $$ s $$ is computed using

$$
s=\sqrt{\frac{\sum_{i=1}^{n}(x_i-\bar{x})^2}{n-1}}.
$$

First, we calculate each squared deviation:

$$
\begin{array}{lclcl}
(4.0-3.92)^2 & = & (0.08)^2 & \approx & 0.0064,\[1mm]
(3.1-3.92)^2 & = & (-0.82)^2 & \approx & 0.6724,\[1mm]
(4.3-3.92)^2 & = & (0.38)^2 & \approx & 0.1444,\[1mm]
(4.9-3.92)^2 & = & (0.98)^2 & \approx & 0.9604,\[1mm]
(3.1-3.92)^2 & = & (-0.82)^2 & \approx & 0.6724,\[1mm]
(4.0-3.92)^2 & = & (0.08)^2 & \approx & 0.0064,\[1mm]
(3.3-3.92)^2 & = & (-0.62)^2 & \approx & 0.3844,\[1mm]
(4.4-3.92)^2 & = & (0.48)^2 & \approx & 0.2304,\[1mm]
(4.4-3.92)^2 & = & (0.48)^2 & \approx & 0.2304,\[1mm]
(4.1-3.92)^2 & = & (0.18)^2 & \approx & 0.0324,\[1mm]
(4.4-3.92)^2 & = & (0.48)^2 & \approx & 0.2304,\[1mm]
(3.7-3.92)^2 & = & (-0.22)^2 & \approx & 0.0484,\[1mm]
(3.1-3.92)^2 & = & (-0.82)^2 & \approx & 0.6724,\[1mm]
(4.1-3.92)^2 & = & (0.18)^2 & \approx & 0.0324,\[1mm]
(3.9-3.92)^2 & = & (-0.02)^2 & \approx & 0.0004.
\end{array}
$$

Summing these squared deviations gives

$$
\sum (x_i-\bar{x})^2\approx 4.3240.
$$

The sample variance is then

$$
s^2=\frac{4.3240}{15-1}=\frac{4.3240}{14}\approx 0.3089,
$$

and therefore the sample standard deviation is

$$
s\approx \sqrt{0.3089}\approx 0.56.
$$

**Step 3. Compute the 95% Confidence Interval for $$\mu$$**

Since the sample size is small and the population standard deviation is unknown, we use the $$ t $$-distribution with $$ n-1=14 $$ degrees of freedom. For a 95% confidence level, the two-tailed critical value (approximately) is

$$
t^*\approx 2.15.
$$

The standard error is

$$
SE=\frac{s}{\sqrt{n}}=\frac{0.56}{\sqrt{15}}\approx\frac{0.56}{3.87}\approx 0.1447.
$$

The margin of error (ME) is

$$
ME=t^*\cdot SE \approx 2.15\times 0.1447 \approx 0.311.
$$

Thus, the 95% confidence interval for $$\mu$$ is

$$
\bar{x}\pm ME=3.92\pm 0.31,
$$

which gives

$$
3.92-0.31\approx 3.61 \quad \text{and} \quad 3.92+0.31\approx 4.23.
$$

We can now write the confidence interval as

$$
3.61<\mu<4.23.
$$

**Interpretation**

This confidence interval tells us that we are 95% confident that the true average student evaluation rating (on the 5-point scale) for all college students in the state falls between $$ 3.61 $$ and $$ 4.23 $$. In other words, if we were to take many samples similar to this one, approximately 95% of the intervals calculated from those samples would contain the true population mean $$ \mu $$.

The final answer is:

$$
\boxed{3.61<\mu<4.23.}
$$
The 95% confidence interval for the population mean rating is $$ 3.61 < \mu < 4.23 $$.

Solución de inteligencia artificial de Upstudy

Responder

Solución

Extra Insights

preguntas relacionadas

Latest Statistics Questions