Studies show that smaller class sizes have a direct impact on student success. Researchers selected a random sample of 67 elementary school classrooms across the United States and found that the mean class size was $$ \bar{x}_{e}=23.2 $$ students with a standard deviation of 1.3 students. They also selected a random sample of 33 high school classrooms across the United States and found that the mean class size was $$ \bar{x}_{h}=23.5 $$ with a standard deviation of 1.2 students. (a) Use the Compare Two Population Means tool to construct a $$ 95 \% $$ confidence interval to estimate the difference between the mean elementary class size and the mean high school class size $$ \mu_{e}-\mu_{h} $$. Round answers to 4 decimal places The confidence interval is

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We first note that the samples have

- Elementary: $$ n_e=67 $$, $$\bar{x}_e=23.2$$, $$s_e=1.3$$.
- High school: $$ n_h=33 $$, $$\bar{x}_h=23.5$$, $$s_h=1.2$$.

We wish to construct a $$95\%$$ confidence interval for
$$
\mu_e-\mu_h.
$$
The point estimate for the difference is
$$
\bar{x}_e-\bar{x}_h=23.2-23.5=-0.3.
$$

Because the two samples are independent and we are using a two-sample $$t$$-interval with possibly unequal variances (Welch’s method), the standard error (SE) is given by
$$
\text{SE}=\sqrt{\frac{s_e^2}{n_e}+\frac{s_h^2}{n_h}}.
$$

Substitute the values:
$$
\text{SE}=\sqrt{\frac{1.3^2}{67}+\frac{1.2^2}{33}}=\sqrt{\frac{1.69}{67}+\frac{1.44}{33}}.
$$

Calculating each term:
$$
\frac{1.69}{67}\approx0.02522,\qquad \frac{1.44}{33}\approx0.04364.
$$
Thus,
$$
\text{SE}=\sqrt{0.02522+0.04364}\approx\sqrt{0.06886}\approx0.2624.
$$

Next, we determine the degrees of freedom using the Welch–Satterthwaite approximation:
$$
\text{df}\approx\frac{\left(\frac{s_e^2}{n_e}+\frac{s_h^2}{n_h}\right)^2}{\frac{\left(\frac{s_e^2}{n_e}\right)^2}{n_e-1}+\frac{\left(\frac{s_h^2}{n_h}\right)^2}{n_h-1}}.
$$

Substituting values:
- For the elementary group:
  $$
  \frac{s_e^2}{n_e}=\frac{1.69}{67}\approx0.02522;\quad \left(0.02522\right)^2\approx0.0006363.
  $$
- For the high school group:
  $$
  \frac{s_h^2}{n_h}=\frac{1.44}{33}\approx0.04364;\quad \left(0.04364\right)^2\approx0.0019059.
  $$

Now,
$$
\text{Numerator}=\left(0.02522+0.04364\right)^2=\left(0.06886\right)^2\approx0.0047441.
$$
And the denominator is:
$$
\frac{0.0006363}{67-1}+\frac{0.0019059}{33-1}=\frac{0.0006363}{66}+\frac{0.0019059}{32}\approx0.00000964+0.00005956\approx0.00006920.
$$
Thus,
$$
\text{df}\approx\frac{0.0047441}{0.00006920}\approx68.52.
$$

We can use $$ \text{df}\approx68 $$ which yields a two-tailed $$ t $$-critical value for a $$95\%$$ confidence level of approximately $$ t^* \approx 2.000 $$ (this value is very close for large df).

Now, the margin of error (ME) is
$$
\text{ME}=t^*\times \text{SE}\approx2.000\times0.2624\approx0.5248.
$$

Thus, the $$95\%$$ confidence interval for $$ \mu_e-\mu_h $$ is:
$$
(-0.3-0.5248,\,-0.3+0.5248)=(-0.8248,\,0.2248).
$$

Rounding to 4 decimal places, the confidence interval is:
$$
\boxed{(-0.8248,\;0.2248)}.
$$
The $$95\%$$ confidence interval for the difference between the mean elementary class size and the mean high school class size is $$(-0.8248, 0.2248)$$.

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