91影视

n $=20$

c $=90\%=0.90$

The formula used: $E=t_{a/2}脳\frac{s}{\sqrt{n}}$

The three conditions are: Random, independent, and Normal/ Large sample

Random: Because the sample of $20$kids represents less than $10\%$of the total student population, I am satisfied.

Independent: pleased, because the sample of $20$students represents less than $10\%$of the total student population.

Normal/ Large sample: Because the pattern in the normal quantile plot is essentially linear, I'm satisfied. This means the distribution is close to Normal.

Because all of the conditions have been met, it is appropriate to calculate a confidence interval for the population mean. Interval of confidence

The mean is

$$\begin{gathered} \overline{x}=\frac{340}{20} \\ \overline{x}=17 \end{gathered}$$

The variance is

$s=5.3680$

In the table of the student's T distribution, look for the t-value in the row starting with degrees of freedom $df=n鈭�1=20鈭�1=19$and in the column with $C=90$percent:

$t_{a/2}=1.729$

The margin of error is

$$\begin{gathered} E=t_{a/2}脳\frac{s}{\sqrt{n}} \\ =1.729脳\frac{5.3680}{\sqrt{20}} \\ =2.0754 \end{gathered}$$

The boundaries of the confidence interval

$$\begin{gathered} \overline{x}鈭�E=17鈭�2.0754=14.9246 \\ \overline{x}+E=17+2.0754=19.0754 \end{gathered}$$

There are $90\%$confident that the true mean vertical jump of students at this school is between$14.9246$inches and $19.0754$inches.

From the Result part (a):

$(14.9246,19.0754)$

The confidence interval contains the value $15$indicating that the mean vertical distance is likely to be $15$and so failing to reject the assertion that the mean is $15$There is insufficient data to support the allegation that the true mean vertical distance of this school's students is less than $15$inches. It is observed that we made the same conclusion as in the previous exercise.