Question

A statistics teacher wanted to see whether there was a significant difference in ages between day students and night students. A random sample of 35 students is selected from each group. The data are given below. Day Students 1 22 1 29 20 20 24 18 24 20 23 24 21 23 27 26 23 21 22 17 30 19 18 22 19 25 19 23 18 25 21 20 18 21 Night Students 18 23 24 27 1925 25 31 - 24 23 24 27 21 20 23 22 20 20 23 23

Answer 1

a)

Answer:

.	Day
Mean	22
Std dev	3.1343
.	.
.	Night
Mean	23.2857
Std dev	3.2681

Explanation:

For calculation purposes, the point estimate of the population mean and the population standard deviation is obtained in excel. The screenshot is shown below,

x fc E3 A 1 Day - B Night =AVERAGE(A2:A36) E C D Mean Std dev Day Excel function 22 AVERAGE(A2:A11) 3.1343 STDEV(A2:A11) Mean Std dev Night 23.2857 AVERAGE(B2:B13) 3.2681 STDEV(B2:B13)

b)

Answer:

$\text{For day students, 95\% CI}=(20.9233,23.0767)$

$\text{For night students, 95\% CI}=(22.1631,24.4084)$

There is no difference in ages between two groups

Explanation:

For day students

The confidence interval for the mean is obtained using the following formula,

95% CI = T +tex

where,

The t critical value is obtained from t critical values table for significance level = 0.05 and degree of freedom = n -1 = 35 - 1 = 34

te = 2.0322

now,

3.1343 95% CI = 22 + 2.0322 x V35

$\text{95\% CI}=(20.9233,23.0767)$

For night students

$\text{95\% CI}=23.2857\pm 2.0322\times \frac{3.2681}{\sqrt{35}}$

$\text{95\% CI}=(22.1631,24.4084)$

Conclusion:

Since both interval overlap, we can conclude that there is no difference in ages between two groups.

c)

Answer:

Explanation:

The 95% Confidence Interval for the difference in means is obtained using the following formula,

$\text{95\% CI}=(\overline{X}_1-\overline{X}_2)\pm t\times SE_{\overline{X}_1-\overline{X}_2}$

Where,

SEY,-12 = ((nı - 1)sî +(n2 - 1)s nu + n2-2

$SE_{\overline{X}_1-\overline{X}_2}=\sqrt{\left ( \frac{(35-1)3.1343^2+(35-1)3.2681^2}{35+35-2} \right )\left ( \frac{1}{35}+\frac{1}{35} \right )}$

$SE_{\overline{X}_1-\overline{X}_2}=0.7654$

The t critical value is obtained from t critical values table for significance level = 0.05 and degree of freedom = n1+n2 -2 = 68.

$t_c=1.9955$

$\text{95\% CI}=22-23.2857\pm 1.9955\times 0.7654$

$\text{95\% CI}=(-2.8130,0.2416)$

Conclusion:

Since the 95% confidence interval for the difference in means includes 0, we can conclude that there is no difference in means.

d)

Answer:

There is no difference in the age between the two groups.

Explanation:

Two sample t-test is used to compare the means assuming equal variance.

The test is performed in the following steps,

Step 1: The Null and Alternative Hypotheses

$H_0:\mu_1=\mu_2$

$H_0:\mu_1\neq\mu_2$

This is a two-tailed test

Step 2: The t statistic is used to compare the two population means and the significance level for the test is

g= 0,05

The decision rules state the conditions that if,

,

P-value <a = 0.05, reject the null hypothesis,

Step 3: The t statistic is obtained using the formula,

( + ) (Trio 2014 (- ) X-TX

$t=\frac{22-3.1343}{0.7654}$

$t=-1.6798$

Step 3:

The P-value for the t statistic is obtained from t distribution table for the degree of freedom = n1+n2-2=35+35-2=68. (In excel use function =T.DIST.2T(1.6798,68))

$\text{P-value}=0.0976$

Step 4:

Since the P-value= 0.0976 is greater than 0.05 at the 5% significance level for the two-sided alternative hypothesis, the null hypothesis is not rejected. Hence, it can be concluded that there is no difference in the age between the two groups.

e)

The 95% confidence intervals for the mean age for the day students and the night students are overlapping hence there is no difference in the mean ages of two groups.

The 95% confidence intervals for the difference in mean age for the day students and the night students include zero which indicates that there is no difference in the mean ages of two groups.

The hypothesis test is also performed to validate the above result which tells that there is no sufficient evidence to conclude that the mean ages of two groups are different.