Question

(Answers are given but could you please work through the problem and show me the steps?)

Chapter 10

Recently, a local newspaper reported that part-time students are older than full-time students. In order to test the validity of its statement, two independent samples of students were selected.

	Full-Time	Part-Time
$iz+QOmzF9rUdovVQAAAABJRU5ErkJggg==$ (in years)	26	24
s	2	3
n	42	31

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a.	Give the hypotheses for the above.
b.	Determine the degrees of freedom.
c.	Compute the test statistic.
d.	Using α = .05, test to determine whether or not the average age of part-time students is significantly more than full-time students.

ANSWER:

​ a. H0: μp - μf ≤ 0 Ha: μp - μf > 0 b. 49 c. test statistic t = -3.221 d. p-value (.0011) is less than .005, reject H0; the average age of part-time students is significantly more.

The following information is gathered from random samples of day and evening students regarding the number of semester hours they take.

​

	Day	Evening
$iz+QOmzF9rUdovVQAAAABJRU5ErkJggg==$	16	12
s	4	2
n	140	160

​

Develop a 95% confidence interval estimate for the difference between the mean semester hours taken by the two groups of students.

ANSWER:

3.269 to 4.731 hours

Answer 1

Solution 1

The provided sample means are shown below: X 24 Also, the provided sample standard deviations are: S1=2 82 = 3 and the sample sizes are n 42 and n2 31. (1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Ho: μι μο На: > н2 Ha: This corresponds to a right-tailed test, for which a t-test for two population means, with two independent samples, with unknown population standard deviations will be used. (2) Rejection Region

The degrees of freedom has been calculated using

2 + n2 (s/72) )2(3/n2) n1 1 n2-1

df = 49

Based on the information provided, the significance level is a 0.05, and the degrees of freedom are df 49.049. In fact, the degrees of freedom are computed as follows, asssuming that the population variances are unequal: Hence, it is found that the critical value for this right-tailed test is te = 1.677, for a = 0.05 and df = 49.049 1.677} The rejection region for this right-tailed test is R {t t (3) Test Statistics Since it is assumed that the population variances are unequal, the t-statistic is computed as follows: t 24 26 3.221 92 (4) Decision about the null hypothesis

p-value corresponding to t = -3.221 and df = 49 is 0.0011 (Obtained using online calculator. Screenshot attached)

-3.221 T Score: 49 DF. Significance Level: 01 05 10 One-tailed or two-tailed hypothesis?: One-tailed Two-tailed The p-value is.001 136.

Since p-value (0.0011) less than α = 0.05, we reject null hypothesis.

(5) Conclusion

Therefore, there is not enough evidence to claim that  the average age of part-time students is significantly more.

Solution 2

We need to construct the 95% confidence interval for the difference between the population means 1-2The following information has been provided about each of the samples: Sample Mean 1 (X, ) = 16 Pop. Standard Deviation 1 (o1) = 4 Sample Size 1 (N1) = 140 Sample Mean 2 (X2) = 12 Pop. Standard Deviation 2 (a)= 2 Sample Size 2 (N2) = 160 0.05 is 21-a/2 The critical value for a 1.96. The corresponding confidence interval is computed as shown below X-X2e CI 1 na na 42 22 16-12 1.96 140 42 22 .16-12 1.96 140 160 160 (3.269, 4.731) Therefore, based on the data provided, the 95% confidence interval for the difference between the population means that the true difference between population means is contained by the interval (3.269,4.731) 4.731, which indicates that we are 95% confident is 3.269 -<