Question

both questions plz

14. Generation Gap: Education Education influences attitude and lifestyle. nowDifferences in education are a big factor in the "generation gap." Is the 2 Yeoyounger generation really better educated! Large surveys of people age 65 and older were taken in n, = 32 U.S. cities. The sample mean for these cities id 15.2% of the older adults had attended college. Large showed that x, surveys of young adults (age 25-34) were taken in n, sample mean for these cities showed that x, 19.7 % of the young adults had attended college. From previous studies, it is known that o 35 U.S. cities. The be 7.2% and ao,= 5.2% (Reference: American Generations, S. Mitchell). nte meb (a) Does this information indicate that the population percentage of 0.05. mean young adults who attended college is higher? Use a (b) Find a 90% confidence interval for u, -. Explain the meaning of the confidence interval in the context of the problem. Education: Tutoring In the article cited in Problem 17, the results of the following experiment Test was administered to both an experimental group and a control group after 6 weeks of instruction during which the experimental group received peer tutoring and the control group did not. For the experimental group with 30 children, the mean score on the vocabulary portion of the test was x, 368.4 with sample standard deviation s the vocabulary portion of the test for the n, 18. reported: Form 2 of the Gates-MacGintie Reading were n. 39.5. The average score on 30 subjects in the control group 349.2, with sample standard deviation s, 56.6. (a) Use a 1% level of significance to test the claim that the experimental group performed better than the control group. (b) Find a 98% confidence interval for u, -. Explain the meaning of the confidence interval in the context of the problem. was Please provide the following information for Problems 11-22, part (a): (i) What is the level of significance? State the null and alternate hypotheses. (ii) Check Requirements What sampling distribution will you use? What assumptions Compute the corresponding z or t value as appropriate. (iii) Find (or estimate) the P-value. Sketch the sampling distribution and show the area corresponding to the P-value. (iv) Based on your answers in parts (i)-(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level a? (v) Interpret your conclusion in the context of the application. Note: For degrees of freedom d.f.not in the Student's table that is smaller. In some situations, this choice of d.f. may increase the P-value a small amount, and therefore produce a slightly Answers may vary due to rounding. are you making? What is the value of the sample test statistic? use the closest d.f. more "conservative" answer.

Answer 1

14)

a) Two sample t test is used to compare the means assuming equal variance. The test is performed in following steps,

Step i: The Null and Alternative Hypotheses

0 1t= T11 : OH

$H_1: \mu_1<\mu_2$

The significance level is for the test .

0.05

Step ii:

The sample follows the normal distribution.

Assumption: the sampling distribution of mean should be approximately normal. And a sample distribution can be approximate to its population if the sample size is sufficiently large such that

n30

Also sampling distribution requires independence of the observations in the sample

test-statistic

The t-statistic is computed as follows:

X- Х2 ( (пу-1):7 +(пz-1)5 (1 1 + п2 п п+n2-2

From the data given,

Age	mean	n	s
>65	15.2	32	7.2
25-34	19.7	35	5.2

15.2 19.7 -2.951 (32-1)7.22 +(35-1)5.22 (1 ) 32 32+35-2 35

Step iii: Since, the P-value is obtained from t distribution table for the t-statistic = -2.951 and

degree of freedom, df = n1+n2-2= 32+35-2=65

P-value 0.0022

Step iv: Since the P-value is less than 0.05 at 5% significant level, it can be concluded that the null hypothesis is rejected.

Step v: There is a significantly better educated younger generation.

b)

The confidence interval for the difference in means is obtained as follow,

90% CI (X1-X2) ±tx SE

Where,

(n1 1)s(n2-1)s31 1 Standard Error, SE - 1.525 + n1n2 2 n2

The critical value for this left-tailed test is obtained from t critical value table for and degree of freedom, df = n1+n2-2= 32+35-2=65

0.05

1.669

$\text{90\% CI}=\left (15.2-19.7 \right )\pm 1.669\times 1.525$

90% CI (-7.044, -1.956]

18)

a) Two sample t test is used to compare the means assuming equal variance. The test is performed in following steps,

Step i: The Null and Alternative Hypotheses

0 1t= T11 : OH

$H_1: \mu_1>\mu_2$

The significance level is for the test $\alpha=0.01$.

Step ii:

The sample follows the normal distribution.

Assumption: the sampling distribution of mean should be approximately normal. And a sample distribution can be approximate to its population if the sample size is sufficiently large such that

n30

Also sampling distribution requires independence of the observations in the sample

test-statistic

The t-statistic is computed as follows:

X- Х2 ( (пу-1):7 +(пz-1)5 (1 1 + п2 п п+n2-2

From the data given,

	mean	n	s
Experimental group	368.4	30	39.5
Control group	349.2	30	56.6

$t=\frac{368.4-349.2}{\sqrt{\frac{(30-1)39.5^2+(30-1)56.6^2}{30+30-2}\left ( \frac{1}{30}+\frac{1}{30} \right )}}=1.524$

Step iii: Since, the P-value is obtained from t distribution table for the t-statistic = 1.524 and

degree of freedom, df = n1+n2-2= 30+30-2=58

$\text{P-value}=0.0665$

Step iv: Since the P-value is greater than 0.01 at 1% significant level, it can be concluded that the null hypothesis is not rejected.

Step v: There is no difference in experimental and control group mean.

b)

The confidence interval for the difference in means is obtained as follow,

$\text{98\% CI}=\left ( \overline{X}_1-\overline{X}_2 \right )\pm t\times SE$

Where,

(n1 1)s(n2-1)s31 1 Standard Error, SE = 12.601 + _ n1n2 2 n2

The critical value for this left-tailed test is obtained from t critical value table for $\alpha=0.02$ and degree of freedom, df = n1+n2-2= 30+30-2=58

$t=2.392$

$\text{98\% CI}=\left (368.4-349.2 \right )\pm 2.392\times 12.601$

$\text{98\% CI}=\left ( -10.947,49.347 \right )$