Question

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level.

Step 1 of 2:

Suppose a sample of 16491649 tenth graders is drawn. Of the students sampled, 379379 read at or below the eighth grade level. Using the data, estimate the proportion of tenth graders reading at or below the eighth grade level. Enter your answer as a fraction or a decimal number rounded to three decimal places.

The state education commission wants to estimate the fraction of tenth grade students that have reading skills at or below the eighth grade level.

Step 2 of 2:

Suppose a sample of 16491649 tenth graders is drawn. Of the students sampled, 379379 read at or below the eighth grade level. Using the data, construct the 95%95% confidence interval for the population proportion of tenth graders reading at or below the eighth grade level. Round your answers to three decimal places.

A researcher compares the effectiveness of two different instructional methods for teaching pharmacology. A sample of 4949 students using Method 1 produces a testing average of 72.872.8. A sample of 3131 students using Method 2 produces a testing average of 72.372.3. Assume that the population standard deviation for Method 1 is 6.296.29, while the population standard deviation for Method 2 is 15.715.7. Determine the 90%90% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Step 1 of 3:

Find the point estimate for the true difference between the population means.

A researcher compares the effectiveness of two different instructional methods for teaching pharmacology. A sample of 4949 students using Method 1 produces a testing average of 72.872.8. A sample of 3131 students using Method 2 produces a testing average of 72.372.3. Assume that the population standard deviation for Method 1 is 6.296.29, while the population standard deviation for Method 2 is 15.715.7. Determine the 90%90% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Step 2 of 3:

Calculate the margin of error of a confidence interval for the difference between the two population means. Round your answer to six decimal places.

A researcher compares the effectiveness of two different instructional methods for teaching pharmacology. A sample of 4949 students using Method 1 produces a testing average of 72.872.8. A sample of 3131 students using Method 2 produces a testing average of 72.372.3. Assume that the population standard deviation for Method 1 is 6.296.29, while the population standard deviation for Method 2 is 15.715.7. Determine the 90%90% confidence interval for the true difference between testing averages for students using Method 1 and students using Method 2.

Step 3 of 3:

Construct the 90%90% confidence interval. Round your answers to one decimal place.

Answer 1

Solution:

Given

n= 1649 sample size of tenth grade students.

X = 379 students read at or below eight grade level.

. Level of significance .

g= 0,05

Step 1) To find the point estimate for the proportion of tenth grade readers reading at or below the eight grade level is

х 379 p= 0.2298362 n 1649

$\hat p= 0.230$. Sample proportion of tenth grade readers reading at or below the eight grade level.

= 1-1 = 0.770

Step 2) The 95% confidence interval for population proportion of tenth grade readers reading at or below the eight grade level is

$\hat p\pm Z_{\alpha/2}{\sqrt \frac{\hat p*(1-\hat p)}{n}}$

At

g= 0,05

$Z_{\alpha/2}=Z_{0.025}=1.96$

(0.230 +1.96 * 0.230 +0.770. 1649

$(0.230\pm 1.96*\sqrt \frac{0.1771}{1649})$

(0.230 +1.96 * V0.0001073

(0.230 +1.96 +0.0103585)

$(0.230\pm 0.0203027)$

(0.209697, 0.2503027)

( 0.210, 0.250)

The 95% confidence interval for population proportion of tenth grade readers reading at or below the eight grade level is ( 0.210, 0.250)

Given

n1= 49 sample size of method 1

n2= 31 sample size of method 2

$\bar x_1=72.8$ sample mean of method 1

$\bar x_2=72.3$. Sample mean of method 2

$\sigma_1= 6.29$. Population standard deviations of method 1

$\sigma_2= 15.7$ population standard deviations of method 2

$\alpha=0.10$. Level of significance.

Step 1 ) To find the point estimate for the true difference between the population mean is

$Point \ estimate =\bar x_1-\bar x_2$

Point estimate = 0.5

Step 2) To find margin of error for the difference between two population means is

$M.E= Z_{\alpha/2}*\sqrt \frac{\sigma_1^2}{n_1}+\frac{\sigma2^2}{n_2}$

At $\alpha=0.10$

$Z_{\alpha/2}=Z_{=0.05}=1.64$. From Z table

$M.E= 1.64 *\sqrt \frac{6.29^2}{49}+\frac{15.7^2}{31}$

$M.E= 1.64 *\sqrt \frac{39.5641}{49}+\frac{246.49}{31}$

$M.E= 1.64 *\sqrt 0.8074306+7.9512903$

$M.E= 1.64 *\sqrt 8.7587209$

Margin of error =M.E= 4.853602

Step 3) The 90% confidence interval for the true difference between testing average for students using method 1 and students using method 2 that is   is

1 112

$Point \ estimate \pm M.E$

$(0.5\pm 4.853602)$

( -4.3536023, 5.3536023)

( -4.4, 5.4)

The 90% confidence interval for the true difference between testing average for students using method 1 and students using method 2 is

( -4.4, 5.4)