Question

(Computational) Applied Statistics
Problem 1: A noted medical researcher has suggested that a heart attack is less likely to occur among adults who actively participate in athletics. A random sample of 300 adults is obtained. Of that total, 100 are found to be athletically active. Within this group, 10 suffered heart attacks; among the 200 athletically inactive adults, 26 had suffered heart attacks.                   
a) Test the hypothesis that the proportion of adults who are active and sufferedheart attacks is different from the proportion of adults who are not active and suffered heart attacks.     
b) Construct a 95% confidence interval for the difference between the proportions of all active and inactive adults who suffered heart attacks.What can you conclude and why?

Problem 2: The data below refer to aluminum contents in soil at two different locations. Summary of the data is provided in the table below. You mayassume that the data are normally distributed.  
Location     n       sample mean    sample standard deviation
       1           5            2935                                235.7657
       2           4            2637                                741.8416
a) Give a 95% confidence interval for the mean aluminum contents at location 1.
b) Give a 95% confidence interval for the mean aluminum contents at location 2.
What are your conclusions from the confidence intervals in (a) and (b)and why?   
c) At , test H0 : =   versus H1 :   
d) Give the approximate p-value for the test in (c).
Problem 3: Nine students were randomly selected who had taken the TOEFL test twice. A researcher would like to test the claim that students who take the TOEFL test a second time score higher than their first test.
Student                            A       B       C        D       E        F       G      H     I
   First TOEFL Score         480   510   530   540   550   560   600   620 660
Second TOEFL Score    460   500   530   520   580   580   560 640 690
Test the claim using a level of significance of 0.05 and construct a 90% confidence interval for µd .
Problem 4: According to reported figures, the average price of used car nationally is $8,000 with a standard deviation of $4,500. A student at annajah national university wants to purchase a used car and wishes to find out if the average used car price in Nablus is less than the national average. The student collected figures on a random sample of 81 used car sales at dealerships across Nablus. The sample mean price was $7,100.
a) State the null and alternative hypotheses, compute the test statistic, find the p-value and what is your conclusion? Use α = 0.05.
b) If the actual mean of the prices is $7500, find the probability of type II error.
c) What value of n is necessary to ensure that β = 0.10 when α = 0.05 and the actual mean is $7500?
Good LuckDr. Ali Barakat

Answer 1

Problem 1:

(a) Let p₁= true proportion of adults who are active and sufferedheart attacks

p₂=true proportion of adults who are not active and suffered heart attacks

Minitab output:

Test and CI for Two Proportions

Sample X N Sample p
1 10 100 0.100000
2 26 200 0.130000

Difference = p (1) - p (2)
Estimate for difference: -0.03
95% CI for difference: (-0.105031, 0.0450310)
Test for difference = 0 (vs not = 0): Z = -0.75 P-Value = 0.451

Fisher's exact test: P-Value = 0.572

Value of Z score=-0.75

P-Value = 0.451

Since P-value>0.05, we fail to reject H₀ at 5% level of significance and conclude that there is insufficient evidence that the proportion of adults who are active and sufferedheart attacks is different from the proportion of adults who are not active and suffered heart attacks.

(b) 95% confidence interval for the difference between the proportions of all active and inactive adults who suffered heart attacks: (-0.1050, 0.04503).

Since the confidence interval contains zero we get same conclusion as Part (a).

Problem 2:

(a) Minitab output:

N Mean StDev SE Mean 95% CI
5 2935 235.7657 105 (2642, 3228)

95% confidence interval for the mean aluminum contents at location 1: (2642, 3228).

(b) Minitab output:

N Mean StDev SE Mean 95% CI
4 2637 741.8416 371 (1457, 3817)

95% confidence interval for the mean aluminum contents at location 2: (1457, 3817).

(c)

Test and CI for Two Variances

Method

Null hypothesis Sigma(1) / Sigma(2) = 1
Alternative hypothesis Sigma(1) / Sigma(2) not = 1
Significance level Alpha = 0.05

Statistics

Sample N StDev Variance
1 5 235.766 55585.465
2 4 741.842 550328.959

Ratio of standard deviations = 0.318
Ratio of variances = 0.101

Test
Method DF1 DF2 Statistic P-Value
F Test (normal) 4 3 0.10 0.051

Since P-value>0.05, we can assume two population variances are equal.

Two-Sample T-Test and CI

Sample N Mean StDev SE Mean
1 5 2935 236 105
2 4 2637 742 371

Difference = mu (1) - mu (2)
Estimate for difference: 298
95% CI for difference: (-523, 1119)
T-Test of difference = 0 (vs not =): T-Value = 0.86 P-Value = 0.419 DF = 7
Both use Pooled StDev = 517.3185

(d) p-value=0.419.

Problem 3:

Minitab output:

Paired T for First TOEFL score - Second TOEFL score

  N Mean StDev SE Mean
First TOEFL score 9 561.1 56.4 18.8
Second TOEFL score 9 562.2 70.8 23.6
Difference 9 -1.11 25.22 8.41

95% upper bound for mean difference: 14.52
T-Test of mean difference = 0 (vs < 0): T-Value = -0.13 P-Value = 0.449

Since P-value>0.05, there is insufficient evidence to accept the claim that students who take the TOEFL test a second time score higher than their first test.

Minitab output:

Paired T-Test and CI: First TOEFL score, Second TOEFL score

Paired T for First TOEFL score - Second TOEFL score

N Mean StDev SE Mean
First TOEFL score 9 561.1 56.4 18.8
Second TOEFL score 9 562.2 70.8 23.6
Difference 9 -1.11 25.22 8.41

90% CI for mean difference: (-16.74, 14.52)
T-Test of mean difference = 0 (vs not = 0): T-Value = -0.13 P-Value = 0.898

construct a 90% confidence interval for µ_d: (-16.74, 14.52).