Question

Are nursing salaries in Tampa, Florida, lower than those in Dallas, Texas? As reported by the Tampa Tribune, salary data show staff nurses in Tampa earn less than staff nurses in Dallas. Suppose that in a follow-up study of 41 staff nurses in Tampa and 51 staff nurses in Dallas you obtain the following results. Enter negative value as negative number.

Tampa		Dallas
n1=41		n2=51
^-x1=58,800		^-x2=62,800
s1=4,800		s2=6,200

a. Formulate hypothesis so that, if the null hypothesis is rejected, we can conclude that salaries for staff nurses in Tampa are significantly lower than for those in Dallas. Use a=.05

	- Select your answer -<>≤≥=≠Item 1
	- Select your answer -<>≤≥=≠Item 2

b. What is the value of the test statistic?

(to 3 decimals)

c. What is the p-value?

Degrees of freedom is ? Round degrees of freedom to previous whole number.

Consider the following results for independent samples taken from two populations.

Sample 1	Sample 2
n1 = 400	n2= 300
p1= 0.49	p2= 0.39

a. What is the point estimate of the difference between the two population proportions (to 2 decimals)?

b. Develop a 90% confidence interval for the difference between the two population proportions (to 4 decimals). Use z-table.
?  to ?  

c. Develop a 95% confidence interval for the difference between the two population proportions (to 4 decimals). Use z-table.
?  to ?

Consider the following hypothesis test.

H0: u1 - u2 ≤ 0
Ha: u1 -u 2 > 0

Consider the following results for two independent random samples taken from two populations.

Sample 1	Sample 2
n 1 = 100	n 2 = 300
p1 = .22	p 2 = .11

Use pooled estimator of p.

1. What is the value of the test statistic (to 2 decimals)?
   
2. What is the p-value (to 4 decimals)? Use z-table.

Answer 1

Question 1

a.

Null hypothesis :

Ho: 1-2

Alternative hypothesis :

HA 12

b.

as follows Since it is assumed that the population variances are unequal, the t-statistic is computed X2 58800 62800 3.487 48002 62002 41 51

c.

P-value = 0.0004

Formula for degrees of freedom is :

df= 1 2 S. 1 n-1n n2-1n

Degrees of freedom = df = 89.891 $\approx$89 ( Rounded to previous whole number.)

Question 2

a.

Point estimate of the difference between the two population proportions = p₁ - p₂ = 0.49 - 0.39 = 0.10

b.

We need to construct the 90% confidence interval for the difference between population proportions p2. We have been provided with the following information about the sample proportions: Pi Sample Proportion 1 (ội) = 0.49 Sample Size 1 (N1) = 400 Sample Proportion 2 (ộ2) = 0.39 Sample Size 2 (N2) = 300 The critical value for a 0.1 is ze = z1-a/2 = 1.645. The corresponding confidence interval is computed as shown below: p. (1 — Р1) Pе (1 - р2) N2 P (1 - р1) N1 P- (1 — Р2) N2 CI Pi - Р2 , Рі — р2 + z. 0.49(1 1 0.49(1 0.39(1-0.39) 0.49) 0.39(1 - 0.39) 0.49 .0.49-0.39+ 1.645 x 1 0.39 1.645 x 0.49 400 300 400 300 (0.038,0.162) Therefore, based on the data provided, the 90% confidence interval for the difference between the population proportions p the true difference between population proportions is contained by the interval (0.038, 0.162). p2 is 0.038<p< 0.162, which indicates that we are 90% confident that

90% confidence interval for the difference between the two population proportions : ( 0.0381 , 0.1619 )

c.

We need to construct the 95% confidence interval for the difference between population proportions pı - p2. We have been provided with the following information about the sample proportions: Sample Proportion 1 (1) = 0.49 Sample Size 1 (N1) = 400 Sample Proportion 2 (p2) = 0.39 Sample Size 2 (N2) = 300 The critical value for a = 0.05 is ze = z1-a/2 = 1.96. The corresponding confidence interval is computed as shown below: (P ре(1 - Р2) , Рі — р2 + 2, N2 рi (1 - р1) N1 р2(1 — Р2) N2 P. (1 - р1) N1 CI = рі — Р2 0.49(1 0.49) 1 0.49(1 -0.49) 0.39 0.39(1 0.39(1 0.39 - 0.39 1.96 x \| 0.391.96 x 0.49 0.49 400 300 400 300 (0.026, 0.174) Therefore, based on the data provided, the 95% confidence interval for the difference between the population proportions pi - p2 is 0.026 < p < 0.174, which indicates that we are 95% confident that the true difference between population proportions is contained by the interval (0.026, 0.174).

95% confidence interval for the difference between the two population proportions : ( 0.0262 , 0.1738 )

Question 3

Х, — 22, For sample 1, we have that the sample size is N = 100, the number of favorable cases is 22 0.22 100 so then the sample proportion is p _ N1 33 For sample 2, we have that the sample size is N2 300, the number of favorable cases is X2 X2 33 = 0.11 300 so then the sample proportion is p N2 Xi+X2 Ni+N2 22+33 The value of the pooled proportion is computed as p = 0.1375 100+300 Also, the given significance level is a = 0.05 (1) Nul| and Alternative Hypotheses The following null and alternative hypotheses need to be tested: Но: рi <Р2 На : рi > Р2 This corresponds to a right-tailed test, for which a z-test for two population proportions needs to be conducted (2) Rejection Region Based on the information provided, the significance level is a 0.05, and the critical value for a right-tailed test is ze 1.64 The rejection region for this right-tailed test is R {z : z > 1.64}

(3) Test Statistics The z-statistic is computed as follows: Р — Р2 УРI -Р) (1/п +1/п2) 0.22 0.11 = 2.766 V0.1375 10.1375)(1/100 1/300) (4) Decision about the null hyYpothesis 2.766 > ze = 1.64, it is then concluded that the null hypothesis is Since it is observed that z rejected Using the P-value approach: The p-value is p 0.0028, and since p concluded that the null hypothesis is rejected 0.0028 <0.05, it is (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that population proportion pi is greater than p2, at the 0.05 significance level.