Question

U 4 5 The data table represents the measure of a variable before and after a treatment. Does the sample evidence suggest Individual that the treatment is effective in decreasing the value of the response variable? Use the c = 0.10 level of significance. Before, Xi Complete parts (a) through (d). 2 3 30 53 40 47 40 After, Yi 37 33 50 46 35 (a) What type of test should be used? Choose the correct answer below. O A. A hypothesis test regarding the difference between two population proportions from independent samples. OB. A hypothesis test regarding the difference of two means using a matched-pairs design. OC. A hypothesis test regarding two population standard deviations. OD. A hypothesis test regarding the difference of two means using Welch's approximate t. (b) Determine the null and alternative hypotheses. Let Hd FHx - Hy. Choose the correct answer below. O A. Ho: Hd = 0; Hy : Hd #0 B. Ho: Hd = 0; H4; Hd>0 OC. Ho Hd #0; H:Hd > OD. Ho Hd = 0; H: Hd < 0 (c) Use technology to calculate the P-value. (Round to three decimal places as needed.) (d) Draw a conclusion based on the hypothesis test. Choose the correct answer below. A. There is not sufficient evidence to reject the null hypothesis because the P-value > . B. There is sufficient evidence to reject the null hypothesis because the P-value < < OC. There is sufficient evidence to reject the null hypothesis because the P-value >a. OD. There is not sufficient evidence to reject the null hypothesis because the P-value <a

Answer 1

1.

a)

Option B is correct.

a t-test for two paired samples be used

b)

The following table is obtained:

	Sample 1	Sample 2	Difference = Sample 1 - Sample 2
	40	37	3
	30	33	-3
	53	50	3
	47	46	1
	40	35	5
Average	42	40.2	1.8
St. Dev.	8.631	7.396	3.033
n	5	5	5

From the sample data, it is found that the corresponding sample means are:

Xˉ1​=42

Xˉ2​=40.2

Also, the provided sample standard deviations are:

s_1 = 8.631

s2​=7.396

and the sample size is n = 5.

For the score differences we have

Dˉ=1.8

sD​=3.033

() Null and Alternative Hypotheses

The following null and alternative hypotheses need to be tested:

Ho: μD​ = 0

Ha: μD​ > 0

This corresponds to a right-tailed test, for which a t-test for two paired samples be used.

() Test Statistics

The t-statistic is computed as shown in the following formula:

D 1.8 t 1.327 sp/n 3.033/5

() Decision about the null hypothesis

Using the P-value approach: The p-value is p = 0.1276,

and since p = 0.1276≥0.10,

it is concluded that the null hypothesis is not rejected.

() Conclusion

It is concluded that the null hypothesis Ho is not rejected.

Therefore, there is not enough evidence to claim that the population mean μ1​ is greater than μ2​, at the 0.10 significance level.

2.

1)

Option B is correct.

2)

Option d is correct.

3.

The provided sample mean is Xˉ=3439

and

the sample standard deviation is s = 2567.

The size of the sample is n = 100

and the required confidence level is 90%.

The number of degrees of freedom are df = 100 - 1 = 99,

and the significance level is α=0.1.

Based on the provided information,

the critical t-value for α=0.1 and df = 99 degrees of freedom is

t_c = 1.66

The 90% confidence for the population mean μ is computed using the following expression

$CI = \left(\bar X - \frac{t_c \times s}{\sqrt{n}} , \bar X + \frac{t_c \times s}{\sqrt{n}} \right)$

Therefore, based on the information provided, the 90 % confidence for the population mean μ is

$CI = \left(3439 - \frac{ 1.66 \times 2567}{\sqrt{ 100}} , 3439 + \frac{ 1.66 \times 2567}{\sqrt{ 100}} \right)$

option c is correct

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