7.82 New computer monitors? The purchasing depart- ment has suggested that all n

Question

7.82 New computer monitors? The purchasing depart- ment has suggested that all new computer monitors for your company should be flat screens. You want data to assure you that employees will like the new screens The next 20 employees needing a new computer are the subjects for an experiment. (a) Label the employees 01 to 20. Randomly choose 10 to receive flat screens. The remaining 10 get standard monitors. (b) After a month of use, employees express their sat- isfaction with their new monitors by responding to the statement "I like my new monitor" on a scale from 1 to 5, where 1 represents “strongly disagree," 2 is "disagree," 3 is "neutral," 4 is "agree," and 5 stands for "strongly agree." The employees with the flat screens have average satisfaction 4.8 with standard deviation 0.7. The employ- ees with the standard monitors have average 3.0 with standard deviation 1.5. Give a 95% confidence interval for the difference in the mean satisfaction scores for all employees (c) Would you reject the null hypothesis that the mean satisfaction for the two types of monitors is the same versus the two-sided alternative at significance level 0.05? Use your confidence interval to answer this question. Explain why you do not need to calculate the test statistic.

Explanation / Answer

Solution:

Part a

We are given

Total number of employees = 20

Number of employees allotted to flat screens = n1 = 10

Number of employees allotted to standard monitors = n2 = 10

Part b

We have to find the 95% confidence interval for the difference between average satisfaction scores of employees getting flat screen computers and standard monitor computers.

Formula for confidence interval is given as below:

Confidence interval = (X1bar – X2bar) -/+ t/2, n1+n2 – 2 *sqrt[(S1^2/n1)+(S2^2/n2)]

We are given

Population 1 Sample (Flat screen)           

Sample Size = n1 = 10

Sample Mean = X1bar = 4.8

Sample Standard Deviation = S1 = 0.7

Population 2 Sample (Standard monitor)              

Sample Size = n2 = 10

Sample Mean = X2bar = 3

Sample Standard Deviation = S2 = 1.5

Confidence level = 95% or c = 0.95

Significance level = = 1 – c = 1 – 0.95 = 0.05, so /2 = 0.05/2 = 0.025

Degrees of freedom = n1 + n2 – 2 = 10 + 10 – 2 = 18

Critical value = t/2, n1+n2 – 2 = t0.025, 18 = 2.1009 (by using t-table or excel)

Difference in means = (X1bar – X2bar) = 4.8 – 3 = 1.8

Confidence interval = (X1bar – X2bar) -/+ t/2, n1+n2 – 2 *sqrt[(S1^2/n1)+(S2^2/n2)]

Confidence interval = 1.8 -/+ 2.1009*sqrt((0.7^2/10)+(1.5^2/10))

Confidence interval = 1.8 -/+ 2.1009* 0.52345

Confidence interval = 1.8 -/+ 1.099728

Lower limit = 1.8 - 1.099728 = 0.700272

Upper limit = 1.8 + 1.099728 = 2.899728

Confidence interval = (0.700272, 2.899728)

Part c

Yes, we would reject the null hypothesis that the mean satisfaction for the two types of monitors is the same versus the two sided alternative at significance level 0.05, because the value of difference 0 is not included in the given confidence interval. This means, the average satisfaction scores for two types of monitors are not same. If scores are same, then the value 0 should be lies within confidence interval. Here, value 0 is not included in the given confidence interval and that’s why we reject the null hypothesis of same satisfaction scores for two types of monitors.

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