Question

A group of researchers at UTPA would like to determine reasons for low turnout in the RGV. They suspect that political trust in the RGV will be significantly different from the entire U.S. population (µ = 7.22). A group of 7 RGV residents’ scores are listed below. Compare this group of scores to the population to determine if there is a significant different (α = .05)

Participant	Political Trust
1	5
2	6
3	8
4	7
5	7
6	7
7	6

Will we need a one- or two-tailed hypothesis test?

State your null hypothesis

State your alternative hypothesis

Provide the SPSS output for your test and identify (circle or highlight) the t-obtained and the p-value

Did you reject or fail to reject the null hypothesis?

What can you conclude?

Calculate the 95 confidence interval for the sample mean

Calculate Cohens D

Answer 1

Solution:

One sample t test for population mean.

Null hypothesis: H₀: A political trust in the RGV from the entire U.S. population is 7.22.

Alternative hypothesis: H_a: A political trust in the RGV from the entire U.S. population is different from 7.22.

We need to use two tailed test.

We are given α = 0.05

SPSS outputs for this test are given as below:

One-Sample Statistics
	N	Mean	Std. Deviation	Std. Error Mean
Political Trust	7	6.5714	.97590	.36886

One-Sample Test
	Test Value = 7.22
	t	df	Sig. (2-tailed)	Mean Difference	95% Confidence Interval of the Difference
					Lower	Upper
Political Trust	-1.758	6	.129	-.64857	-1.5511	.2540

Test statistic = t = -1.758

P-value = 0.129

P-value > α = 0.05

So, we do not reject the null hypothesis

There is insufficient evidence to conclude that A political trust in the RGV from the entire U.S. population is different from 7.22.

A 95% confidence interval for sample mean is given as below:

Confidence interval = Xbar ± t*S/sqrt(n)

t = 2.4469

(by using t-table or excel)

Confidence interval = 6.5714 ± 2.4469*.97590/sqrt(7)

Confidence interval = 6.5714 ± 0.9026

Lower limit = 6.5714 - 0.9026 = 5.67

Upper limit = 6.5714 + 0.9026 = 7.47

Confidence interval = (5.67, 7.47)

Cohen’ D = (Xbar - µ) / S = (6.5714 – 7.22) / .97590 = -0.66462