Question

A political candidate wishes to determine if endorsing increased social spending is likely to decrease her popularity (α = .05). She has access to data on the popularity of several other candidates who have endorsed increased spending. The data were available both before and after the candidates announced their positions on the issue. The data are as follows:

Popularity rating
Participant	Before	After
1	42	43
2	41	45
3	50	56
4	52	54
5	58	65
6	32	29
7	39	46
8	42	48

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 “Before” mean

Calculate Cohens D

Answer 1

We need to use a one-tailed hypothesis test.

Null hypothesis: H₀: Endorsing increased social spending is not likely to decrease her popularity.

Alternative hypothesis: H_a: Endorsing increased social spending is likely to decrease her popularity.

H₀: µ_D = 0 versus H_a: µ_D < 0

Required SPSS output is given as below:

Paired Samples Statistics
		Mean	N	Std. Deviation	Std. Error Mean
Pair 1	Before	44.5000	8	8.28079	2.92770
	After	48.2500	8	10.60660	3.75000

Paired Samples Correlations
		N	Correlation	Sig.
Pair 1	Before & After	8	.960	.000

Paired Samples Test
		Paired Differences					t	df	Sig. (2-tailed)
		Mean	Std. Deviation	Std. Error Mean	95% Confidence Interval of the Difference
					Lower	Upper
Pair 1	Before - After	-3.75000	3.53553	1.25000	-6.70578	-.79422	-3.000	7	.020

Test statistic = t = -3.00

P-value = 0.020

P-value < α = 0.05

So, we reject the null hypothesis

There is sufficient evidence to conclude that Endorsing increased social spending is likely to decrease her popularity.

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

From given data, we have

Xbar = 44.5

S = 8.280786712

n = 8

df = 7

confidence level = 95%

Critical t value = 2.3646 (by using t-table)

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

Confidence interval = 44.5 ± 8.280786712/sqrt(8)

Confidence interval = 44.5 ± 6.9229

Lower limit = 44.5 - 6.9229 = 37.58

Upper limit = 44.5 + 6.9229 = 51.42

Cohen’s D = Dbar/S_D = -3.75 / 3.5355 = -1.060670344