Question

A Michigan study concerning preference for outdoor activities used a questionnaire with a six-point Likert-type response in which 1 designated “not important” and 6 designated “extremely important.”

A random sample of n1 = 46 adults were asked about fishing as an outdoor activity. The mean response was 4.9. Another random sample of n2 = 51 adults were asked about camping as an outdoor activity. For this group, the mean response was 4.3. From previous studies, it is known that s1 = 1.5 and s2 = 1.2.

Does these data indicate a difference (either way) regarding preference for camping versus preference for fishing as an outdoor activity? Use a = 0.05.

Find a 95% confidence interval for m1 – m2. Explain the meaning of the confidence interval in the context of the problem.

Answer 1

H₀:

H₁:

The test statistic z = ()/sqrt()

                              = (4.9 - 4.3)/sqrt((1.5)^2/46 + (1.2)^2/51)

                              = 2.16

At alpha = 0.05 the critical values are z_0.025 = +/- 1.96

As the test statistic value is greater than the upper critical value (2.16 > 1.96), we should reject H₀.

So these data indiate that there is a difference regarding preference for camping versus preference for fishing as an outdoor activity.

At 95% confidence interval the critical value is z_0.025 = 1.96

The 95% confidence interval is

(m1 - m2) +/- z_0.025 * sqrt()

= (4.9 - 4.3) +/- 1.96 * sqrt((1.5)^2/46 + (1.2)^2/51)

= 0.6 +/- 0.54

= 0.06, 1.14

We are 95% confident that the difference between the two means lies in the above confidence interval.

Since the confidence interval does not cointain 0, so there is sufficient evidence to support that there is a difference regarding preference for camping versus preference for fishing as an outdoor activity.